Answer:
a) 4.8*10^-19 C
b) 3
Explanation:
We know that
F = ma
Also, recall that a = v²/r and F = qvB
Substituting this in the first equation, we have
qvB = mv²/r, since we're looking for q, we make it the subject of the formula and solve for it.
q = mv²/rvB
q = mv/rB
q = [2.66*10^-26 * (5*10^6)] / 0.231 * 1.2
q = 1.33*10^-19 / 0.2772
q = 4.8*10^-19 C
Ratio between the charge and the electron is q/c =
4.8*10^-19 / 1.6*10^-19 = 3
Therefore, the ratio between the charge and electron is 3
Question 2: If a runner travels 50 m in 5 s, his acceleration is
1 point
250 m/s/s
50 m/s/s
10 m/s/s
2 m/s/s
This is the motion of an object when the only force acting on it is the force of gravity.
A) Newton's Third Law
B)Newton's First Law
C)Free fall
D)Terminal velocity
Answer:
c it is free fall when an object force act on the force of gravity
A jogger runs at a speed of 4 m/s for 500]s, slows to 2.5 m/s for the next
300 s and then travels the final 900 s at a speed of 3.5 m/s. What is the
average speed of the jogger in m/s?
O 10 m/s
O 3.47 m/s
O 5 m/s
O 3.33 m/s
Answer:
3.33 m/s
Explanation:
Just add up all of the speeds and divide it by 3 (since there are 3 different speeds)
(4 m/s + 2.5 m/s + 3.5 m/s) / 3 = 3.33 m/s
A 54.0-kg ice skater is moving at 3.94 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.900 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. .931416 Correct: Your answer is correct. kN (b) Compare this force with her weight. Frope W
Answer:
Explanation:
a ) magnitude of force = centripetal force = m v² / R , m is mass , v is velocity of skater , R is radius of circular path .
= 54 x 3.94² / .9
= 931.41 N
= .93141 kN.
b ) Her weight = mg = 54 x 9.8 = 529.2 N
= .5292 kN .
Ratio = .93141 / .5292 = 1.76
A particle starts from rest and is acted on by a net force that does work at a rate that is proportional to the time t. The speed of the particle is proportional to:_____
a. √t
b. t
c. t^2
d.1/√t
e. 1/t
Answer:
The speed of the particle is proportional to:
c. t²
Explanation:
Given;
initial velocity of the particle, u = 0
let the net force on the object = F
Work done on the particle is given by;
W = F x d
W ∝ t
[tex]Fd \ \alpha \ t\\\\\frac{mv}{t}d \ \ \alpha \ t\\\\mvd \ \ \alpha \ t^2\\\\v \ \ \alpha \ t^2[/tex]
Therefore, the speed of the particle is proportional to t²
The speed of the particle is proportional to t.
It should be noted that work is a change in kinetic energy. Therefore, power will be calculated as:
Power = Work / Time = force × velocity
Power is proportional to t. Since force is a constant, then the velocity will be proportional to t. Therefore, the speed of the particle is proportional to t.
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If 2000 ft³ of air is crossing an evaporator coil and is cooled from 80°F to 60°F, what would be the volume of air in ft³ exiting the evaporator coil?
Given :
If 2000 ft³ of air is crossing an evaporator coil and is cooled from 80°F to 60°F.
To Find :
The volume of air in ft³ exiting the evaporator coil.
Solution :
We know, relation between initial and final temperature and volume is given by :
[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}[/tex]
Here, temperature is in Kelvin .
We know, Kelvin temperature is given by :
[tex]K=(F-32)\times \dfrac{5}{9} + 273[/tex]
Putting F = 60° F,
[tex]K=(60-32)\times \dfrac{5}{9} + 273\\\\K = 288.56\ K[/tex]
Putting F = 80° F,
[tex]K=(80-32)\times \dfrac{5}{9} + 273\\\\K = 299.67\ K[/tex]
Putting all values, we get :
[tex]\dfrac{V_1}{2000}=\dfrac{60}{80}\\\\V_1=\dfrac{60\times 2000}{80}\ ft^3\\\\V_1=1500\ ft^3[/tex]
Hence, this is the required solution.
A dog with a mass of 57.0 kg slides down a wet slope with negligible friction. The dog starts from rest and has a speed of 2.10 m/s at the bottom. What is the height of the slope (in m)?
The dog starts from rest and has a speed of 2.10 m/s at the bottom. The height of the slope will be 0.225 m
What is friction?Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.
The friction force prevents any two surfaces of objects from easily sliding over each other or slipping across one another. It depends upon the force applied to the object.
As there is almost zero friction between the total potential energy height is equal to the kinetic energy
m*g*h = 1/2* m*v²
Where m is the mass of the dog
g is the acceleration due to gravity,h is the height of the slope
v is the velocity of the sliding dog
g*h = 1/2* v²
9.81*h = 0.5* 2.1²
=0.225 m
The height of the slope would be 0.225 m.
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The earth travels at 68,000 miles/hour as it moved around the sun. How many miles does the earth travel in one trip around the sun
Answer:
595,680,000 miles
(= 5.96 x 10⁸ miles)
Explanation:
we are given the speed of travel as 68,000 mi/hr
if we assume one common year
= 365 days
= 365 days x 24 hours/day
= 8760 hours
Then the distance travelled
= speed x time
= 68,000 x 8760
= 595,680,000 miles
= 5.96 x 10⁸ miles (rounded to 3 sig. fig and expressed in scientific form)
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Hope this helps!
What type of energy is possessed by a pear falling from a tree,
just before it touches the ground?
zero energy
kinetic energy
potential energy
kinetic and potential energy
Answer:
Kinetic Energy
Explanation:
I can't really explain it, but that's the correct answer.
9. A 50 kg halfback is in the process of making a turn on a football field.
The halfback makes 1/4 of a turn with a radius of 15 meters in 2.1 seconds
before being tackled. What is the net force acting on the halfback before
he is tackled?
Answer:
Net force = 419.5N
Explanation:
Given the following data;
Mass = 50kg
Radius = 15m
Time = 2.1 secs
Turns = 1/4 = 0.25
In order to find the net force, we would first of all solve for the speed and acceleration of the halfback.
To find speed;
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the equation;
[tex]Speed = \frac{distance}{time}[/tex]
But the distance traveled is given by the circumference of a circle = [tex] 2\pi r[/tex]
Since he covered 1/4th of a turn;
[tex] Distance = 0.25 * 2 \pi *r[/tex]
Substituting into the equation;
[tex] Speed, v = \frac {(0.25*2*3.142 * 15)}{2.1}[/tex]
[tex] Speed, v = \frac {23.565}{2.1}[/tex]
Speed, v = 11.22m/s
To find acceleration;
[tex] Acceleration, a = \frac {v^{2}}{r}[/tex]
Where, v = 11.22m/s and r = 15m
Substituting into the equation, we have;
[tex] Acceleration, a = \frac {11.22^{2}}{15}[/tex]
[tex] Acceleration, a = \frac {125.8884}{15}[/tex]
Acceleration, a = 8.39m/s²
To find the net force;
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
[tex] F = ma[/tex]
Where;
F represents force.m represents the mass of an object.a represents acceleration.Substituting into the equation, we have;
[tex] F = 50 * 8.39[/tex]
F = 419.5N
Therefore, the net force acting upon the halfback is 419.5 Newton.
What is centripetal force?Define it with example.
Answer:
Centripetal force is defined as, "the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation,"
Explanation:
so if you put a small ball por grape and put it on a table and you take a wine glass and spin it around the grape the grape will continue spinning even if you stop spinning the glass hope this helped
;)
Answer:
A force acting on a moving body at an angle to the direction of motion, tending to make the body follow a circular or curved path. The force of gravity acting on a satellite in orbit is an example of a centripetal force; the friction of the tires of a car making a turn similarly provides centripetal force on the car.
Explanation:
Car is going from 0 to 60 miles per hour in 6 seconds what is the advertised acceleration in mph squared
Answer:
10 m/h^2
Explanation:
a=v/t
a=60/6
a=10 m/h^2
4. From home a car drives 16 km east and then 24 km south. What is the resultant vector for
the car?
What constant acceleration is required to bring a rocket to an altitude of 650 m [up] from rest in 7.2 s? What is the rocket’s velocity at that point?
Answer:
a = 15.68 [m/s²]; v = 112.92 [m/s]
Explanation:
To solve this problem we must use the combination of two interesting topics of physics, the principle of energy conservation and then the application of kinematics.
The principle of Energy Conservation tells us that kinetic energy is transformed into potential energy or vice versa. For this particular problem, we must imagine the rocket at a height above 650 [m], suddenly this rocket falls to the ground. We will propose the reference point of potential energy at ground level, at this point the potential energy is equal to zero.
[tex]E_{pot}=E_{kin}[/tex]
[tex]E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}[/tex]
where:
m = mass of the rocket [kg]
g = gravity acceleration = 9.81[m/s²]
v = velocity of the rocket [m/s]
m*g*h = 0.5*m*v²
9.81*650 = 0.5*v²
v = √((6376.5)/0.5)
v = 112.92 [m/s]
Now using the following kinematic equation we have:
[tex]v_{f} = v_{o} + (a*t)[/tex]
where:
Vf = final velocity = 112.92 [m/s]
Vo = initial velocity = 0 (at the begining the rocket is at rest)
a = rocket acceleration [m/s²]
t = time = 7.2 [s]
Note: in the kinematics equation the positive sign of acceleration means that the rocket accelerates in the direction of motion, i.e. activates its thrusters to descend and descend to 650 [m] in 7.2 [s].
112.92 = 0 + (a*7.2)
a = 15.68 [m/s²]
An Indy 500 race car's velocity increases from 2 mis to 40 mis over a 40 s time interval. What is its
acceleration ?
Answer:8
Explanation:
In 2 to 3 sentences explain the roles of chloroplast and mitochondria in the process of plant cells gaining energy.
Answer:
Explanation:
Mitochondria is the powerhouse of the cell. And Chroloplast helps plants capture energy
A cat is running at 8.0 m/s and trying to catch a mouse, who has a tenth of the mass of the cat. If the cat speeds up by 2.0 m/s, the cat has ten times of the kinetic energy of the mouse. What is the speed of the mouse
Answer:
The speed of the mouse is 10 m/s
Explanation:
Given;
speed of the cat, u = 8 m/s
let the mass of the cat = m
then, mass of the mouse, = m / 10
let the speed of the mouse = v
Let the kinetic energy of the mouse = K.E
When the cat speeds up by 2 m/s, its new speed = 10 m/s,
then its kinetic energy is given as;
¹/₂m(10)² = 10K.E
[tex]50 m = 10[\frac{1}{2} *\frac{m}{10}.v^2][/tex]
50 = ¹/₂v²
v² = 100
v = √100
v = 10 m/s
Therefore, the speed of the mouse is 10 m/s
1. A balloon drifts 100 m towards the west in 10 seconds, then the wind suddenly changes
and the balloon flies 60m towards the east in the next 6 seconds.
a) What total distance did it travel? 160 meters
b) What was the average speed during the first 10 seconds?
c) What was its average speed during the next 6 seconds?
d) What was its average speed for the entire trip?
e) What was its total displacement?
1) What was its average velocity during the first 10 seconds?
g) What was its average velocity during the next 6 seconds?
h) What was its average velocity for the entire trip?
Answer:
Explanation:
a) The total distance traveled by the balloon = (100 + 60) m
= 160 metes
b) speed = [tex]\frac{distance covered}{time taken}[/tex]
The average speed for the first flight = [tex]\frac{100}{10}[/tex]
= 10 m/s
c) The speed during its next flight = [tex]\frac{60}{6}[/tex]
= 10 m/s
d) Average speed for the entire trip = [tex]\frac{(10 + 10)}{2}[/tex]
= 10 m/s
e) Total displacement = [tex]\sqrt{(100)^{2} + (60)^{2} }[/tex]
= [tex]\sqrt{13600}[/tex]
= 116.62 m
f) velocity = [tex]\frac{displacement}{time}[/tex]
Average velocity during the first 10 seconds = [tex]\frac{100}{10}[/tex]
= 10 m/s
g) Average velocity during the next 6 seconds = [tex]\frac{60}{6}[/tex]
= 10 m/s
h) Average velocity for the entire trip = [tex]\sqrt{(10)^{2} + (10)^{2} }[/tex]
= [tex]\sqrt{200}[/tex]
= 14.14 m/s
how much heat is needed to convert 1.0kg of ice at -10°C to steam at 100°C?
Q= Q1 +Q2 +Q3+Q4+ Q5
3066 J
In a popular demonstration, vinegar and baking soda are combined in a bottle as shown. Sodium carbonate, water as well as carbon dioxide gas are formed. The balloon traps the carbon dioxide.
According to the law of conservation of mass, the total mass of the reactants equals the total mass of the products. How could you alter the demonstration above so that the mass of products would not equal the total mass of the reactants?
A) Use less of the reactants.
B) Do not use a clean bottle.
C) Take the balloon off the bottle.
D) Shake the bottle while the chemicals are reacting.
Answer:
Shake the bottle while the chemicals are reacting.
Explanation:
Answer:
C
Explanation:
If you take the balloon off, the product will leak.
A car going around a curve of radius R at a speed V experiences a centripetal acceleration ac. What should be the radius of the path that leads to same centripetal acceleration while it travels at speed 2V
Answer:
4R
Explanation:
The centripetal acceleration a is given by a = v²/r where V = speed of object and r = radius of circle. Now, it is given that V = speed of car = V and radius of circle = R, so its centripetal acceleration a = V²/R.
Now, if the speed is doubled to V' = 2V with the same centripetal acceleration, we find the new radius of the curve R' from
a = V'²/R'
R' = V'²/a
R' = (2V)² ÷ V²/R
R' = 4V²R/V²
R' = 4R
So, the new radius is 4 times the initial radius.
2. An aluminum plate has a circular hole cut in it. A copper ball (solid sphere) has exactly the same diameter as the hole when both are at room temperature, and hence can just barely be pushed through it. If both the plate and the ball are now heated up to a few hundred degrees, the ball will fit more easily. Why
Answer:
Thermal expansion
Explanation:
One of the effects of heat is that it causes expansion of materials.
When the copper ball and aluminium plate are both heated, the both materials expand. When they expand, the copper ball can is found to fit more easily into the aluminium plate after both metals have become wider than they were at room temperature.
The coefficient of thermal expansion for aluminium is 35% greater than that of copper. This is the reason why the copper ball can fit into the aluminum plate more easily at higher temperature. The aluminum has expanded 35% more than copper making it easier and leaving a lot of space for the copper to fit into the aluminum plate.
What is conserved in physical changes?
O shape
O energy
O mass
Odensity
4. A 5.0 kg block moving to the right at 12.0 m/sec collides with a 4.0 kg block moving to the right at 2 m/sec. If the 4.0 kg moves to the right at 10 m/sec after the collision, what will be the speed of the 5.0 kg block? How much energy was lost during the collision?
Answer:
Energy lost = 55.28Joules
Explanation:
Using the law of momentum:
m1u1+m2u2 = m1v1+m2v2
m1 and m2 are the masses of the objects
u1 and u2 are the initial velocities.
v1 and v2 are final velocities
Given
m1 = 5kg
m2 = 4kg
u1 = 12m/s
u2 = 2m/s
v1 = 10m/s
v2 = ?
Substitute
5(12)+4(2)=4(10)+5v2
60+8 = 40+5v2
68-40 = 5v2
28 = 5v2
v2 = 28/5
v2 = 5.6m/sec
Hence the speed of the 5kg block is 5.6m/s
Energy lost = Kinetic energy after collision - kinetic energy before collision
KE after collision = 1/2m1v1²+1/2m2v2²
KE after collision = 1/2(5)10²+1/2(4)(5.6)²
= 250+62.72
= 312.72Joules
KE before collision = 1/2m1u1²+1/2m2u2²
KE before collision = 1/2(5)12²+1/2(4)(2)²
= 360+8
= 368Joules
Energy lost = 368-312.72
Energy lost = 55.28Joules
An athlete swings a 6.50-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.900 m at an angular speed of 0.700 rev/s. (a) What is the tangential speed of the ball
Answer:
v = 3.951 m/s
Explanation:
Given that,
Mass of a ball, m = 6.5 kg
Radius of the circle, r = 0.9 m
Angular speed of the ball, [tex]\omega=0.7\ rev/s=4.39\ rad/s[/tex]
Let v is the tangential speed of the ball. It is given in terms of angular speed is follows :
[tex]v=r\omega\\\\v=0.9\times 4.39\\\\v=3.951\ m/s[/tex]
So, the tangential speed of the ball is 3.951 m/s.
A roller coaster stops at the top of a hill. What force brings the roller coaster back down to the ground?
Answer:
Simple enough that it seems like a trick question: gravity.
Very Urgent!!!!!!! Please!!!!!!!!
A 35kg child standing on ice (frictionless) pushes a 7kg bowling ball with a force of 100N. What is the force acting on the 35kg child? What is the acceleration of the 35kg child?
Please real answers!! Gives many points!!!!!!
Answer:
a = 2.85714 m/s^2
Explanation:
Fnet=ma
eaqual and oppisate forces means that the 100N pushed on the ball comes back to the child.
100 = 35*a
a= 100/35
An object thrown straight upward reaches a height of 10 meters. If the object is thrown upward with twice this speed, it will reach a height of ____ m.
Answer:
If the object is thrown upward with twice the initial speed, it will reach a height of 40 meters.
Explanation:
Based on the Principle of Energy Conservation we find that the square of the initial speed of the object ([tex]v[/tex]), measured in meters per second, is directly proportional to its change in height ([tex]\Delta h[/tex]), measured in meters. That is:
[tex]v^{2} \propto \Delta h[/tex]
[tex]v^{2} = k\cdot \Delta h[/tex] (Eq. 1)
Where [tex]k[/tex] is the proportionality constant, measured in meters per square second.
Such constant is eliminated by using the following relationship:
[tex]\left(\frac{v_{2}}{v_{1}} \right)^{2} = \frac{\Delta h_{2}}{\Delta h_{1}}[/tex]
[tex]\Delta h_{2} = \left(\frac{v_{2}}{v_{1}} \right)^{2}\cdot \Delta h_{1}[/tex] (Eq. 2)
Where:
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the object, measured in meters per second.
[tex]\Delta h_{1}[/tex], [tex]\Delta h_{2}[/tex] - Initial and final changes in height, measured in meters.
If we know that [tex]\frac{v_{2}}{v_{1}} = 2[/tex] and [tex]\Delta h_{1} = 10\,m[/tex], then the change in height is:
[tex]\Delta h_{2} = 2^{2}\cdot (10\,m)[/tex]
[tex]\Delta h_{2} = 40\,m[/tex]
If the object is thrown upward with twice the initial speed, it will reach a height of 40 meters.
When the object is thrown with double the speed, it reaches a height of 40m.
Laws of motion:Given that the height attained by the object is h = 10m.
Let the initial speed of the object be u
and its final speed will be v = 0 at the highest point.
Then from the third equation of motion, we get,
[tex]v^2=u^2-2gh\\\\0=u^2-2gh\\\\h=\frac{u^2}{2g}[/tex]
if the object is thrown at double the speed, that is u' = 2u, then the height attained will be:
[tex]h'=\frac{u'^2}{2g}\\\\h'=\frac{(2u)^2}{2g}\\\\h'=\frac{4u^2}{2g}\\\\h'=4h[/tex]
h' = 4 × 10m
h' = 40m
So, the height attained will be 40 m.
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PLEASE HELP!!!!
An object is launched horizontally from a cliff. The cliff is 80 m high and the object has an initial launch velocity of 50 m/s.
What is the initial horizontal velocity?
What is the initial vertical velocity?
What is the final horizontal velocity?
How much time did it take for the object to hit the ground?
What is the final vertical velocity?
What is the final resultant speed?
How far from the base of the cliff will the projectile land?
Answer:
a. 50 m/s b. 0 m/s c. 50 m/s d. 4.04 s e. -39.6 m/s f. 63.78 m/s g. 202 m
Explanation:
a. What is the initial horizontal velocity?
Since the object is launched horizontally, it initial horizontal velocity is 50 m/s
b. What is the initial vertical velocity?
Since the object is launched horizontally, it has no initial vertical component. So, its initial vertical velocity is 0 m/s
c. What is the final horizontal velocity?
Its final horizontal velocity is 50 m/s since no force acts on it in the horizontal direction to change its value.
d. How much time did it take for the object to hit the ground?
We use the equation s = ut - 1/gt² since the object is falling under gravity where u = initial vertical velocity = 0 m/s, s = height of cliff = 80 m, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the object to hit the ground.
s = ut - 1/2gt²
80 m = 0 × t - 1/2 × -9.8 m/s² × t²
80 m = 4.9 m/s² × t²
t² = 80 m ÷ 4.9 m/s²
t² = 16.33 s²
t = √(16.33 s²)
t = 4.04 s
e. What is the final vertical velocity?
Using v = u + at where u = initial vertical velocity = 0 m/s, v = final vertical velocity, a = acceleration = -g = -9.8 m/s²and t = time it takes object to reach the ground = 4.04 s.
Substituting these values into the equation, we have
v = u + at
v = 0 m/s + (-9.8 m/s²) × 4.04 s
v = -39.6 m/s
f. What is the final resultant speed?
The final resultant speed v' is the resultant of the final horizontal velocity and the final vertical velocity. Let u' = final vertical velocity = 50 m/s.
v' = √(u'² + v²)
v' = √((50 m/s)² + (-39.6 m/s)²)
v' = √(2500 m²/s² + 1568.16 m²/s²)
v' = √(4068.16 m²/s²)
v' = 63.78 m/s
g. How far from the base of the cliff will the projectile land?
The distance from the base of the cliff, d where the projectile lands is
d = u't where u' = horizontal velocity = 50 m/s and t = time it takes object to land = 4.04 s
d = 50 m/s × 4.04 s
d = 202 m
A 4.33-kg soccer ball rolling eastward at a speed of 2.74 m/s is kicked so that it reverses direction and attains a speed of 6.35 m/s. If the duration of the interaction is 55.33 ms, what is the average force on the ball by the player's foot, in N
Answer:
0.71 N
Explanation:
The inertia of a body, I, is the product of the force applied on it and the time which it acts. Also, it is equal to the change in the momentum, P, of a body.
So that:
I = Ft
I = ΔP
Ft = m(v - u)
Where F is the force, t is the time, m is the mass of object, v is the final velocity and u is the initial velocity.
Given that: m = 4.33 kg, u = -2.74 m/s, v = 6.35 m/s and t = 55.33 s
F x 55.33 = 4.33(6.35 - (-2.74))
F x 55.33 = 4.33(6.35 + 2.74)
55.33 F = 4.33 x 9.09
55.33 F = 39.3597
F = [tex]\frac{39.3597}{55.33}[/tex]
= 0.711363
The average force on the ball by the player is 0.71 N.