When the plastic coating around a cord is damaged, why does the cord become hazardous?

How much gravitational potential energy does a 45.2 kg object have when it is 21.9m above the ground?

Blizzard [7]

Answer:

Explanation:

The formula for gravitational potential energy is

Ep = m · g · h Assuming that the acceleration is g = 10m/s²

Ep = 45.4 · 10 · 21.9 = 9,942.6 J

God is with you!!!

6 0

3 years ago

Why are simple everyday actions considered

Oxana [17]

They create energy ……………….

8 0

3 years ago

Read 2 more answers

a constant force acts on an object of mass 10 kg for a duration of 2 seconds it increases the object velocity from 4metre per se

inn [45]

Answer:

F=ma

a=v/t

a=vf-vi/t

a=8-4/2

a=4÷2

a=2msec*2

F=10×2

F=20N

3 0

2 years ago

At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She

Citrus2011 [14]

Answer:

a

The height is $H = 6.74 \ m$

b

The horizontal distance is $D = 23.74 \ m$

Explanation:

From the question we are told that

The speed is $v = 15 \ m/s$

The angle is $\theta = 40^o$

The height of the cannon from the ground is h = 2 m

The distance of the net from the ground is k = 1 m

Generally the maximum height she reaches is mathematically represented as

$H = \frac{v^2 sin^2 \theta }{2 * g } + h$

=> $H = \frac{(15)^2 [sin (40)]^2 }{2 * 9.8} + 2$

=> $H = 6.74 \ m$

Generally from kinematic equation

$s = ut + \frac{1}{2} at^2$

Here s is the displacement which is mathematically represented as

s = [-(h-k)]

=> s = -(2-1)

=> s = -1 m

There reason why s = -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

a = -g = -9.8

$u = v sin (\theta)$

So

$-1 = (vsin 40 )t + \frac{1}{2} * (-9.8) t^2$

=> $-4.9t^2 + 9.6418t + 1 = 0$

using quadratic formula to solve the equation we have

$t = 2.07 \ s$

Generally distance covered along the horizontal is

$D = v cos (40) * 2.07$

=> $D = 15 cos (40) * 2.07$

=> $D = 23.74 \ m$

7 0

3 years ago

NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low- mass sail and the ene

RUDIKE [14]

Answer:

d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.

Explanation:

Let us take the momentum of a photon unit as u

we know that the rate of change of momentum is proportional to the force exerted.

For a absorbing surface, the photon is absorbed, therefore the final momentum is zero. From this we can say that

F = (u - 0)/t = u/t

for a unit time, the force is proportional to the momentum of the wave due to its energy density. Therefore,

F = u

For a reflecting surface, the momentum of the wave strikes the sail and changes direction. Since we know that the speed of light does not change, then the force is proportional to

F = (u - (-u))/t = 2u/t

just as the we did above, it becomes

F = 2u.

From this we can see that the force for a reflective sail is twice of that for an absorbing sail, and we know that the pressure is proportional to the force for a given area. From these, we conclude that the sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.

3 0

3 years ago