What best describes gravity?

Suppose a 4,000-kg elephant is hoisted 20 m above Earth’s surface. Use a calculator and follow the steps below to find the eleph

stiv31 [10]

GPE = 78,380 J
w = 39,240 N

First list what you know. You know the elephants mass and it’s height. You also know gravity on Earth. I will use g = 9.81.
m = 4,000 kg
h = 20 m
g = 9.81 m/s^2

You need to find the elephants weight. Weight = mass x gravity
w = mg
w = (4000 kg)(9.81 m/s^2)
w = 39,240 N (N = newtons)

Now, knowing the elephants weight, you can calculate its GPE.
Gravitational Potential Energy = weight x height

GPE = wh
GPE = (39,240N)(20m)
GPE = 78,380 J (J = joules)

4 0

2 years ago

If you want to know how energy will move between two objects, what do you need to know about the objects?

Inessa [10]

You need to know how much friction that object.

8 0

2 years ago

A particle is moving with (SHM) of period 8.0s and amplitude5.0m

nadezda [96]

Answer:

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$

Explanation:

Given the description of period and amplitude, the SHM could be described by:

$f(x)=5\,sin(\frac{\pi}{4}x)$

and its angular velocity can be calculated doing the derivative:

$f(x)=5\, \,sin(\frac{\pi}{4}x)\\f'(x)=5\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

And therefore, the tangential velocity is calculated by multiplying this expression times the radius of the movement (3 m):

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$ and is given in m/s.

Then the maximum speed is obtained when the cosine function becomes "1", and that gives:

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

The acceleration is found from the derivative of the velocity expression, and therefore given by:

$acceleraton(x)=-15\,\frac{\pi^2}{16}\,sin(\frac{\pi}{4}x)$

and the maximum of the function will be obtained when the sine expression becomes "-1", which will render:

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$

6 0

2 years ago

Water at the top of a slope has potential energy. true or false

ASHA 777 [7]

<span>The statement is TRUE. Water does have potential energy at the top of a slope. The reason why is that potential energy is energy possessed by a body based on its position relative to a zero point. In this case, water at the top of the slope is at an elevation above ground (zero point). The energy is not kinetic (moving) energy since the water is not moving.</span>

8 0

2 years ago

Read 2 more answers

Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person

sertanlavr [38]

Answer:

v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

f ’= f (v + v₀) / (v- $v_{s}$ )

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

v₀ = v_{s} = v ’

we substitute

f ’= f (v + v’) / (v - v ’)

f ’/ f (v-v’) = v + v ’

v (f ’/ f -1) = v’ (1 + f ’/ f)

v ’= (f’ / f-1) / (1 + f ’/ f) v

v ’= (f’-f) / (f + f’) v

let's calculate

v ’= (3400 -3000) / (3000 +3400) 343

v ’= 400/6400 343

v ’= 21.44 m / s

3 0

2 years ago