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Effectus [21]
3 years ago
12

A 1000-kg space probe is motionless in space. to start moving, its main engine is fired for 5 s during which time it ejects exha

ust gases at 5000 m/s. at the end of this process it is moving at 20 m/s. the approximate mass of the ejected gas is:
Physics
1 answer:
photoshop1234 [79]3 years ago
8 0
<span>When a collision occurs, the change in momentum, is the difference in the momentums of the two objects after the collision minus the momentum of the object before the collision. From the aforementioned , we have change in momentum to be final momentum after collision - momentum before collision. But then the impulse of the fird bullet is just the change in momentum. Impulse = Ft where F = Ma. Where F= force and a is the acceleration. F = 1000* a; since a = V/t; i. e change in velocity by time. So we have 5000/5 =1000ms^2. Hence F = 1000 * 1000 = 1000000N. Then our Impulse , Ft = 1000000 * 5 = 5000000N. Momentum just before collision = 1000 * 500 = 500000. After collision we have final momentum = m * 20 where m is the mass after collsion. So we have 20M - 500000 = 5000000. 20M = 5000000-500000 = 4500000. M = 225000Kg.</span>
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Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

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At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

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r = 0.25m

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Replacing at the previous equation we have that the angular velocity is

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Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

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