The energy required to start a reaction is called the activation energy. true false

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

2 years ago

A ball is suspended by a lightweight string, as shown in the figure above.

Irina18 [472]

<u>Answer </u>

A. 1 and 2

<u>Explanation </u>

At point 1 we have the highest potential energy and the kinetic energy is zero.

At 2 the potential energy is minimum and the kinetic energy is maximum.

The law of conservation of energy says that energy cannot be created nor destroyed. So, the change in P.E = Change in K.E.

P.E = height × gravity × mass. The height referred here is the perpendicular height. Gravity and mass are constant in this case.

From the diagram it can be seen clearly that the vertical height from 2 to 1 is much greater than from 4 to 3.

This shows that the change in P.E is greater between 1 and 2 and so is kinetic energy.

6 0

2 years ago

Read 2 more answers

Suppose the coefficient of static friction between the road and the tires on a car is 0.638 and the car has no negative lift. Wh

larisa [96]

Answer:

12.6332454263 m/s

Explanation:

m = Mass of car

v = Velocity of the car

$\mu$ = Coefficient of static friction = 0.638

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of turn = 25.5 m

When the car is on the verge of sliding we have the force equation

$\dfrac{mv^2}{r}=\mu mg\\\Rightarrow v=\sqrt{\mu gr}\\\Rightarrow v=\sqrt{0.638\times 9.81\times 25.5}\\\Rightarrow v=12.6332454263\ m/s$

The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s

4 0

2 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him

nataly862011 [7]

Answer:

a) $u_c=0\ m.s^{-1}$ & $m_c.v_c=m_b.v_b\times \cos\theta$

b) $v_c=0.0566\ m.s^{-1}$

c) $p_e=2.9218\ kg.m.s^{-1}$

Explanation:

Given:

mass of the book, $m_b=1.35\ kg$

combined mass of the student and the skateboard, $m_c=97\ kg$

initial velocity of the book, $v_b=4.61\ m.s^{-1}$

angle of projection of the book from the horizontal, $\theta=28^{\circ}$

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

$u_c=0\ m.s^{-1}$

where:

$u_c=$ initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

$m_c.v_c=m_b.v_b\times \cos\theta$

where:

$v_c=$ final velcotiy of the student after throwing the book

b)

$m_c.v_c=m_b.v_b\times \cos\theta$

$97\times v_c=1.35\times 4.61\cos28$

$v_c=0.0566\ m.s^{-1}$

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

$p_e=m_b.v_b\sin\theta$

$p_e=1.35\times 4.61\times \sin28^{\circ}$

$p_e=2.9218\ kg.m.s^{-1}$

6 0

2 years ago

What causes an object to move, change direction or change speed?

Wittaler [7]

I believe it would be an unbalanced force. Because the forces are unbalanced, one side is stronger and, therefore, the object will move.

7 0

2 years ago

Read 2 more answers