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Mnenie [13.5K]
2 years ago
10

 explain why earths acceleration is usually very small compared to the acceleration of the object the earth interact with

Physics
1 answer:
Marianna [84]2 years ago
6 0

Answer:

F-ma

Explanation:

If you are speaking of objects like satellites, etc. then their mass is much less than that of the Earth. A good approximation is Newton's first law of motion:

Force = Mass ×  Acceleration

often written:

F = m a

The gravitational force is the same between the Earth and the object - only the mass differs. So the acceleration is inversely proportional to the mass.

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5 0
3 years ago
8) A racecar accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the
Vlada [557]

8)Acceleration of the car is 11.17 m/s².

9) Acceleration of skater is 0.57 m/s².

10) The deceleration is 1.8 m/s².

11) The average acceleration of the sprinter is 2 m/s².

12) The velocity of the stroller after 4.75 minutes is 171 m/s.

13) The skier will be moving as 11 m/s speed.

14) 0.56 seconds is required by the rattle snake to reach the speed of 28 m/s from rest.

Answer:

Explanation:

8) Acceleration of the car is determined by finding the ratio of change in velocity to the time taken for the change in velocity. As here the initial velocity is 18.5 m/s and the final velocity is 46.1 m/s, then in 2.47 seconds, the acceleration is

Acceleration = (46.1-18.5)/2.47 = 11.17 m/s².

Acceleration of the car is 11.17 m/s².

9)Similarly, the acceleration of the skater with initial velocity zero to final velocity as 5.7 m/s in 10 seconds is

Acceleration = (5.7-0)/10=0.57 m/s².

So acceleration of skater is 0.57 m/s².

10) In this case, the shuttle bus is stopping so its speed is decreasing from high value to low value. This kind of acceleration related to the decrease in velocity is termed as deceleration as the value of acceleration will be coming as negative.

Acceleration = (0-9)/5=-1.8 m/s².

So the deceleration is 1.8 m/s².

11) Average acceleration of sprinter = (Final velocity-Initial velocity)/Time

Average acceleration = (7.5-5)/1.25=2 m/s².

So the average acceleration of the sprinter is 2 m/s².

12)Since the acceleration of the stroller is given as 0.6 m/s². And the initial velocity is given as zero. So for the time of 4.75 minutes, the velocity will be equal to the final velocity.

As per equations of motion,

v = u +at

As u =0, a = 0.6 m/s² and t = 4.75 ×60 s = 285 s

So v = 0.6×285 = 171 m/s

Thus, the velocity of the stroller after 4.75 minutes is 171 m/s.

13) Similarly, in this case, u = 0, a = 2.2 m/s² and t = 5 s

Then v = u+at=0+(2.2×5)=11 m/s

So the skier will be moving as 11 m/s speed.

14) Here a = 50 m/s² and v = 28 m/s with u = 0 so time taken to reach the speed of 28 m/s is

v = u +at = 0+ (50 t)

28 = 50 t

t = 28/50 = 0.56 s

So 0.56 seconds is required by the rattle snake to reach the speed of 28 m/s from rest.

4 0
3 years ago
The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -41 oC and 900 Pa, respect
Fittoniya [83]

Answer:

Part a)

\rho = 0.0205 kg/m^3

Part b)

\rho = 1.22 kg/m^3

So density of atmosphere at Martian Surface is very less than the density at Earth.

Explanation:

Part a)

As per ideal gas equation we know that

PM = \rho RT

here we know that Martian atmosphere is equivalent to that of carbon

so we will have

P = 900 Pa

T = 273 - 41 = 232 K

now we will have

(900)(0.044) = \rho (8.31)(232)

\rho = 0.0205 kg/m^3

Part b)

Now for the earth surface the density of air is given for

P = 101.6 kPa

T = 18 ^oC

so we will have

PM = \rho RT

(101.6\times 10^3)(0.029) = \rho(8.31)(273 + 18)

\rho = 1.22 kg/m^3

So density of atmosphere at Martian Surface is very less than the density at Earth.

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2 years ago
The proper time between two events is measured by clocks at rest in a reference frame in which the two events: The proper time b
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Answer:

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2 years ago
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The derived unit for density are g/cm3.<br><br> True<br> False
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True, the measurement shown is a derived unit.
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2 years ago
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