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2 years ago

* PLEASE HELP EVEN FOR JUST ONE OF THE QUESTION WOULD REALLY HELP ME*

Physics

1 answer:

kondaur [170]2 years ago

3 0

The distance covered in one orbit is equal to the circumference of earth.

Given that radius of earth is 3959 miles.

Circumference of earth= 2πr = 2*3.14*3959 = 24875.13 miles

Speed is 26,000 miles per hour.

Since, Velocity= Distance/Time

Time= Distance/Velocity = 24875.13/26000

Time = 0.9567 hours

Therefore,

Average hour per orbit is 0.957 approximately.

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If the mass of the Jupiter is 1.9*10 kg and radius is fbx loom what is acclerallon due to gravity on the supiter ?

Bond [772]

Answer:

$g=25.92\ m/s^2$

Explanation:

Given that,

The mass of the Jupiter, $M=1.9\times 10^{27}\ kg$

The radius of Jupiter, $r=69,911\ km=69911\times 10^3 km$

We need to find the acceleration due to gravity on Jupiter. The formula is

$g=G\dfrac{M}{r^2}$

Put all the values,

$g=6.67\times 10^{-11}\times \dfrac{1.9\times 10^{27}}{(69911\times 10^3)^2}\\\\g=25.92\ m/s^2$

So, the value of acceleration due to gravity on the Jupiter is $25.92\ m/s^2$ .

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2 years ago

An object accelerating may be changing in what two ways

Oksana_A [137]

Velocity and direction

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2 years ago

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

joja [24]

Answer:

Approximately $9.7\; \rm m \cdot s^{-2}$ .

Explanation:

Assuming that there is no other force on this vehicle, the $16000\; \rm N$ force from the road would be the only force on this vehicle. The net force would then be equal to this $16000\; \rm N\!$ force. The size of the net force would be $16000\; \rm N\!\!$ .

Let $m$ denote the mass of this vehicle and let $\Sigma F$ denote the net force on this vehicle.

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, $a$ , would be:

$\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}$ .

For this vehicle, $\Sigma F = 16000\; \rm N$ whereas $m = 1650\; \rm kg$ . The acceleration of this vehicle would be:

$\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}$ .