All categories

3 years ago

What must occur for work to be done on an object?

1 answer:

katrin2010 [14]3 years ago

8 0

The answer is D) the object must move. The definition of work is movement (in any form).

Hope this helped! If it did, would you mind marking my answer as Brainliest? Thanks :D

You might be interested in

A man drops a penny V=0 off the top of the Golden Gate Bridge how fast will the penny be moving when it hits the ground? The gol

BARSIC [14]

Answer:

Vf = 73.4 m/s

Explanation:

This is the case of vertical motion where we have to find the final velocity of the penny when it hits the ground. We can use 3rd equation of motion to find the final velocity:

2gh = Vf² - Vi²

where,

g = 9.8 m/s²

h = height = 275 m

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

2(9.8 m/s²)(275 m) = Vf² - (0 m/s)²

Vf = √5390 m²/s²

Vf = 73.4 m/s

7 0

2 years ago

A 65-cm segment of conducting wire carries a current of 0.35 A. The wire is placed in a uniform magnetic field that has a magnit

Artyom0805 [142]

Answer: The angle between the wire segment and the magnetic field 66.42°

Explanation:

Please see the attachment below

8 0

3 years ago

Read 2 more answers

A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an

Roman55 [17]

Answer:

$v_{2f} = \frac{2vm_1}{m_2 + m_1}$

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

Before the collision

$P_i = m_1v$

After the collision

$P_f = m_1v_{1f} + m_2v_{2f}$

So using the law of momentum conservation

$P_i = P_f$

$m_1v = m_1v_{1f} + m_2v_{2f}$

We can solve for the speed of ball 1 post collision in terms of others:

$v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

Their kinetic energy is also conserved before and after collision

$m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2$

$m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2$

From here we can plug in $v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

$m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2$

$m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2$

$m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2$

$v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0$

$v_{2f}(1 + \frac{m_2}{m_1}) = 2v$

$v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}$

8 0

3 years ago

Read 2 more answers

9. Calculate the distance (in km) that Charlie runs if he maintains the average

Karo-lina-s [1.5K]

Correct Question:

Calculate the distance (in km) charlie runs if he maintains an average speed of 8 km/hr for 1 hour

Answer:

The total distance covered by Charlie is 8 km in 1 hour.

Explanation:

The average velocity as given in the question is,

v = 8 km/hr

Total time taken,

$$t=1 hour$

As we know the formula to evaluate the total distance d when the average velocity and time is given;

$v=\frac{d}{t}$

$d=v \times t$

$d=8 \times 1$

$d=8 k m$

Hence, the total distance covered by Charlie in 1 hour will be 8 km.

5 0

2 years ago

A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

Harman [31]

Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

coefficient of kinetic friction, $\mu_k=0.4$

So, the energy dissipated during the skidding of car:

Frictional force:

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

$f=5880\ N$

Now from the work-energy equivalence:

$E=f.d$

$E=5880\times 15$

$E=88200\ J$ is the dissipated energy.

3 0

3 years ago