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professor190 [17]
3 years ago
7

The moon orbits the earth once every 27 days at a distance of 384400 km. The international space station orbits the earth at an

altitude of 400km. Determine the period of orbit of the space station
Physics
1 answer:
Leno4ka [110]3 years ago
7 0

Answer:

ive never done this befor but i think its 36 times aroud earth a day

Explanation:

384400/400=961  

961% of 27 days is 40minuets

1440/40=36

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The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2
viktelen [127]

Answer:

Distance, d = 778.05 m                          

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2

Let d is the distance covered by car. Using second equation of motion as :

d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m

So, the car will cover a distance of 778.05 meters.

5 0
3 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
4 years ago
If a person is walking at 1.2 m/s and 60 seconds later the person is running at 10 m/s, what was the acceleration rate?​
Marina CMI [18]
The acceleration rate would be .14667 m/s^2
6 0
3 years ago
A sprinter has a mass of 80 kg and a KE of 4000 J. What is the sprinter’s speed?
djyliett [7]
There you go.

Hope this helps!

8 0
3 years ago
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6.Convert 22 °C to K. (°C = K - 273)
DIA [1.3K]

Answer:

295

Explanation:

if C = k - 273

k = C + 273

substitute 22 for C

K = 22 + 273

k = 295

3 0
3 years ago
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