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Phantasy [73]
3 years ago
12

A fuel-filled rocket is at rest. It burns its fuel and expels hot gas. The gas has a momentum of 1,500 kg x m/s backward. What i

s the
momentum of the rocket?
a
0
b
-1,500 kg x m/s forward
1,500 kg x m/s forward
с
Physics
1 answer:
storchak [24]3 years ago
3 0

Answer:

a

Explanation:

Its mass times m/s there is no m/s so 1500x0 = 0

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What parameters affects the inductance of a coil​
Kazeer [188]

Answer:

Number of Wire Turns in the Coil.

Explanation:

The greater the number of turns of wire in the coil, the greater the inductance. Fewer turns of wire in the coil results in lesser inductance. More coils of wires indicate a greater amount of magnetic field force for a given amount of coil current.

4 0
2 years ago
a golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0. if the hang time of the golf ball is 33.4 s
Kobotan [32]
Solving this using the time, we know that range = horizontal velocity x time of flight 

since there are no horizontal forces acting on the ball, there are no horizontal accelerations and the initial horizontal velocity of 36 cos 28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have 

range = 36 cos 28 x 3.44 s = 109.3 m

6 0
2 years ago
The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 sec
marin [14]

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

\epsilon = \frac{\sigma}{\sqrt{n} }

To reduce the standard error to 0.03 s, let the additional number of trials = x

0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials

Therefore, the additional trials needed is 48 trials.

6 0
3 years ago
A 10 Kg ball is rolling at 2.5 m/s. It is then hit from behind with a bat that puts a 300 N force on the ball for a quick .3 sec
vazorg [7]

Answer: g. gg g rfrcdv

Explanation:

vfv g bygyb

7 0
2 years ago
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
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