Because light travels in a straight line and casts a shadow, Isaac Newton hypothesized that light is

A 2.5 g bullet traveling at 350 m/s hits a tree and slows uniformly to a stop while penetrating a distance of 12 cm into the tre

Cloud [144]

Answer: Work done = 153.125Joules, Work done = 0.003Nm

Explanation:

Kinetic energy of a body is the energy possessed by a body by virtue of its motion.

Mathematically,

K.E = 1/2MV²

Where;

M = mass of the body = 2.5g = 0.0025kg

V = velocity of the body = 350m/s

Substituting this values in the formula, we have;

K.E = 1/2× 0.0025×350²

K.E = 153.125Joules

Work done is the force applied to body to cause it to move through a distance.

Work = Force × distance

Force = ma = 0.0025 × 10

Force = 0.025N

Distance = 12cm = 0.12m

Work = 0.025×0.12

Work = 0.003Nm

work done by the tree in stopping the bullet is 0.003N

4 0

3 years ago

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Mr Ndlovu works in centurion and works at Rosebank.It takes him 13minutes to get to work and he needs to be there at 08h15

VARVARA [1.3K]

Answer:

8:02 no less

Explanation:

8:15 - 13mins = 8:02

6 0

3 years ago

How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the

AnnyKZ [126]

Answer:

Option C - 39.2 J

Explanation:

We are given that;

Mass; m = 2 kg.

Distance moved off the floor;d = 10 m.

Acceleration due to gravity;g = 9.8 m/s².

We want to find the work done.

Now, the Formula for work done is given by;

Work = Force × displacement.

In this case, it's force of gravity to lift up the boots, thus;

Formula for this force is;

Force = mass x acceleration due to gravity

Force = 2 × 9.8 = 19.2 N

∴ Work done = 19.6 × 2

Work done = 39.2 J.

Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.

7 0

3 years ago

Read 2 more answers

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h

hram777 [196]

Answer:

$v_{2}=3.5 m/s$

Explanation:

Using the conservation of energy we have:

$\frac{1}{2}mv^{2}=mgh$

Let's solve it for v:

$v=\sqrt{2gh}$

So the speed at the lowest point is $v=7 m/s$

Now, using the conservation of momentum we have:

$m_{1}v_{1}=m_{2}v_{2}$

$v_{2}=\frac{1*7}{2}$

Therefore the speed of the block after the collision is $v_{2}=3.5 m/s$

I hope it helps you!

8 0

3 years ago

Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of

morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner $(v_1)$ =3.1 m/s

speed of second runner $(v_2)$ =4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner $t=\frac{295.6}{4.65}=63.56 s$

time required by first runner to reach finish line $=\frac{250}{3.1}=80.64 s$

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0

4 years ago