If a marble is released from a height of 10 meters, how long would it take for it to hit the ground?

A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

8 0

3 years ago

A 150kg person stands on a compression spring with spring constant 10000n/m and nominal length of 0.50.what is the total length

Ivahew [28]

Answer:

<em>The total length of the spring would be 0.65 m</em>

Explanation:

The Concept

Hooke's law evaluates the increment of spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;

F = k x ..................1

where F is the force applied to the spring

k is the spring constant

x is the spring stretch or extension

Step by Step Calculations

We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;

x = F/k

but F = m x g

so, x = (m x g)/k

given that, the mass of the person m =150 kg

g is the acceleration due to gravity = 9.81 m/ $s^{2}$

k is the spring constant = 10000 N/m

then x = (9.81 m/ $s^{2}$ x 150 kg)/10000 N/m

x = 0.14715 m

the extension experienced by the spring after the compression is 0.14715 m

The total length of the spring would be;

L = 0.14715 m + 0.5 m = 0.64715

L ≈ 0.65 m

Therefore the total length of the spring would be 0.65 m

4 0

3 years ago

Describe each class of lever and explain to characteristics of each

Nataly [62]

-- Class I lever

The fulcrum is between the effort and the load.

The Mechanical Advantage can be anything, more or less than 1 .

Example: a see-saw

-- Class II lever

The load is between the fulcrum and the effort.

The Mechanical Advantage is always greater than 1 .

Example: a nut-cracker, a garlic press

-- Class III lever

The effort is between the fulcrum and the load.

The Mechanical Advantage is always less than 1 .

I can't think of an example right now.

8 0

3 years ago

Read 2 more answers

What happened to the speed of light if it travels from air into glass?

Ghella [55]

Answer:

it will disperse into many different colors

6 0

3 years ago

Read 2 more answers

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>

Let's draw the free body diagram of the system using the given data.
From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

$N_x=86.62N$

We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

$N_y=F_V=mg-Tsin59\\$