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KatRina [158]
3 years ago
11

Give an example of unbalanced force?​

Physics
1 answer:
Simora [160]3 years ago
5 0

Answer:

If you kick a football and it moves from one place to another, it means that unbalanced forces are acting upon it. Ball moves from one place to another after kicking it. This is an example of unbalanced force.

Explanation:

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SCORPION-xisa [38]

Answer:

46

Explanation:

HOPE IT'S HELPFUL FOR U MATE AND PLZ MARK AS BRAINLIEST I REALLY NEED IT

5 0
3 years ago
Read 2 more answers
(e) For photons of energy 7.10 eV, what stopping potential would be required to arrest the current of photoelectrons
Masja [62]

2.37eV stopping potential would be required to arrest the current of photoelectrons.

<h3 /><h3>What is stopping potential ?</h3>

The minimal negative voltage that must be provided to the anode to halt the photocurrent is known as stopping potential. When expressed in electron volts, the maximal kinetic energy of the electrons is equal to the stopping voltage.

Kmax = eV₀

2.37eV = eV₀

V₀ = 2.37eV

to learn more about stopping potential go to - brainly.com/question/4655588

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8 0
2 years ago
Why do we eat today
earnstyle [38]
Because if we don't you'll be hungry, if you don't eat for about a month you'll die
3 0
3 years ago
Read 2 more answers
Hot air balloon ( look at pic)
kramer

Answer:

I think that the answer is convection.

Explanation:

Hope this helps.

6 0
3 years ago
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
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