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Irina-Kira [14]
3 years ago
5

What type of energy does an object gain as it is lifted at a constant speed?

Physics
1 answer:
marshall27 [118]3 years ago
7 0
D. Gravitational potential
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In 0.60 seconds, a projectile goes from 0 to 610 m/s. What is the acceleration of the projectile?
IceJOKER [234]

Answer: a=1016.66 m/s^{2}

Explanation:

Acceleration a is expressed in the following formula:

a=\frac{V_{f}-V_{o}}{t}

Where:

V_{f}=610 m/s is the final velocity of the projectile

V_{o}=0 m/s  is the initial velocity of the projectile

t=0.6 s is the time

Solving:

a=\frac{610 m/s-0 m/s}{0.6 s}

a=1016.66 m/s^{2} This is the acceleration of the projectile

6 0
3 years ago
A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.
mash [69]

a) The acceleration of gravity is 4.96 m/s^2

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

g=\frac{GM}{(R+h)^2}

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2

b)

The critical velocity for a satellite orbiting around a planet is given by

v=\sqrt{\frac{GM}{R+h}}

where we have again:

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s

Converting into km/h,

v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

T=\frac{2\pi (R+h)}{v}

where

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

v = 6668 m/s

Substituting,

T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s

d)

One day consists of:

t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

n=\frac{t}{T}=\frac{86400}{8452}=10.2

e)

The escape velocity for an object in the gravitational field of a planet is given by

v=\sqrt{\frac{2GM}{R+h}}

where here we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

Substituting, we find

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s

f)

We can apply again the formula to find the escape velocity for the rocket:

v=\sqrt{\frac{2GM}{R+h}}

Where this time we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=0, because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0
3 years ago
Kinetic energy is the energy of . It increases with and .Potential energy is the energy of or . It increases with , and
lorasvet [3.4K]

Answer:

k is the energy of motion it increase with four of load you carry

8 0
2 years ago
Which statement is true regarding coppers ability to conduct electricity
Alecsey [184]
Copper is the second best known electrical conducting substance. The first best one is silver. We almost always use copper because silver costs too much.
6 0
3 years ago
Two balls of mass m1 and m2, with velocities v1 and v2 collide head on. Is there any way for both balls to have zero velocity af
nordsb [41]

Answer:

Explanation:

As the final Kinetic energy is zero or less than initial kinetic energy, the collision must be inelastic.  

In Inelastic collision both the bodies must stick together as final velocity is zero for both the bodies.

To conserve the momentum, momentum associated before the collision of first must be equal and opposite to the momentum associated with the second ball.

i.e.

m_1v_1=m_2v_2

8 0
3 years ago
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