All categories

3 years ago

A force of 85 N is used to push a box along the floor a distance of 15 m. How much work was done?

Physics

2 answers:

mario62 [17]3 years ago

7 0

A :-) work = force x distance
W = 85 x 15
W = 1275 joules

.:. The work done is 1275 joules

denpristay [2]3 years ago

4 0

Answer:

1275J

Explanation:

Given parameters:

Force on box = 85N

Distance moved = 15m

Unknown:

Work done = ?

Solution:

Work done is the amount of force applied on a body to move it through a specific distance.

Work done = Force x distance

Now insert the parameters and solve;

Work done = 85 x 15 = 1275J

You might be interested in

If Siobhan hits a 0.25 kg volleyball with 0.5 N of force, what is the acceleration of the ball?

Alekssandra [29.7K]

Answer:

2 meters per second²

Explanation:

8 0

3 years ago

Read 2 more answers

Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support

alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

$\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)$

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

$r_{M\perp}(F_M)-r_{W\perp}(W)=0$

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

$r_{M\perp}(F_M)=r_{W\perp}(W)$

Now, we divide by the distance of the muscle:

$(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}$

Now, we substitute the values:

$F_M=\frac{(2.4cm)(50N)}{5.1cm}$

Now, we solve the operations:

$F_M=23.53N$

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

$\Sigma F_v=F_j-F_M-W$

Since there is no vertical movement the sum of vertical forces is zero:

$F_j-F_M-W=0$

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

$F_j=F_M+W$

Now, we plug in the values:

$F_j=23.53N+50N$

Solving the operations:

$F_j=73.53N$

Therefore, the force is 73.53 Newtons.

8 0

1 year ago

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi

melomori [17]

Complete Question:

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experiences the smallest drop in temperature, and which one experiences the largest drop? Sample A: 4.0 kg of water [c = 4186 J/(kg·C°)] Sample B: 2.0 kg of oil [c = 2700 J/(kg·C°)] Sample C: 9.0 kg of dirt [c = 1050 J/(kg·C°)]

Answer:

A. Smallest B. Largest.

Explanation:

Assuming no heat exchange except for the heat removed from any sample (which we know is the same for the three ones), and that the process is done using only conduction, we can use the equation that relates the heat lost or gained by one object, with the mass of the object and the consequent change in temperature, as follows:

Q = c*m*ΔT, where c, is a proportionality constant called specific heat, which is different for each material.

As we know that the heat removed is the same for the three samples, we can equate the right sides of the equation for each sample, as follows:

cw*mw*ΔTw = co*mo*ΔTo = cd*md*ΔTd

Replacing by the givens, we have:

4.0 kg. 4,186 J/kgºC*ΔT(ºC) = 2.0 kg*2,700 J/kgºC*ΔT(ºC) =9.0kg*1,050J/kgºC*ΔT(ºC)

As the three expressions must be equal each other, it's clear that the unknown term (the drop in temperature) must compensate the product of the mass times the specific heat.

This product is the following for the three samples:

Water: 4.0 kg*4,186 J/kgºC = 16,744 J/ºC

Oil : 2.0 kg*2,700 J/kgºC = 5,400 J/ºC

Dirt: 9.0 * 1,050 J/kgºC = 9,450 J/ºC

Clearly, we see that in order to keep the heat exchange equations equal each other, the water must suffer the smallest drop in temperature, and the oil must experience the largest one.

So, the sample A experiencies the smallest drop in temperature, and sample B does the largest one.

5 0

3 years ago

3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of

Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

$n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555$

Again using Snell's law for blue light as :

$n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283$

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0

4 years ago

The internal energy of material is determined by

Mnenie [13.5K]

The combined amount of kinetic and potential energy of its molecules

6 0

4 years ago