Answer:
Explanation:
Units are measure of a quantity. There are two types of units,
The fundamental units
The derived units
The fundamental units are units that are independent i.e they can stand alone while derived units are derived from fundamental units (They are dependent)
The fundamental units are the units of fundamental quantities (Length, mass and time)
Length is measured in metres (m)
Mass is measured in kilograms (kg)
Time is measured in seconds(s)
This are called the standard units where other units are derived.
Other units of mass are grams
Weight is the product of mass and acceleration due to gravity
Weight = mass×acc due to gravity
W = mg
Using units
W = kg × m.s²
Since acceleration is measured in m/s²
W = kgm/s²
Hence the metric unit of weight is kgm/s² or Newtons
An 800 N table is slid across the kitchen floor by pushing with a force of 100 N. If the table moves at a constant speed, what is the friction force with the floor?
A. 100 N
B. 8 N
C. 800 N
D. over 800 N
Answer:
A. 100 N
Explanation:
Given
Table = 800N
Pushing Force = 100N
Required
Determine the friction force with the floor
From the question, we understand that the table moves at a constant speed.
In this kind of situation, the friction force equals to the pushing force.
Hence:
Friction Force = Pushing Force
Friction Force = 100N
A golf ball rolls up a hill toward a miniature-golf hole. Assume the direction toward the hole is positive. If the golf ball starts with a speed of 2.0 m/s and slows at a constant rate of 0.50 m/s2. What is the golf ball’s velocity if the constant acceleration continues for 6.0 s? *
Answer:
0.2533333334
Explanation:
It is simple, just do the math
The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 621 nm in vacuum. This light propagates in a liquid whose index of refraction at this wavelength is 1.18.What is the speed of this light in the liquid
Answer:
2.54*10^8 m/s
Explanation:
Given that
Speed of light in space, c = 299792458 m/s
Wavelength of the beam of light, λ = 621 nm or 621*10^-9 m
Index of refraction of the liquid, n = 1.18
Speed of light in the liquid, v = ?
The index of refraction of an optical material, n is given by the following formula.
n = c/v, and if we substitute directly, we have
1.18 = 299792458 / v, if we make v subject of formula, we have
v = 299792458 / 1.18
v = 2.54*10^8 m/s
the difference between mechanical and chemical weathering
Answer:
Chemical weathering demands chemical reactions with minerals inside the rock and causes changes in rock composition. Sometimes this process will produce a different kind of product due to the reaction. Mechanical weathering only involves the physical breakage of rocks to smaller pieces of fragments.
Explanation:
a man in a gym is holding an 8.0 kg weight at arm's length, a distance of 0.55 m from his shoulder joint What is the torque about his shoulder joint due to the weight if his arm is horizontal
Answer:
[tex]the \: torque= 44 \: Nm.[/tex]
Explanation:
[tex] torque \: \boxed{t} \: is \: the \: product \: of \:one \: of \: the \: force \: - raduis \: \\ and \: sine \: of \: the \: angle \: of \: inclination : \\ torque \: \boxed{t} = fr \sin( \alpha ) \\ f = w = mg = 80. \times 10 = 80 \\ r = 0.55 \\ \alpha = 90 \\ \sin( \alpha ) = 1 \\ therefore \: the \: torque \: \boxed{t} \: is : \\ 80 \times 0.55 \times 1 = 44 \: Nm.[/tex]
♨Rage♨
A man is holding a weight at arm's length, a distance apart from his shoulder joint, it will experience torque. The torque about his shoulder joint due to the weight if his arm is horizontal is 43.12 N.m .
What is torque?Torque is the rotation caused about a point when a force is applied at a distance from the point.
The mass of man is 8.0 kg and the distance from the shoulder joint is 0.55 m. The torque applied will be the product of force and distance.
τ =F x r
and Force = weight of the body = mass x acceleration due to gravity
Substitute the values to get the torque.
τ = 8 kg x 9.8 m/s² x 0.55 m
τ = 43.12 N.m
Hence, the torque about his shoulder joint due to the weight if his arm is horizontal is 43.12 N.m
Learn more about torque.
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What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side?
Answer:
sciencephysicsphysics questions and answersA 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. ...
Question: A 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. The Block Is Released From Rest And Slides Across A Track As Shown In (Figure 1). While Most Of The Track Is Frictionless, There Is A 55-cm Section Of Track That Has A Coefficient Of Friction With The Block Of 0.4. A Bit Further On, The Track Ascends ...
A 12-kg block is pressed against a spring (spring constant 870 N/m ), compressing it some distance. The block is released from rest and slides across a track as shown in (Figure 1). While most of the track is frictionless, there is a 55-cm section of track that has a coefficient of friction with the block of 0.4. A bit further on, the track ascends into a hill that is 40-cm tall.
What is the minimum compression of the spring necessary for the block to slide past the section with friction?
What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side? 40cm 12kg Ww.w.wwwww 55cm
Explanation:
A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignore and using conservation of mechanical energy, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.
Answer:
a) The initial speed of the rock is approximately 14.607 meters per second.
b) The greatest height of the rock from the base of the cliff is 42.878 meters.
Explanation:
a) The rock experiments a free-fall motion, that is a vertical uniform accelerated motion due to gravity, in which air friction and effects of Earth's rotation. By Principle of Energy Conservation we have the following model:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}[/tex] (Eq. 1)
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energies, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energies, measured in joules.
By definitions of gravitational potential and translational kinetic energies we expand and simplify the equation above:
[tex]m\cdot g\cdot (y_{1}-y_{2})= \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})[/tex]
[tex]g\cdot (y_{1}-y_{2}) = \frac{1}{2}\cdot (v_{2}^{2}-v_{1}^{2})[/tex] (Eq. 2)
Where:
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and final height, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the rock, measured in meters per second.
If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 32\,m[/tex], [tex]y_{2} = 0\,m[/tex] and [tex]v_{2} = 29\,\frac{m}{s}[/tex], then the equation is:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (32\,m-0\,m) = \frac{1}{2}\cdot \left[\left(29\,\frac{m}{s} \right)^{2}-v_{1}^{2}\right][/tex]
[tex]313.824 = 420.5-0.5\cdot v_{1}^{2}[/tex]
[tex]0.5\cdot v_{1}^{2} = 106.676[/tex]
[tex]v_{1} \approx 14.607\,\frac{m}{s}[/tex]
The initial speed of the rock is approximately 14.607 meters per second.
b) We use (Eq. 1) once again and if we know that [tex]g =9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 32\,m[/tex], [tex]v_{1} \approx 14.607\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], then the equation is:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (32\,m-y_{2}) = \frac{1}{2}\cdot \left[\left(0\,\frac{m}{s} \right)^{2}-\left(14.607\,\frac{m}{s} \right)^{2}\right][/tex]
[tex]313.824-9.807\cdot y_{2} = -106.682[/tex]
[tex]9.807\cdot y_{2} = 420.506[/tex]
[tex]y_{2} = 42.878\,m[/tex]
The greatest height of the rock from the base of the cliff is 42.878 meters.
A block with mass 0.50 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m . When released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The spring constant k is 100 N/m. What is the coefficient of kinetic friction between the block and the tabletop
Answer:
μ = 0.41
Explanation:
[tex]\frac{1}{2} kx^2 = \mu mgd[/tex]
[tex]\mu = (\frac{1}{2} kx^2)/mgd[/tex]
[tex]\mu = (\frac{1}{2} (100)(0.2)^2)/(0.5)(9.81)(1)[/tex]
[tex]\mu = 0.41[/tex]
The coefficient of kinetic friction between the block and the tabletop is 0.41.
The given parameters;
mass of the block, m = 0.5 kgextension of the spring, x = 0.2 mspring constant, K = 100 N/mThe coefficient of kinetic friction between the block and the tabletop by applying the principle of conservation of energy as shown below.
Fd = Uₓ
μmgd = ¹/₂kx²
where;
μ is the coefficient of kinetic frictionThe coefficient of kinetic friction between the block and the tabletop;
[tex]\mu_k = \frac{kx^2}{mgd} \\\\\mu_k = \frac{100\times 0.2^2}{2\times 0.5\times9.8 \times 1 } \\\\\mu_k = 0.41[/tex]
Thus, the coefficient of kinetic friction between the block and the tabletop is 0.41.
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What is the answer to this question?
Answer:buttt
It mean b
Explanation:
Additional Practice on Speed
1. A group of campers and one group leader left a campsite in a canoe.
They traveled at an average rate of 10 km/h. Two hours later, the other
group leader left the campsite in a motorboat. He traveled at an average
rate of 22 km/h. How long had each boat traveled when the motorboat
caught up with the canoe?
a.
canoe: 2.2 h; motorboat: 1 h
b. canoe: 2.2 h; motorboat: 10 h
c. O canoe: 3 2 h; motorboat: 12 h
3
3
d. O canoe: 4.4 h; motorboat: 2 h
Answer:
canoe = 3 h 40 min. motorboat = 1 h 40 min
Explanation:
As both boats travel at a constant speed, we can apply the definition of average velocity to both, so we can get:[tex]v_{b} = \frac{\Delta x_{b} }{\Delta t_{b} } (1)[/tex]
[tex]v_{mb} = \frac{\Delta x_{mb} }{\Delta t_{mb} } (2)[/tex]
Now, when the motorboat caught up with the canoe, distance travelled by both boats were the same, so, solving (1) and (2) for Δx, we get:[tex]v_{b} *\Delta t_{b} = v_{mb} *\Delta t_{mb} (3)[/tex]
if we take the initial time for the canoe to be t₀ =0, the initial time for the motorboat will be tmb₀ = 2 Hs.So, replacing in (3) we have:[tex]v_{b} *t_{b} = v_{mb} *(t_{b} - 2 hs.) = 10 km/h* t_{b} = 22 km/h* (t_{b} - 2 hs.)[/tex]
Solving for tb:[tex]t_{b} = \frac{-44 }{-12} hs = 3h 40 min.[/tex]
tmb = tb - 2 hs. = 1h 40 min.
You are given two vectors A⃗ =−3.00i^+7.00j^ and B⃗ =5.00i^+2.00j^. Let counterclockwise angles be positive.
1- What angle θA, where 0∘≤θA<360∘, does A⃗ make with the +x-axis?
2-What angle θB, where 0∘≤θB<360∘, does B⃗ make with the +x-axis?
Express your answer in degrees.
3-Vector C⃗ is the sum of A⃗ and B⃗ , so C⃗ =A⃗ +B⃗ . What angle θC, where 0∘≤θC<360∘, does C⃗ make with the +x-axis?
Express your answer in degrees
Answer:
Ax = -3 By= 7 tan phi = -7 / 3 phi = -66.8
Since theta is in the second quadrant theta = 180 - 66.8 = 113.2 deg
Bx = 5 By = 2 tan theta = .4 theta = 21.8 deg
Cx = Ax + Bx = 2 Cy = Ay + By = 9
tan theta = 9 / 2 = 4.5 theta = 77.5 deg
1) The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].
2) The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].
3) The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].
1) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec A\,\bullet \, \vec u}{\|\vec A\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec A\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec A[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec A = (-3, 7)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec A\| = \sqrt{(-3)^{2}+7^{2}}[/tex]
[tex]\|\vec A\| = \sqrt{58}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(-3)\cdot (1)}{\sqrt{58} \cdot 1}[/tex]
[tex]\cos \theta \approx -0.393[/tex]
The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].
2) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec B\,\bullet \, \vec u}{\|\vec B\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec B\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec B[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec B = (5, 2)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec B\| = \sqrt{5^{2}+2^{2}}[/tex]
[tex]\|\vec B\| = \sqrt{29}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(5)\cdot (1)}{\sqrt{29} \cdot 1}[/tex]
[tex]\cos \theta \approx 0.928[/tex]
The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].
3) In this case, we need to find the value of [tex]\vec C[/tex] by vectorial sum before calculating the angles with respect to x-axis:
[tex]\vec C = \vec A + \vec B = (-3, 7) + (5, 2)[/tex]
[tex]\vec C = (2, 9)[/tex]
In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec C\,\bullet \, \vec u}{\|\vec C\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec C\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec C[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec B = (2, 9)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec C\| = \sqrt{2^{2}+9^{2}}[/tex]
[tex]\|\vec C\| = \sqrt{85}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(2)\cdot (1)}{\sqrt{85} \cdot 1}[/tex]
[tex]\cos \theta \approx 0.217[/tex]
The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].
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I HAVE A QUESTION ABOVE CAN ANYONE PLEASE HELP ME I WOULD REALLY APPRECIATE THE HELP I HAVE BEEN ON THIS QUESTION FOREVER i will give you the brainlist !!!!!
Answer:
answer is C
Explanation:
Answer: Monetary policy used during times of recession
Explanation:
A bus accelerates from rest at 2.25 m/s2 for 12.3 seconds.
Answer:
6.15m for 12.3 seconds
Explanation:
All living creatures have some form of sleep that they engage in at regular periods, making sleep a(n) __________ function. A. environmental B. biological C. species-specific D. psychological
The answer is B. biological
Answer:
ans: a biological function
Answer:
B
Explanation:
EDGE2020
A solenoid with 2,781 turns has a radius of 74.7 mm and is 38.4 cm long. If this solenoid carries a current of 68.4 A, what is the magnitude of the magnetic field near the center of the solenoid
Answer:
B = 0.62 T
Explanation:
Assuming that the solenoid can be treated as of infinite length (as its radius is much smaller than the length) we can consider that the magnetic field created by the current, near the center of it, is constant, and that it's almost zero outside it.Applying Ampere's law to a closed surface that encloses the wire carrying the current, we get:[tex]B= \mu_{o} *n*I (1)[/tex]
where μ₀ = 4*π*10⁻⁷ T*m/A, n = turns per unit length = 2,781/0.384 m and I = 68.4 AReplacing by the givens in (1) we get:B= 4*π*10⁻⁷ T*m/A* 7242.2 (1 /m) * 68.4 A = 0.62 TB = 0.62 Tif a cross country runner covers 347 meters in 134 seconds what is her speed
A truck is traveling at an initial velocity
of +39m/s and it starts slowing down smoothly for 3.3s. It covers a distance of 45m while slowing down. What is the acceleration?
Answer:
Approximately [tex]-15.37m/s^2[/tex]
Explanation:
The acceleration can be found using the formula:
[tex]x_{f} = x_{i} + v_{i} (t)+\frac{1}{2} (a)(t^{2} )[/tex]
The work is as shown:
[tex]45m = 0 + 39 m/s (3.3s)+\frac{1}{2} a(3.3s)^2[/tex]
[tex]45m=128.7m + 5.445s^2 a[/tex]
[tex]-83.7m=5.445s^2a[/tex]
[tex]a = -15.37190083m/s^2[/tex]
a is about -15.37[tex]m/s^2[/tex]
A dart is loaded into a compressed spring. Starting from rest the spring is released and shoots the dart up into the air. From the instant the dart is released (before it starts to move) until it reaches its maximum height, what is the Net work done on the dart
Answer:
Wnet = 0
Explanation:
The work-energy theorem says that the total net work done on an object, is equal to the change in the kinetic energy of that object.In this case, the total change in kinetic energy of the dart is zero, due to the initial status is at rest (loaded into a compressed spring) and the final is also at rest (when it reaches to its maximum height).The total work done on the dart can be written as follows:[tex]W_{net} = \frac{1}{2} k* \Delta x^{2} - m*g*h_{max} (1)[/tex]
Now, in absence of friction, all the elastic potential energy becomes gravitational potential energy, so the following is true:[tex]\frac{1}{2} k* \Delta x^{2} = m*g*h_{max} (2)[/tex]
From (1) and (2) we conclude that Wnet = 0.what vehicle are runaway ramps designed for
Answer:truck
Explanation:
truck
A 3.05 kg block is sliding across level ground and feels a 5.27 N friction force. What is the coefficient of friction?
Answer:
0.176
Explanation:
(5.27)/(3.05*9.8)
It is indeed 0.176 for the answer.
A half-eaten apple turns brown which of these is evidence a chemical change took place A.th3e apple is softer b.the apple changed color C.half the apple was eaten D.the apple is warmer \
Answer:
b the apple changed color
What is the mass of an object if a force of 17N causes it to accelerate at 1.5 m/sec/sec
A. 11.38
B. 11.3 kg
C. 12.3 kg
D. 14.3 kg
Answer:
B.11.3 kg
Explanation :
F=ma here force is given as 17N and acceleration as 1.5 m/sec/sec using the formula mass can be calculated
∴ F=ma
17N =m × 1.5 m[tex]s^{-2}[/tex]
[tex]m=\frac{F}{a}[/tex]
[tex]m= \frac{17 kg m/sec/sec}{1.5 m/sec/sec}[/tex]
∴ m=11.3 kg
A current of 4sin4t A flows through a 5-F capacitor. Find the voltage v(t) across the capacitor given that v(0)
Answer:
[tex]v(t) = \frac{-cos4t}{5} + \frac{6}{5} Volts[/tex]
Explanation:
Given
current I = 4sin4t
capacitance C = 5F
v(0) = 1
The voltage across the capacitor is expressed as:
[tex]v(t) = \frac{1}{C} \int\limits^t_{t_0} {i(t)} \, dt + v(t_0)[/tex]
Given v(0) = 1, means at t = 0, v(t0) = 1
Substitute
[tex]v(t) = \frac{1}{5} \int\limits^t_{0} {4sin4t} \, dt + 1\\v(t) = \frac{1}{5} [\frac{-4cos4t}{4} ] + 1\\\\v(t) = \frac{1}{5} (-cos4t)+1 + \frac{1}{5}\\v(t) = \frac{-cos4t}{5} + \frac{6}{5}[/tex]
Which answer is it
0.32 m/s
3.2 m/s
9.8 m/s
40 m/s
Answer:
3.2 m/s
Explanation:
you take the amount of blocks height wise, and the number of blocks with wise, and divide. height over with, so in this case 16/5. 16 divided by 5 is 3.2
Besides being asleep and being awake list the different states of consciousness
Answer:
bias.
Consciousness.
Hypnosis.
Priming.
Sleep.
Trance.
Explanation:
a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy
Answer:
E = 1.76 J
Explanation:
Given that,
Mass of an object, m = 0.4 kg
It moves by a vertical distance of 0.45 m in the Earth's gravitational field.
We need to find the change in its gravitational potential energy. It can be given by the formula as follow :
[tex]E=mgh\\\\E=0.4\times 9.8\times 0.45\\\\E=1.76\ J[/tex]
So, the change in its gravitational potential energy is 1.76 J.
Help please it’s urgent
A fisherman notices that one wave passes the bow of his anchored boat
every 3 seconds. He measured the wavelength to be 8.5 meters. How fast
are the waves traveling?
1.)2.83m/s
2.)0.283m/s
3.)28.3m/s
An animation is being created with a thousand tennis balls falling out of a truck, bouncing off the ground and each other. The animator wants to make the scene realistic but also wants to be efficient in the program. Which of these characteristics could be ignored when establishing boundary conditions for the system?
aerodynamic properties
gravitational forces
Newton's laws of motion
the diameter of the tennis balls
Answer:
aerodynamic properties
Explanation:
For this test the correct answers are:
1. Initial conditions
2. the speed of a character at the moment of contact
3. the hardness of the sidewalk where the mug landed
4. the location where potential energy is zero
5. aerodynamic properties
I just took this test and these were the correct answers.
Which is the right one?
(Show work)
What average mechanical power (in W) must a 69.0 kg mountain climber generate to climb to the summit of a hill of height 305 m in 48.0 min
Answer: Power= 71.61W
Explanation:
POWER = Work done / Time
But Work done = Gravitational Potential energy
= Mgh
Therefore Power = Mgh/ t
g= 9.8m/s
Where t in mins to secs becomes 48.0x 60=2880 secs
Power= 69x 9.8 x 305/ 2880
=206,241/2880
71.61W