All categories

2 years ago

How much work is done when a very large force is applied to an object but does not move the object?

Physics

2 answers:

yarga [219]2 years ago

7 0

Answer:

A. None

(took the test)

Ierofanga [76]2 years ago

4 0

Work = (force) x (distance)

If the distance is zero (the object doesn't move), then it
doesn't matter how great the force is. The work is still zero.

You might be interested in

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$

Let the speed of car B before collision = $v_{B1}$

Momentum of car B before collision = $m_{B} v_{B1}$

Momentum after collision = $m_{A} v_{A} + m_{B} v_{B2}$

Applying the law of conservation of momentum:

$m_{B} v_{B1} = m_{A} v_{A} +m_{B} v_{B2}$

$m_{A} = 1100 kg\\m_{B} = 1400 kg$

$(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s$

3 0

3 years ago

Read 2 more answers

The superheroine Xanaxa, who has a mass of 68.1 kg , is pursuing the 75.3 kg archvillain Lexlax. She leaps from the ground to th

Alexandra [31]

The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

$U = mg\Delta h$

Here,

$\Delta h$ = Change in height

m = mass of super heroine

g = Acceleration due to gravity

The change in height will be,

$\Delta h = h_f - h_i$

The final position of the heroin is below the ground level,

$h_f = -16.1m$

The initial height will be the zero point of our system of reference,

$\Delta h = -16.1m-0m$

$\Delta h = -16.1m$

Replacing all this values we have,

$U = mg\Delta h$

$U = (68.1kg)(9.8m/s^2)(-16.1m)$

$U = -10744.81J$

Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J

4 0

3 years ago

What do radio waves and microwaves have in common?

Tamiku [17]

Answer:

I Will say the Answer is A

Explanation:

5 0

2 years ago

Read 2 more answers

Question 3 of 15

omeli [17]

Answer:

B) Degrees

Explanation:

The directions of the vectors are often defined in terms of due East, due North, due West and due South. A direction exactly in between of North and East can be described as Northeast, similarly we can describe directions in terms of Northwest, Southeast and South west.

From these, the direction of a vector can be easily expressed in degrees, which is measured counter clockwise about its tail from due East. Considering that we can say that East is at 0° , North is at 90° , West is at 180 and South is at 270° counter clockwise rotation from due East.

So, we know that the direction of a vector lying somewhere between due East i.e 0° and due North i.e 90°, will be measured in degrees, which will have a value between 0°-90°

4 0

2 years ago

Calculate the wavelengths of the first five emission lines of the Balmer series for hydrogen

In-s [12.5K]

Answer:

Explanation:

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe. In quantum physics, when electrons transition between different energy levels around the atom (described by the principal quantum number, n) they either release or absorb a photon. The Balmer series describes the transitions from higher energy levels to the second energy level and the wavelengths of the emitted photons. You can calculate this using the Rydberg formula.

5 0

2 years ago