Answer:
The normal force is the force that the floor does as a reaction of the gravitational force that an object does against the floor (is the resistance that objects have when other objects want to move trhough them, and the force comes by the 3rd Newton's law, and this is specially used in cases where the first object is fixed, like walls or the floor). With this in mind, the point in where the normal force will be greater is the point that is closer to the center of mass of the object (the point with more mass)
If the wheels are in the extremes of the object, and the center of mass is in the middle of the object, the normal force will be equal. Now if for example, you put a little mass in one end of the object, now the center of weight displaces a little bit and is not centered, and the side is where you put the weight on will receive a bigger normal force from the floor than the other side.
The force on the front and rear wheels can be determined by finding the
force acting under equilibrium conditions.
The normal reaction between the floor and the front wheel is larger. The
correct option is c) The force on the front wheels must be greater than the
force on the rear wheels.
Reasons:
The forces acting on the wheels are;
Vertical forces acting on the cart:
The weight of the cart acting on the floor
The normal reaction of the floor on the wheels
The horizontal acting on the cart:
Pushing force acting horizontally
Friction force acting in the reverse direction
At equilibrium, the cart does not tip over
Sum of moment about the rear wheel = [tex]\mathbf{W \times \dfrac{d}{2} + F \times h - N_{front} \times d} = 0[/tex]
Where;
h = The height of the cart
d = depth of the cart
[tex]N_{front}[/tex] = The normal reaction at the front wheels
Therefore;
[tex]\displaystyle N_{front} = \frac{W \times \dfrac{d}{2} + F \times h}{d} = \mathbf{ \frac{W}{2} + \frac{F \times h}{d}}[/tex]
Sum of moment about the front wheel = [tex]\mathbf{ F \times h + N_{rear} \times d - W \times \dfrac{d}{2} } = 0[/tex]
Therefore;
[tex]\displaystyle N_{rear} = \frac{W \times \dfrac{d}{2} - F \times h}{d} = \mathbf{ \frac{W}{2} - \frac{F \times h}{d}}[/tex]
Which gives;
[tex]\displaystyle N_{front} = \mathbf{ \displaystyle N_{rear} + 2 \times \frac{F \times h}{d}}[/tex]
[tex]\displaystyle 2 \times \frac{F \times h}{d} > 0[/tex]
Therefore;
[tex]\displaystyle N_{front} > \displaystyle N_{rear}[/tex]
The correct option is c) The force on the front wheels must be greater than
the force on the rear wheels.
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A natural force of attraction exerted by the earth upon objects, that pulls
objects towards earth's center is called
A natural force of attraction exerted by the earth upon objects, that pulls objects towards earth's center is called Gravitational force .
Suppose there is a uniform magnetic field, B, pointing into the page (so your index finger will point into the page). If the velocityof a proton is straight up(thumb pointing up) then RHR2 shows that the force points to the left. What would the direction of the force be if the velocitywere a) down b) to the rightc) to the leftd) into the pagee) out of the page
Answer:
Explanation:
We shall show all given data in vector form and calculate the direction of force with the help of following formula
force F = q ( v x B )
q is charge , v is velocity and B is magnetic field.
Given B = - Bk ( i is right , j is upwards and k is straight up the page )
v = v j
F = q ( vj x - Bk )
= -Bqvi
The direction is towards left .
a ) If velocity is down
v = - v j
F = q ( - vj x - bk )
= qvB i
Direction is right .
b ) v = v i
F = q ( vi x - Bk )
= qvB j
force is upwards
c ) v = - vi
F = q ( -vi x - Bk )
= -qvBj
force is downwards
d ) v = - v k
F = q( - vk x -Bk )
= 0
No force will be created
e ) v = v k
F = q( vk x -Bk )
= 0
No force will be created
You are to drive to an interview in another town, at a distance of 310 km on an expressway. The interview is at 11:15 a.m. You plan to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 42.0 km/h for 42.0 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview
Answer:
Explanation:
Time to cover first 100 km = 1 hour.
time remaining = 3.15 - 1 = 2.15 hour .
Time to cover next 42 km = 1 hour .
Time remaining = 2.15-1 = 1.15 hour.
Distance to be covered = 310 - 142
= 168 km
least speed needed = distance remaining / time remaining
= 168 / 1.15
= 146.08 km / h .
An object on a number line moved from x = 15 cm to x = 165 cm and then
moved back to x = 25 cm, all in a time of 100 seconds.
What was the average velocity of the object?
Answer:
v_avg = 2.9 cm/s
Explanation:
The average velocity of the object is the sum of the distance of all its trajectories divided the time:
[tex]v_{avg}=\frac{x_{all}}{t}[/tex]
x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm
Then, x_all = 150cm + 140cm = 290cm
The average velocity is, for t = 100s
[tex]v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}[/tex]
hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s
Answer:
The average velocity of the object is 0.1cm/s
Explanation:
Given that the object travels from point 15cm to 165cm and back to 25cm within 100 seconds
The average velocity is calculated as thus.
Average Velocity = ∆D/t
Where ∆D represent the displacement.
The displacement is calculated as follows.
∆D = End point - Start Point.
From the question, the end and start point are 25cm and 15cm respectively.
Hence,
∆D = 25cm - 15cm
∆D = 10cm.
t = 100 seconds
So, Average Velocity = 10cm/100s
Average Velocity = 0.1cm/s
Hence, the average velocity of the object is 0.1cm/s
A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.540 m/s. The block has a head-on elastic collision with a 40.0 gram block that is initially at rest. Since the collision is head-on, all velocities lie along the same line, both before and after the collision. (a) What is the speed of the 20.0 gram block after the collision
Answer:
0.180 m/s
Explanation:
Solving the equations for conservation of momentum and energy for an elastic collision gives ...
v₁' = ((m₁ -m2)v₁ +2m₂v₂)/(m₁ +m₂) . . . . v₁' is the velocity of m₁ after collision
__
Here, we have (m₁, m₂, v₁, v₂) = (20 g, 40 g, 0.540 m/s, 0 m/s).
Substituting these values in to the equation for v₁', we have ...
v₁' = ((20 -40)(0.540) +2(40)(0))/(20 +40) = (-20/60)(0.540)
v₁' = -0.180 . . . m/s
The speed of the 20 g block after the collision is 0.180 m/s to the left.
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a specified tile surface. A typical athletic shoe has a coefficient of 0.870. In an emergency, what is the minimum time interval in which a person starting from rest can move 3.20 m on a tile surface if she is wearing the following footwear?
(a) footwear meeting the Postal Service minimum
(b) a typical athletic shoe
Answer:
(a) 1.14 s
(b) 0.87 s
Explanation:
A person moves by the help of frictional force, as a result of gtound reaction. So, the formula for frictional force is:
F = μR
where,
F = frictional force
μ = coefficient of friction
R = Normal Reaction = Weight of Body = W = mg
Therefore,
F = μmg
but, from Newton's 2nd Law of Motion:
F = ma
Comparing both equations, we get:
μmg = ma
a = μg ---------- equation (1)
Now, to calculate the distance moved by a body, we use 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
using equation (1), we get:
s = (Vi)(t) + (0.5)μgt²
where,
s = distance moved by body
Vi = initial velocity of body
t = time taken to cover the distance
g = acceleration due to gravity
(a)
Vi = 0 m/s
g = 9.8 m/s²
s = 3.2 m
μ = 0.5
t = ?
Therefore,
3.2 m = (0 m/s)(t) + (0.5)(0.5)(9.8 m/s²)t²
t² = 3.2 m/(0.5)(0.5)(9.8 m/s²)
t = √1.30612 s²
t = 1.14 s
(a)
Vi = 0 m/s
g = 9.8 m/s²
s = 3.2 m
μ = 0.5
t = ?
Therefore,
3.2 m = (0 m/s)(t) + (0.5)(0.5)(9.8 m/s²)t²
t² = 3.2 m/(0.5)(0.5)(9.8 m/s²)
t = √1.30612 s²
t = 1.14 s
(a)
Vi = 0 m/s
g = 9.8 m/s²
s = 3.2 m
μ = 0.87
t = ?
Therefore,
3.2 m = (0 m/s)(t) + (0.5)(0.87)(9.8 m/s²)t²
t² = 3.2 m/(0.5)(0.87)(9.8 m/s²)
t = √0.75 s²
t = 0.87 s
A projectile is fired with an initial speed of 37.7 m/s at an angle of 41.2° above the horizontal on a long flat firing range. Determine the maximum height reached by the projectile.
Answer:
h = 31.46 m
Explanation:
We have,
Initial speed of a projectile is 37.7 m/s
It was projected at an angle of 41.2° above the horizontal on a long flat firing range.
It is required to find the maximum height reached by the projectile. The formula used to find it is given by :
[tex]h=\dfrac{u^2\sin^2\theta}{2g}[/tex]
Plugging all the known values,
[tex]h=\dfrac{(37.7)^2\times \sin^2(41.2)}{2\times 9.8}\\\\h=31.46\ m[/tex]
So, the maximum height reached by the projectile is 31.46 m.
Bats are capable of navigating using the earth's field-a plus for an animal that may fly great distances from its roost at night. If, while sleeping during the day, bats are exposed to a field of a similar magnitude but different direction than the earth's field, they are more likely to lose their way during their next lengthy night flight. Suppose you are a researcher doing such an experiment in a location where the earth's field is 50muT at a 60 degree angle below horizontal. You make a 70 cm diameter, 100-turn coil around a roosting box; the sleeping bats are at the center of the coil. You wish to pass a current through the coil to produce a field that, when combined with the earth's field, creates a net field with the same strength and dip angle (60 degree below horizontal) as the earth's field but with a horizontal component that points south rather than north.What is the proper orientation of the coil? What is the necessary current? Express your answer to two significant figures and include the appropriate units.
Answer:
The correct magnitude of the coil's magnetic field= =50μT
Explanation :
The magnetic field takes place as a result of movement of charge I e current, can also occurs from magnetised material, magnetic field as a result of charge movement can be deducted using right hand grip rule.
At the equator magnetic field lines are parallel towards the earth's surface and the angle of inclination of the magnetic lines of force at the horizontal position is referred to as the angle of dip at the point.
As the current is produced then the varying magnetic field is opposed ,then there is induced current when the coil is positioned at varying magnetic field.
Given from the question,
angle below horizontal θ=60-degree
The Earth's magnetic field B=50μT
The horizontal magnetic field can be expressed in terms of the formula below;
BH=Bcosθ
B(H) = earth's horizontal component of magnetic field
Ø is the angle between coil's field
B=the magnetic field in Tesla
Then,
BH=50μT×cos60∘
=50μT× 0.5
As the current is passed through the coil to produce a field , when combined with the earth's field, which creates a net field with the same strength and dip angle (60 degrees below horizontal) as the earth's field.
It can be deducted that B has the same magnitude and angle which makes the those vertical component to cancel each other since they are the same.
For the magnetic field to be pointed out at North direction, we can calculate the corrected magnetic field using the formula below
Bc=2BH
Bc=2×50μT× 0.5=50μT
The correct magnitude of the coil's magnetic field= =50μT
The kinetic energy of a ball with a mass of 0.5 kg in a velocity of 10 meterss
Answer:
The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is J.
Explanation:
Thus the kinetic energy of the ball is 25 J. The unit Joule (J) is the same as kgm^2/s^2 and is the SI unit for kinetic energy.
Hope this helps.
At the same temperature, two wires made of pure copper have different resistances. The same voltage is applied at the ends of each wire. The wires may differ in:________
a. length.b. cross-sectional area.c. resistivity.d. amount of electric current passing through them
Answer:
A and B are true
Explanation:
C and D are false
Answer:
The answer is the length, cross-sectional area, and the amount of electric current passing through them.
So,
a
b
d
are the correct answers.
Explanation:
I hope this helps future Physics students who are also struggling...
If you have any questions let me know!!
This answer is 100% correct
One of the advantages of skates is that they allow you to push against ice, which typically has low friction. Let’s consider an hockey player on frictionless ice who is going to shoot a slap shot. The player weighs Mp = 100kg. He is going to fire the puck, which weighs Mpuck = 0.5kg with a velocity of v = 50(m/s). How fast does the hockey player recoil in the direction opposite to the puck’s trajectory. If the force he exerts on the puck occurs over a time, ∆t = 0.1s what is the average force needed to create the impulse that accelerates the puck to that velocity?
Answer:
a) v1f = 0.25 m/s
b) F = 50000N
Explanation:
a) In order to calculate the speed of the player after he fires the puck, you use the conservation of momentum law. Before the puck is in motion and after the total momentum of both player and puck must conserve:
[tex]m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}[/tex] (1)
m1: mass of the player = 100kg
v1i: initial velocity of the player = 0m/s
v2f: final velocity of the puck = ?
m2: mass of the puck = 0.5kg
v2i: initial velocity of the puck = 0 m/s
v2f: final velocity of the puck = 50 m/s
You replace these values into equation (1) and you solve for v1f (final velocity of player):
[tex]0kgm/s+0kgm/s=(100kg)v_{1f}+(0.5kg)(50m/s)\\\\v_{1f}=-\frac{(0.5kg)(50m/s)}{100kg}\\\\v_{1f}=-0.25\frac{m}{s}[/tex]
The minus sign means that player moves in a opposite direction to the motion of the puck
The velocity of the player after he fires the puck is 0.25 m/s
b) The force needed is given by the change in time , of the momentum of the player, which is given by:
[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}[/tex]
The change on the velocity of the puck is 50m/s, and the time interval is 0.1s.
[tex]F=(100kg)\frac{50m/s}{0.1s}=50000N[/tex]
The force needed to create the needed impulse, to accelerate the puck is 50000N
g 16.0 kg sled is pulled on a horizontal frictionless ice rink with a light spring of force constant 220 N/m. If the sled has an acceleration of 2.0 when pulled horizontally, how much does the spring stretch (in meters)
Answer:
Spring stretches by 0.145m
Explanation:
We are given;
Mass of sled;m = 16 kg
Force constant;k = 220 N/m
Acceleration;a = 2 m/s²
Now, we know that from Newton's second law of motion, Force is given by the expression ; F = ma
Thus,
F = 16 x 2
F = 32 N
Now from Hooke's law, we know that this force is expressed as;
F = -kx
Where x is the distance that the spring stretches.
The minus sign shows that this force is in the opposite direction of the force that’s stretching or compressing the spring. But we'll take the magnitude and therefore ignore the negative sign to get F = Kx
Thus,
32 = 220x
x = 32/220
x = 0.145 m
Answer:
[tex]x=-0.145m[/tex]
Explanation:
Information we have:
Mass of the sled: [tex]m=16kg[/tex]
Spring constant: [tex]k=220N/m[/tex]
acceleration of the sled: [tex]a=2m/s^2[/tex]
We use the formula for the force on a spring (Hooke's Law)
[tex]F=-kx[/tex]
where k is the spring constant and x the distance the mass moved from the equilibrium
and we also use Newton's second Law of motion:
[tex]F=ma[/tex]
we combine these two equations:
[tex]-kx=ma[/tex]
we solve the last equation for the distance x:
[tex]x=-\frac{ma}{k}[/tex]
and substitute the values:
[tex]x=-\frac{(16kg)(2m/s^2)}{220N/m}\\ x=-0.145m[/tex]
the negative number means that the mass was moved in the opposite direction that the force, this because the force in a spring is restorative and points towards the equilibrium point
If an object is in equilibrium, which of the following statements is not true? A) The velocity is constant. B) The object must be at rest. C) The net force acting on the object is zero. D) The acceleration of the object is zero. E) The speed of the object remains constant.
When the object is in equilibrium so the statement i.e. not true should be that object should be at rest.
Newton second law of motion:
In the case when the net force of an object should be zero at the time when it is in equilibrium. In the case when the net force should be zero the acceleration should also be zero. And, if the acceleration if zero so the speed and velocity should remain the same. So for maintaining the equilibrium, the object should not have to be only at rest.
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The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 10.6 V and b = -4.90 V/m. (a) Determine the potential at x = 0. 10.6 Correct: Your answer is correct. V Determine the potential at x = 3.00 m. V (b) Determine the magnitude and direction of the electric field at x = 0. magnitude 4.9 Correct: Your answer is correct. V/m direction +x Correct: Your answer is correct. Determine the magnitude and direction of the electric field at x = 3.00 m. magnitude V/m direction
Answer:
a) 10.6V
b) E = 4.9V/m, +x direction
c) E = 4.9V/m, +x direction
Explanation:
You have the following function:
[tex]V=a+bx[/tex] (1)
for the potential in a region between x=0 and x=6.00 m
a = 10.6 V
b = -4.90V/m
[tex]V=10.6V-4.90\frac{V}{m}x[/tex]
a) for x=0 you obtain for V:
[tex]V=10.6V-4.90\frac{V}{m}(0m)=10.6V[/tex]
b) The relation between the potential difference and the electric field can be written as:
[tex]E=-\frac{dV}{dx}=-b=-(-4.9V/m)=4.9V/m[/tex] (2)
the direction is +x
c) The electric field is the same for any value of x between x=0 and x=6m.
Hence,
E = 4.9V/m, +x direction
A ball is dropped from the top of a 91-m-high building. What speed does the ball have in falling 3.2 s?
Answer:
The speed of the ball is 28.4m/s.
Explanation:
Given that the formula of speed is Speed = Distance/Time. So you have to substitute the values into the formula :
[tex]speed = distance \div time[/tex]
Let distance = 91m,
Let time = 3.2s,
[tex]speed = 91 \div 3.2[/tex]
[tex]speed = 28.4 \: metrepersecond[/tex]
A. Suppose the laser light has wavelength 400 nm, and that the two small slits, which act as point sources of light, are separated by 0.1 mm. (Recall that 1 nm = 10-9 meters.) Approximately how many nodal lines would be present in the pattern of overlapping light waves when the laser light emerges from the two slits in the mask?
Answer:
200 nodal lines
Explanation:
To find the number of lines you first use the following formula for the condition of constructive interference:
[tex]dsin\theta=m\lambda[/tex] (1)
d: distance between slits = 0.1mm = 0.1*10^-3 m
θ: angle between the axis of the slits and the m-th fringe of interference
λ: wavelength of light = 400 nm = 400*10^-9 m
You obtain the max number of lines when he angle is 90°. Then, you replace the angle by 90° and solve the equation (1) for m:
[tex]dsin90\°=m\lambda\\\\d=m\lambda\\\\m=\frac{d}{\lambda}=\frac{0.1*10^{-3}m}{500*10^{-9}m}=200[/tex]
hence, the number of lines in the interference pattern are 200
A ball is fired from a toy gun with an initial speed of 7.00 m/s with the gun aimed at 300 upward with respect to the horizontal. The point of firing is 1.32 m above the horizontal floor. What is the horizontal component of the ball's initial velocity?
Answer:
6.0621 m/s
Explanation:
we all know that horizontal component of the velocity in projectile motion is
= u cosα
u= initial speed of 7.00 m/s of the gun.
α = angle of the initial velocity with the horizontal = 30°
therefore, horizontal component of the ball's initial velocity = 7×cos30°
[tex]7\times\frac{\sqrt{3} }{2} \\=6.0621 \text{m/s}[/tex]
When time is measured in days, the decay constant for a particular radioactive isotope is 0.16. Determine the time required for a confined sample of the isotope to fall to 80% of its original value.
Answer:
The time take is [tex]t = 1.3964 \ days[/tex]
Explanation:
From the question we are told that
The decay constant is [tex]\lambda = 0.16[/tex]
The percentage fall is [tex]c = 0.80[/tex]
The equation for radioactive decay is mathematically represented as
[tex]N(t) = N_o * e^{- \lambda t }[/tex]
Where is [tex]N(t)[/tex] is the new amount of the new the isotope while [tex]N_o[/tex] is the original
At initial [tex]N_o = 100[/tex]% = 1
At [tex]N(t ) = 80[/tex]% = 0.80
[tex]0.80 = 1 * e^{- 0.16 t }[/tex]
=> [tex]-0.223 = -0.16 t[/tex]
=> [tex]t = 1.3964 \ days[/tex]
Answer:
t = 1.4 days
Explanation:
The law of radioactive decay gives the amount of radioactive substance, left after a certain amount of time has passed. The formula of law of radioactive decay is given as follows:
N = N₀ (e)^-λt
where,
λ = decay constant = 0.16
N₀ = Initial Amount of the Substance
N = The Amount of Substance Left after Decay = 80% of N₀ = 0.8 N₀
t = Time Required by the Substance to decay to final value = ?
Substituting these values in the law of radioactive decay formula, we get:
0.8 N₀ = N₀ (e)^-0.16 t
0.8 = (e)^-0.16 t
ln (0.8) = -0.16 t
t = - 0.2231/-0.16
t = 1.4 days
Four cubes of the same volume are made of different materials: lead (density 11,300 kg/m3), aluminum (density 2700 kg/m3), wood (density 800 kg/m3), and Styrofoam (density 50 kg/m3). You place the cubes in a large container filled with water.Rank the buoyant forces that the water exerts on the cubes from largest to smallest.Rank from largest to smallest. To rank items as equivalent, overlap them.
Answer:
Lead > Aluminium > Wood > Styrofoam
Explanation:
Buoyant force is described by the Archimedes's Principle, which states that buoyant force is equal to the weight of the fluid displaced by the submerged object. By Newton's Laws, the buoyant force is represented by the following equation of equilibrium:
[tex]\Sigma F = F_{D} - W_{cube} = 0[/tex]
[tex]F_{D} = W_{cube}[/tex]
[tex]F_{D} = \rho_{cube} \cdot g \cdot V_{cube}[/tex]
Where:
[tex]\rho_{cube}[/tex] - Density of the cube, measured in kilograms per cubic meter.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]V_{cube}[/tex] - Volume of the cube, measured in cubic meters.
Let suppose that volume of the cube is known. Given that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the buoyant force is computed for each material:
Lead ([tex]\rho_{cube} = 11,300\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(11,300\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 110,819.1V_{cube}[/tex]
Aluminium ([tex]\rho_{cube} = 2,700\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(2,700\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 26478.9V_{cube}[/tex]
Wood ([tex]\rho_{cube} = 800\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(800\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 7845.6V_{cube}[/tex]
Styrofoam ([tex]\rho_{cube} = 50\,\frac{kg}{m^{3}}[/tex])
[tex]F_{D} = \left(50\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot V_{cube}[/tex]
[tex]F_{D} = 490.35V_{cube}[/tex]
Therefore, the buoyant forces that the water exerts on the cubes from largest to smallest corresponds to: Lead > Aluminium > Wood > Styrofoam.
As part of an exercise program, a woman walks south at a speed of 2.00 m/s for 60.0 minutes. She then turns around and walks north a distance 3000 m in 25.0 minutes (a) What is the woman's average velocity during her entire motion?
Answer:
0.824m/s
Explanation:
To calculate the average velocity we have to find the total distance and the total time
We have to find the distance and time in each motions
FIRST MOTION
The values given are
Speed= 2m/s , t = 60minutes
The time has to be converted to seconds. 60×60 = 3600seconds
Distance= speed×time
= 2× 3600
= 7200m
In the first motion the distance is 7200m and the time is 3600seconds
SECOND MOTION
The values given are
Distance= 3000m
Time= 25 mins to seconds
= 25×60
= 1500 seconds
In the second motion distance is 3000m and the time is 1500 seconds
The total distance can be calculated by applying the formular ( d1-d2) since she moved in an opposite direction
Total distance= 7200-3000
= 4200m
The total time (t1+t2) = 1500+3600
= 5100 seconds
Therefore, average velocity is calculated by applying the formular
Total distance/ Total time
= 4200/5100
= 0.824m/s South
Hence the average velocity is 0.824m/s South.
Answer:
The woman's average velocity during her entire motion is 2 m/s
Explanation:
Given;
initial speed of the woman, u = 2.00 m/s
initial time taken, t₁ = 60 minutes = 3600 seconds
initial displacement of the woman, x₁ = ?
final displacement of the woman, x₂ = 3000 m north
final time taken , t₂ = 25.0 minutes = 1500 seconds
The woman's average velocity during her entire motion:
initial displacement of the woman, x₁ = u x t₁ = 2.00 m/s x 3600 seconds
= 7200 m South
[tex]Average \ velocity = \frac{\delta X}{\delta t} = \frac{X_1-X_2}{t_1-t_2} \\\\V_{avg.} = \frac{7200-3000}{3600-1500} = \frac{4200}{2100} = 2 \ m/s[/tex]
Therefore, the woman's average velocity during her entire motion is 2 m/s
A 58.0 kg snow skier is on the top of 351 m high hill. After she has gone down a vertical distance of 142 m, what is her mechanical energy? Explain your answer (CER)!
Answer:
Explanation:
Initially skier is at a height of 351 m . Her kinetic enery will be nil because she is at rest . Her potential energy will be calculated as follows
potential energy = mgh where m is mass , h is height and g is acceleration due to gravity
potential energy = 58 x 9.8 x 351
= 199508 .4 J
Total mehanical energy = potential energy + kinetic energy
= 199508.4 J
According to conservation of mechanical energy , at the height of 142 m also total mechanical energy will be same . At this height some potential energy will be converted into kinetic energy but total of potential and kinetic energy will be same.
Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form a charge of −7.50 nC? electrons (b) How many electrons must be removed from a neutral object to leave a net charge of 0.580 µC? electrons
Answer:
a) 4.681*10^10 electrons
b) 3.67*10^12 electrons
Explanation:
The amount of electrons in a charge of 1C is:
[tex]1C=6.2415*10^{18}\ electrons[/tex]
You use the previous equality as a conversion factor.
a) The sing of the charge is not important in the calculation of the number electrons, so, you use the absolute value of the charge
[tex]7.50nC=7.50*10^{-9}C*\frac{6.2415*10^{18}}{1C}=4.681*10^{10}\ electrons[/tex]
In 7.50nC there are 4.61*10^18 electrons
b)
[tex]0.580\mu C=0.580*10^{-6}C*\frac{6.2415*10^{18}}{1C}=3.67*10^{12}\ electrons[/tex]
To obtain a charge of 0.580 µC in a neutral object you need to take out 3.67*10^12 electrons
A bullet of mass 0.0021 kg initially moving at 497 m/s embeds itself in a large fixed piece of wood and travels 0.65 m before coming to rest. Assume that the acceleration of the bullet is constant. What force is exerted by the wood on the bullet
Answer:
The force exerted by the wood on the bullet is 399.01 N
Explanation:
Given;
mass of bullet, m = 0.0021 kg
initial velocity of the bullet, u = 497 m/s
final velocity of the bullet, v = 0
distance traveled by the bullet, S = 0.65 m
Determine the acceleration of the bullet which is the deceleration.
Apply kinematic equation;
V² = U² + 2aS
0 = 497² - (2 x 0.65)a
0 = 247009 - 1.3a
1.3a = 247009
a = 247009 / 1.3
a = 190006.92 m/s²
Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;
F = ma
F = 0.0021 x 190006.92
F = 399.01 N
Therefore, the force exerted by the wood on the bullet is 399.01 N
Which best describes earth's magnetic field lines?
Answer:
The field lines go out of Earth near Antarctica, enter Earth in northern Canada, and are not aligned with the geographic poles.
Explanation:
Answer: the field lines go out of earth near the north pole, enter earth in the south pole, and are not aligned with the geographic poles.
Explanation: just took the test on edge 2020.
with one example each , state and explain five outputs derieved from a business policy
A plane travels 2.5 KM at an angle of 35 degrees to the ground, then changes direction and travels 5.2 km at an angle of 22 degrees to the ground. What is the magnitude and direction of the planes total displacement??
Answer:
7.7 km 26°
Explanation:
The total x component is:
x = 2.5 cos(35°) + 5.2 cos(22°) = 6.87
The total y component is:
y = 2.5 sin(35°) + 5.2 sin(22°) = 3.38
The magnitude is:
d = √(x² + y²)
d = 7.7 km
The direction is:
θ = atan(y/x)
θ = 26°
Sometimes earthquakes occur at a boundary between the Indian and Eurasian Plates. The collisions of these two continental plates formed the Himalayan mountain range.
Based on this information, which type of boundary exists between the Indian and Eurasian Plates?
continental-continental convergent
continental-continental divergent
oceanic-continental divergent
oceanic-continental convergent
Answer:
Continental Continental convergent.
Explanation:
Continental Continental convergent is a type of earthquakes that occur between boundaries of two continents in which the tectonic plates move closer to each other or converge.
Eurasian plates refers to tectonic plates that is found in the Continent of Eurasia which include Asian and Europe excluding India subcontinent.
From the question, the earthquake occur between boundaries of two continents India and Eurasia and this converging and collision of the two continental plates formed the Himalayan mountain range.
Therefore, it is continental continental convergent.
Answer:
A
Explanation:
As you are waiting at the Willard airport for a friend coming from Chicago, suddenly something goes wrong with the loudspeakers, and they start making a low pitch hum. You realized that you are standing exactly in the middle between the two loudspeakers on the opposite walls. Since the sound from both loudspeakers has the same intensity, you are standing in the interference maximum As you move toward one of the loudspeakers, the intensity decreases. When you have moved by 0.5 m , you hit the first point of zero intensity. Assuming that the speed of sound is v = 330 m/s , what is the frequency f of the sound in Hz ?
Answer:
The frequency is [tex]f = 165 Hz[/tex]
Explanation:
From the question we are told that
The position of zero intensity is [tex]L = 0.5 \ m[/tex] from the center
Now the wavelength of the sound is mathematical represented as
[tex]\lambda = 4 L[/tex]
[tex]\lambda = 4 * 0.5[/tex]
[tex]\lambda = 2 \ m[/tex]
Now the frequency of the sound is mathematically represented as
[tex]f = \frac{v}{\lambda}[/tex]
substituting values
[tex]f = \frac{330}{ 2}[/tex]
[tex]f = 165 Hz[/tex]
A 12 cm diameter piston-cylinder device contains air at a pressure of 100 kPa at 24oC. The piston is initially 20 cm from the base of the cylinder. The gas is now compressed and 0.1 kJ of boundary work is added to the gas. The temperature of the gas remains constant during this process.
a. How much heat was transferred to/from the gas?
b. What is the final volume and pressure in the cylinder?
c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?
Answer:
ΔQ = 0.1 kJ
[tex]\mathbf{v_f = 1.445*10^{-3} m^3}[/tex]
[tex]\mathbf{P_f = 156.5 \ kPa}[/tex]
ΔS = -0.337 J/K
The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.
Explanation:
GIven that:
Diameter of the piston-cylinder = 12 cm
Pressure of the piston-cylinder = 100 kPa
Temperature =24 °C
Length of the piston = 20 cm
Boundary work ΔW = 0.1 kJ
The gas is compressed and The temperature of the gas remains constant during this process.
We are to find ;
a. How much heat was transferred to/from the gas?
According to the first law of thermodynamics ;
ΔQ = ΔU + ΔW
Given that the temperature of the gas remains constant during this process; the isothermal process at this condition ΔU = 0.
Now
ΔQ = ΔU + ΔW
ΔQ = 0 + 0.1 kJ
ΔQ = 0.1 kJ
Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ
b. What is the final volume and pressure in the cylinder?
In an isothermal process;
Workdone W = [tex]\int dW[/tex]
[tex]W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int \dfrac{dv}{V} \\ \\ \\ W = nRT In V |^{V_f} __{V_i}} \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}[/tex]
Since the gas is compressed ; then [tex]v_f< v_i[/tex]
However;
[tex]W =- nRT \ In \dfrac{V_f}{V_i}[/tex]
[tex]W =- P_1V_1 \ In \dfrac{V_f}{V_i}[/tex]
The initial volume for the cylinder is calculated as ;
[tex]v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3[/tex]
Replacing over values into the above equation; we have :
[tex]100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}[/tex]
The final pressure can be calculated by using :
[tex]P_1V_1 = P_2V_2 \\ \\ P_iV_i = P_fV_f[/tex]
[tex]P_f =\dfrac{P_iV_i}{V_f}[/tex]
[tex]P_f =\dfrac{100*2.261*10^{-3}}{1.445*10^{-3}}[/tex]
[tex]P_f = 1.565*10^2 \ kPa[/tex]
[tex]\mathbf{P_f = 156.5 \ kPa}[/tex]
c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?
The change in entropy of the gas is given by the formula:
[tex]\Delta S=\dfrac{\Delta Q}{T}[/tex]
where
T = 24 °C = (24+273)K
T = 297 K
[tex]\Delta S=\dfrac{-100 \ J}{297 \ K}[/tex]
ΔS = -0.337 J/K
The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.
A transmission line consisting of two concentric circular cylinders of metal, with radii a, and b, with b > a, is filled with a uniform dielectric material with permittivity ε and permeability μ0. A TEM mode is propagated along the line and the peak value of magnetic field when rho = a is B0. The cylinders are symmetric about the z-axis. Calculate the time-averaged power flow, P, along the line.
Answer:
P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.
Explanation:
So, we will be making use of the data or parameters given in the question above;
=>" radii a, and b, with b > a, is filled with a uniform dielectric material with permittivity ε and permeability μ0."
=> " A TEM mode is propagated along the line and the peak value of magnetic field when rho = a is B0."
So, we will be making use of the two equations below;
Ë = ( λ/ 2πEP) × P'. --------------------------(1)..
B' = √ μE × ( λ/ 2πEP) × P' --------------(2).
Where equation (1) and (2) represent Gauss' law and magnetic field equation respectively.
Jo = 1/√ μE × λ/2 × π × a.
When we solve for charge per unit length, we have;
λ = 2 × π × Jo × a × √ μE.
The energy flux,s = E' × J'= √μE× |Jo(Z) |^2 × a^2/b^2 { cos^2 kz - wt + Avg Jo} Z'.
Hence, the time. Average power flux = 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z'.
Therefore, P = ∫z' . <s> da
P = ∫ ∫ 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z' pd pd p Θ.
(Take limit on the first at second integration as : 2π,0 and b,a).
P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.