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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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A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at â7.0 meters per second2 to a
Harrizon [31]
<span>By algebra, d = [(v_f^2) - (v_i^2)]/2a. Thus, d = [(0^2)-(15^2)]/(2*-7) d = [0-(225)]/(-14) d = 225/14 d = 16.0714 m With 2 significant figures in the problem, the car travels 16 meters during deceleration.</span>
8 0
2 years ago
2 boxes connected by a plus sign hold Wave 1 on top and Wave 2 on bottom. The crests of Wave 1 line up with the troughs of Wave
loris [4]

Answer:

No, the resulting wave in the diagram does not demonstrate destructive interference. The resulting wave in the diagram shows a bigger wave than Wave 1 or Wave 2. If it demonstrated destructive interference, it would be a smaller wave or a horizontal line. With destructive interference, waves break down to form a smaller wave, or cancel each other out, resulting in no wave formation.

6 0
2 years ago
Read 2 more answers
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
2 years ago
A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen
Svetllana [295]

Answer:

h_p = 30.46\ m

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

v_f=g\cdot t

Where g = 9.8 m/s^2

Similarly, the distance y the object has traveled is calculated as follows:

\displaystyle y=\frac{g\cdot t^2}{2}

If we know the height h from which the object was dropped, we can solve the above equation for t:

\displaystyle t=\sqrt{\frac{2\cdot y}{g}}

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

t_2=2.56 - 2 = 0.56\ sec

Therefore, it has traveled down a distance:

\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m

Thus, the height of the pen is:

h_p = 32 - 1.54\Rightarrow h_p=30.46\ m

8 0
2 years ago
If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
Gnom [1K]

The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2


and the average current density is

J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2

8 0
2 years ago
Read 2 more answers
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