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RideAnS [48]
4 years ago
15

9. An astronaut of mass 90 kg walks in space outside her spaceship and receives a 30-N force from a nitrogen spurt gun. What acc

eleration does she experience?
Physics
2 answers:
Leokris [45]4 years ago
7 0

Answer:

Acceleration of the spurt gun is 0.33 m/s^2

Explanation:

Given that,

Mass of the astronaut is 90 kg

Force received from a nitrogen spurt gun is 30 N

To find :

Acceleration received to the astronaut.

Solve :

Let a is the acceleration. It can be found using the second law of motion. The formula of force is given by :

F=ma

a=\dfrac{F}{m}

a=\dfrac{30\ N}{90\ kg}

a = 0.33 m/s^2

Therefore, the acceleration experienced by the astronaut is 0.33 m/s^2

VashaNatasha [74]4 years ago
5 0
Acceleration is equal to force / mass
thus 30/90 is 0.33 m/s^2
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A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters
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Answer:

Part a)

F = 7.76 N

Part b)

\theta = -137.7 degree

Part c)

\theta = -127.7 degree

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

x = -19 + t - 3t^3

y = 20 + 7t - 9t^2

Part a)

differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

a_x = \frac{d}{dt}(0 +1 - 9t^2)

a_x = -18t

a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)

a_y = \frac{d}{dt}(0 +7 - 18t)

a_y = -18

Now the acceleration of the object is given as

\vec a = (-18t)\hat i + (-18)\hat j

at t= 1.1 s we have

\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

\vec F = m\vec a

\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)

\vec F = -5.74 \hat i - 5.22 \hat j

now magnitude of the force will be

F = \sqrt{5.74^2 + 5.22^2} = 7.76 N

Part b)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-5.22}{-5.74}

\theta = -137.7 degree

Part c)

For velocity of the particle we have

v_x = \frac{dx}[dt}

v_x = (0 +1 - 9t^2)

v_y = \frac{dy}{dt}

v_y = (0 +7 - 18t)

now at t = 1.1 s

\vec v = -9.89\hat i - 12.8 \hat j

now the direction of the velocity is given as

\theta = tan^{-1}(\frac{v_y}{v_x})

\theta = tan^{-1}(\frac{-12.8}{-9.89})

\theta = -127.7 degree

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A watermelon is dropped off of a 50 ft bridge, and it explodes upon impact with the ground. How fast was it traveling in mph upo
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Answer: 56.72 ft/s

Explanation:

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Answer:

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