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Anna11 [10]
4 years ago
13

Can anyone tell me the ans of this question also please

Physics
2 answers:
Dmitry_Shevchenko [17]4 years ago
6 0

Answer:

Hey there!

Stopwatch X recorded 40 seconds, and stopwatch Y recorded 50 seconds.

Stopwatch Y recorded 10 seconds longer than stopwatch X.

Hope this helps :)

vladimir2022 [97]4 years ago
3 0

Answer:

Explanation:

Stopwatch X recorded 40 seconds, and stopwatch Y recorded 50 seconds.

Stopwatch Y recorded 10 seconds longer than stopwatch X.

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If an object undergoes a change in momentum of 10 kg m/s in 3 s ,then the force acting on it is
Paha777 [63]

Answer:

Force = 3.333 Newton

Explanation:

Given the following data;

Change in momentum = 10 Kgm/s

Time = 3 seconds

To find the force acting on it;

In Physics, the change in momentum of a physical object is equal to the impulse experienced by the physical object.

Mathematically, it is given by the formula;

Force * time = mass * change in velocity

Impulse = force * time

Substituting into the formula, we have;

10 = force * 3

Force = 10/3

Force = 3.333 Newton

8 0
3 years ago
What is the magnitude of the gravitational force of attraction to Jupiter exerts on IO
kotegsom [21]

The gravity force between Jupiter and Io will be 6.343 × 10²² N.

<h3>What is Newton's law of gravitation?</h3>

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

Given data;

Mass of Jupiter,\rm m_j = 1.9 \times 10^{27} \ kg

Mass of moon of Jupiter,\rm m_{i_0}= 8.9 \times 10^{22} \ kg]

The gravitational constant is,\rm G =  6.67 \times 10^{-11 } \ m^3  kg^{-1}  s^{-2}

Distance between Jupiter and Io, R = 421,700 km = 4,217,00,000 m

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

The gravitational force is found as;

\rm F = G \frac{ m_J m_{I_0}}{R^2} \\\\\ F = (6.67\times 10^{-11}) \frac{( (1.9\times 10^{27})\times (8.9\times 10^{22} )} { (421700000)^2}\\\\ F_g = 6.343 \times 10^{22} \  N

Hence, the gravity force between Jupiter and Io will be 6.343 × 10²² N.

The complete question is

"Jupiter has a mass of 1.9 × 1027 kg, and its moon Io has a mass of 8.9 × 1022 kg. Their centers are separated by a distance of 421,700 km what is the force of gravity acting on Io? "

To learn more about Newton's law of gravitation, refer to the link.

brainly.com/question/9699135.

#SPJ1

8 0
2 years ago
In lightning , light is seen first and sound is heard later it is due to
Maksim231197 [3]

Answer:

option 4

Explanation:

Light's velocity in air ( 3 × 10^8 m/s ) is much greater than sound's velocity in air ( 343 m/s )

Hence due to difference in velocities , during lightning light is seen first & sound is heard later

8 0
3 years ago
Read 2 more answers
An astronaut, of total mass 86.0 kg including her suit, stands on a spherical satellite of mass 360 kg, both at rest relative a
Afina-wow [57]

Answer: 0.56 m/s

Explanation:

Hi, to answer this question we have to apply the formula of the conservation of momentum.

m1 v1 = m2 v2 (because the system is stationary at the beginning)

Where:

m1 = mass of the astronaut

v1= velocity of the astronaut

m2= mass of the satellite

v2= velocity of the satellite

Replacing with the values given and solving:

86 kg (2.35m/s) = 360 kg v2

202.1 kgm/s=360kg v2

202.1kgm/s /360kg =v2

v2 = 0.56 m/s

Feel free to ask for more if needed or if you did not understand something.

5 0
3 years ago
A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force e
Natali5045456 [20]

Answer:

a) F = 660.576\,N, b) a_{c} = 6.953\,\frac{m}{s^{2}}, c) v \approx 7255.423\,\frac{m}{s}, \omega = 9.583\times 10^{-4}\,\frac{rad}{s}, d) T \approx 1.821\,h

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

F = G\cdot \frac{m\cdot M}{r^{2}}

F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}

F = 660.576\,N

b) The centripetal acceleration of the satellite is:

a_{c} = \frac{660.576\,N}{95\,kg}

a_{c} = 6.953\,\frac{m}{s^{2}}

c) The speed of the satellite is:

v = \sqrt{a_{c}\cdot R}

v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}

v \approx 7255.423\,\frac{m}{s}

Likewise, the angular speed is:

\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}

\omega = 9.583\times 10^{-4}\,\frac{rad}{s}

d) The period of the satellite's rotation around the Earth is:

T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)

T \approx 1.821\,h

6 0
3 years ago
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