All categories

3 years ago

A force moves an object along an x axis. how is the work done by the force related to the force?

2 answers:

6 0

When an object moves along the x axis through a particular distance (d) due to a force (F) applied on it, then the work done (W) and the applied force (F) are related by the formula below

W = Fd

Mnenie [13.5K]3 years ago

3 0

The work done is equal to the product of the force applied and the displacement produced in the direction of the force.

You might be interested in

2. A common physics experiment involves lowering an open tube into a cylinder of water and moving the tube up and down to adjust

notsponge [240]

Answer:

Explanation:

This question pertains to resonance in air column. It is the case of closed air column in which fundamental note is formed at a length which is as follows

l = λ / 4 where l is length of tube and λ is wave length.

here l = .26 m

λ = .26 x 4 = 1.04 m

frequency of sound = 330 Hz

velocity of sound = frequency x wave length

= 330 x 1.04

= 343.2 m /s

b )

Next overtone will be produced at 3 times the length

so next length of air column = 3 x 26

= 78 cm

c )

If frequency of sound = 256 Hz

wavelength = velocity / frequency

= 343.2 / 256

= 1.34 m

= 134 cm

length of air column for resonance

= wavelength / 4

134/4

= 33.5 cm

7 0

3 years ago

When the electric field strength is large, electric field lines are: a) close together.

LekaFEV [45]

Answer: a) close together

Explanation: The electric field lines also represent the intensity of the field, in this sense for strong electric fields it is usually draw the lines close to each other. In constrast when they are far apart the electric field is weak.

6 0

3 years ago

A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each

irinina [24]

Answer:

1.24 m/s

Explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

$\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s$

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

$\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s$

So the flow speed of each of the narrower pipe is:

$v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}$

$v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s$

8 0

4 years ago

a 230-newton box rests on a plank 5 meters long. One end is 1.2 meters higher than the other end. Find the components of the for

NISA [10]

Answer:

55.3 N, 223.3 N

Explanation:

First of all, we can find the angle of the inclined plane.

We have:

L = 5 m the length of the incline

h = 1.2 m is the height

We also have the relationship

$h = L sin \theta$

where $\theta$ is the angle of the incline. Solving for the angle,

$\theta= sin^{-1} (\frac{h}{L})=sin^{-1} (\frac{1.2 m}{5 m})=13.9^{\circ}$

Now we can find the components of the weight of the box, which is the force that the box exerts on the plank. Calling W = 230 N the weight of the box, we have:

- Component parallel to the incline:

$W_{par} = W sin \theta = (230 N)(sin 13.9^{\circ}) =55.3 N$