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2 years ago

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You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3

,500 N across the carpet to a spot 5 m away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dresser?

1 answer:

LenaWriter [7]2 years ago

6 0

Answer:

0 Joules

Explanation:

The work done is given by

$W=F\times s\times cos\theta$

where,

F = Force applied

s = Displacement of the object = 0 m

$\theta$ = Angle between the force applied and the horizontal = 0

$W=F\times 0\times cos0\\\Rightarrow W=0\ J$

Work is only observed when there is a displacement.

The work done by me is 0 Joules as I was unable to move it.

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While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

$at=V_{f}-V_{0}$

and finally I can add $V_{0}$ to both sides so I get:

$V_{f}=at+V_{0}$

and now I can proceed and substitute the known values:

$V_{f}=at+V_{0}$

$V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0$

which solves to:

$V_{f}=28.8m/s$

8 0

3 years ago

Read 2 more answers

A cannon is fired horizontally at 243 m/s off of a 62 meter tall, shear vertical cliff. How far in meters from the base of the c

Alekssandra [29.7K]

Answer:

865.08 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 243 m/s

Height (h) of the cliff = 62 m

Horizontal distance (s) =?

Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:

Height (h) of the cliff = 62 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

62 = ½ × 9.8 × t²

62 = 4.9 × t²

Divide both side by 4.9

t² = 62/4.9

Take the square root of both side.

t = √(62/4.9)

t = 3.56 s

Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:

Initial velocity (u) = 243 m/s

Time (t) = 3.56 s

Horizontal distance (s) =?

s = ut

s = 243 × 3.56 s

s = 865.08 m

Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.

6 0

3 years ago

Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$

2)

The term $2\gamma \Phi$ must have the same units of $\Psi$ in order to be added to it. Therefore,

$[\gamma \Phi] = [\frac{mol}{m\cdot s^2}]$

We also know that the units of $\gamma$ are $[\frac{J}{kg}]$ , therefore

$[\frac{J}{kg}][\Phi]= [\frac{mol}{m\cdot s^2}]$

And so, the fundamental units of $\Phi$ are

$[\Phi]= [\frac{mol\cdot kg}{J\cdot m\cdot s^2}]$

However, the Joules can be written as

$[J]=[kg][\frac{m^2}{s^2}]$

Therefore

$[\Phi]= [\frac{mol\cdot kg}{(kg \frac{m^2}{s^2})\cdot m\cdot s^2}]=[\Phi]= [\frac{mol}{m^3}]$

#LearnwithBrainly

7 0

3 years ago

A train has an acceleration of magnitude 0.90 m/s2 while stopping. A pendulum with a 0.55-kg bob is attached to a ceiling of one

anygoal [31]

The angle of the pendulum with the vertical is $5.2^{\circ}$

Explanation:

As the train decelerates, the bob of the pendulum will feel a force given by

$F=ma$

where

m = 0.55 kg is the mass of the bob

$a=0.9 m/s^2$ is the magnitude of the acceleration

In the horizontal direction.

The pendulum will be inclined at an angle $\theta$ from the vertical, so it will be in equilibrium, and therefore the horizontal component of the tension in the string must be equal to the net force F of the previous equation:

$T sin \theta = ma$ (1)

where T is the tension in the string.

We also know that the bob is in equilibrium along the vertical direction: so the vertical component of the tension must be equal to the weight of the bob,

$T cos \theta = mg$ (2)

where $g=9.8 m/s^2$ is the acceleration of gravity.

Dividing eq.(1) by eq(2), we get:

$tan \theta = \frac{a}{g}$

And therefore, we find the angle:

$\theta=tan^{-1}(\frac{a}{g})=tan^{-1}(\frac{0.90}{9.8})=5.2^{\circ}$

Learn more about forces and acceleration:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

3 0

3 years ago

If gravity did not affect the pain of a horizontally thrown ball the ball would

Drupady [299]

If gravity had no effect on a ball after you threw it ... and there also
were no air to slow it down ... then the ball would continue traveling
in a straight line, in whatever direction you threw it.

That's the heart and soul of Newton's laws of motion ... any object
keeps moving at the same speed, and in a straight line in the same
direction, until a force acts on it to change its speed or direction.\

If you threw the ball horizontally, then it would keep moving in the
same direction you threw it. But don't forget: The Earth is not flat.
The Earth is a sphere. So, as the ball kept going farther and farther
in the same straight line, the Earth would curve away from it, and it
would look like the ball is getting farther and farther from the ground.

4 0

3 years ago

Read 2 more answers