1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
natta225 [31]
3 years ago
9

If you exert a force of 10.0 n to lift a box a distance of 0.9 m how much work do you do

Physics
1 answer:
dalvyx [7]3 years ago
5 0

A bit vague since it's not clear whether the applied force is in the same direction as the box's displacement. But let's assume it is. Then the work done is:

W = Fd

W = work, F = force, d = displacement

Given values:

F = 10.0N, d = 0.9m

Plug in and solve for W:

W = 10.0(0.9)

W = 9J

You might be interested in
A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,
fomenos

Answer:

The momentum would be doubled

Explanation:

The magnitude of the momentum of the freight train is given by:

p=mv

where

m is the mass of the train

v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

m'=2m

therefore, the new momentum is

p'=m'v=(2m)v=2(mv)=2p

so, the momentum has also doubled.

7 0
3 years ago
A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic
Ilia_Sergeevich [38]

Answer:

\mu_k=0.27

Explanation:

According to the free body diagram, in this case, we have:

\sum F_x:-F_k=ma\\\sum F_y:N=mg

Recall that the force of friction is given by:

F_k=\mu_k N

Replacing and solving for the coefficient of kinetic friction:

-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}

Finally, we calculate \mu_k:

\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27

4 0
3 years ago
When two continental crusts converge what land form is created?
mixas84 [53]

The compressional forces stemming from a convergent plate boundary.

There will also be earthquakes along the plate margin. This is also referred to as a collision boundary.


Hope this helps

6 0
3 years ago
A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2
Ahat [919]

Answer:

Approximately 0.608 (assuming that g = 9.81\; \rm N\cdot kg^{-1}.)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

  • Let m represent the mass of this car.
  • Let r represent the radius of the circular track.

This answer will approach this question in two steps:

  • Step one: determine the centripetal force when the car is about to skid.
  • Step two: calculate the coefficient of static friction.

For simplicity, let a_{T} represent the tangential acceleration (1.90\; \rm m \cdot s^{-2}) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to 90^\circ or \displaystyle \frac{\pi}{2} radians.

The angular acceleration of this car can be found as \displaystyle \alpha = \frac{a_{T}}{r}. (a_T is the tangential acceleration of the car, and r is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity (u and v) to (tangential) acceleration a_{T} and displacement x:

v^2 - u^2 = 2\, a_{T}\cdot x.

The idea is to solve for the final angular velocity using the angular analogy of that equation:

\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta.

In this equation, \theta represents angular displacement. For this motion in particular:

  • \omega(\text{initial}) = 0 since the car was initially not moving.
  • \theta = \displaystyle \frac{\pi}{2} since the car travelled one-quarter of the circle.

Solve this equation for \omega(\text{final}) in terms of a_T and r:

\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}.

Let m represent the mass of this car. The centripetal force at this moment would be:

\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}.

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, m\, g.

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let \mu_s denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g.

The size of this force should be equal to that of the centripetal force when the car is about to skid:

\mu_s\, m\, g = \pi\, m\, a_{T}.

Solve this equation for \mu_s:

\mu_s = \displaystyle \frac{\pi\, a_T}{g}.

Indeed, the expression for \mu_s does not include any unknown letter. Let g = 9.81\; \rm N\cdot kg^{-1}. Evaluate this expression for a_T = 1.90\;\rm m \cdot s^{-2}:

\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608.

(Three significant figures.)

7 0
3 years ago
In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 2.2 rev/
Hunter-Best [27]

Answer:

n_{T} = 31.68\,rev

Explanation:

The angular acceleration is:

\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}

\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}

And the angular deceleration is:

\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}

\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}

The total number of revolutions is:

n_{T} = n_{1} + n_{2}

n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}

n_{T} = 31.68\,rev

4 0
3 years ago
Other questions:
  • Jim runs off a diving board and lands in the water 3 meters from the end of the board. When he runs at the same speed on a highe
    8·1 answer
  • Which system protects us from radiation and the vacuum of space
    11·2 answers
  • Which characteristics is common in mature rivers river
    7·2 answers
  • Which form of energy does a plant store when light is transformed during photosynthesis
    15·1 answer
  • Which of the following statements best explains why a book resting on a table is in equilibrium?
    8·1 answer
  • Bill and Karen go to the ice-skating rink. Bill has twice the mass of Karen. While they are standing still and talking, suddenly
    6·1 answer
  • In a cell, the amount nutrition coming in equals the amount of waste going out. This is an example of _____.
    12·1 answer
  • Which types of light are absorbed by genetic material?
    15·1 answer
  • boat travels through a river at a speed of 25 m/s, passing through schools and opening the soundtrack. One student measured freq
    6·1 answer
  • Is it possible to code the impact force using an accelerometer?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!