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Effectus [21]
2 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]2 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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dopasuj wartości pracy z ramki do przedstawionych sytuacji ,a nastepnie wyraż te prace w dżulach uwaga jedna wartośc pracy nie b
Eduardwww [97]

Answer:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Explanation:

Całą część pytania można znaleźć na poniższym schemacie.

Z diagramu załączonego poniżej; mamy

A - samolot lotniczy

B - dźwig

C - podnoszenie ciężarów

Wszyscy to wiemy ;

1kJ = 1000 J

1MJ = 1000000 J

Mamy cztery opcje; i.e 200 MJ, 800 kJ, 1.6 kJ  and 250 mJ

Z czterech opcji można wykluczyć 250 mJ, ponieważ jest to 0,25 J, co przedstawia bardzo niską energię w porównaniu z trzema warunkami pokazanymi na schemacie.

Więc:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Największą pracę wykona samolot. Jest tak, ponieważ ma bardzo dużą masę i bardzo dużą prędkość. W związku z tym istnieje potrzeba wytworzenia ogromnej ilości ciepła i energii.

Z drugiej strony żuraw może podnieść ładunek o wiele większy i przewyższa ciężar ciężaru, więc praca wykonywana przez dźwig musi być zdecydowanie większa niż praca ciężarka.

3 0
2 years ago
What is the average acceleration of a car that is initially at rest at a stoplight and then accelerates to 24 m/s in 9.4 s?
andrezito [222]

brainly.com/question/11542618?answering=true&answeringSource=greatJob%2FquestionPage

8 0
2 years ago
Read 2 more answers
A newly discovered planet has a mean radius of 7380 km. A vehicle on the planet\'s surface is moving in the same direction as th
Butoxors [25]

Answer:

292796435 seconds ≈ 300 million seconds

Explanation:

First of all, the speed of the car is 121km/h = 33.6111 m/s

The radius of the planet is given to be 7380 km = 7380000 m

From the relationship between linear velocity and angular velocity i.e., v=rw, the angular velocity of the car will be w=v/r = 33.6111/7380000 = 0.000000455 rad/s = 4.55 x 10⁻⁶ rad/sec

If the angular velocity of the vehicle about the planet's center is 9.78 times as large as the angular velocity of the planet then we have

w(vehicle) = 9.78 x w(planet)

w(planet) = w(vehicle)/9.78 = 4.55 x 10⁻⁶ / 9.78 = 4.66 x 10⁻⁷ rad/sec

To find the period of the planet's rotation; we use the equation

w(planet) = 2π÷T

Where w(planet) is the angular velocity of the planet and T is the period

From the equation T = 2π÷w = 2×(22/7) ÷  4.66 x 10⁻⁷ = 292796435 seconds

Therefore the period of the planet's motion is 292796435 seconds which is approximately 300, 000, 000 (300 million) seconds

8 0
2 years ago
Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th
DENIUS [597]

Answer:

a) vboat = 5.95 m/s  b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As  we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w =  \frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

3 0
2 years ago
If you wished to warm 100 kg of water by 15 degrees celsius for your bath, how much heat would be required? (give your answer in
Anit [1.1K]
For the answer to the question above, 
<span>Q = amount of heat (kJ) </span>
<span>cp = specific heat capacity (kJ/kg.K) = 4.187 kJ/kgK </span>
<span>m = mass (kg) </span>
<span>dT = temperature difference between hot and cold side (K). Note: dt in °C = dt in Kelvin </span>

<span>Q = 100kg * (4.187 kJ/kgK) * 15 K </span>
<span>Q = 6,280.5 KJ = 6,280,500 J = 1,501,075.5 cal</span>
6 0
2 years ago
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