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2 years ago

13

Why do we feel the force of gravity from the Earth but we don’t feel the force of gravity from the moon?

2 answers:

Zigmanuir [339]2 years ago

8 0

Answer:

The gravitational force on the moon is less than on Earth because the strength of gravity is determined by an object's mass. The bigger the object, the bigger the gravitational force. Gravity is pretty much everywhere. We just feel it in different ways depending on our state of motion.

Explanation:

Hope this helped!!

vaieri [72.5K]2 years ago

8 0

The gravitational force on the moon is less than on Earth, because the strength of gravity is determined by an objects' mass. The bigger the object, the bigger the gravitational force.

:)

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In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe

Nata [24]

Answer:

Explanation:

Given

Each student exert a force of $F=400 N$

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students $F=18\times 400=7200 N$

therefore weight of car $W=7200$

mass of car $m=\frac{W}{g}$

$m=\frac{7200}{9.8}=734.69 kg$

(b) $7200 N \approx 1618.624\ Pound-force$

$734.69 kg\approx 1619.71 Pounds$

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3 years ago

a car going 22 m/s accelerates to pass a truck. Five seconds later the car is going 35 m/s. Calculate the acceleration of the ca

Maksim231197 [3]

Using the formula

$a\: = \frac{v - u}{t}$

$a = \frac{35 m/s \: - \: 22 m/s}{5s}$

$a = 2.6 {m/s^2}$

5 0

3 years ago

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

as

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

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3 years ago

Acceleration is a derived unit

aleksklad [387]

Acceleration is a derived unit because it is derived from two quantities : Velocity and time.

We know, acceleration = Change in velocity/Time

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2 years ago

Read 2 more answers

Given this graph plotting velocity versus time, estimate the acceleration of object A at points X and Y respectively.

Nikolay [14]

I think answer should be d. Please give me brainlest let me know if it’s correct

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3 years ago