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Debora [2.8K]
3 years ago
14

An ideal spring obeys hooke's law: f = −bikx. a mass of m = 0.3 kg hung vertically from this spring stretches the spring 0.13 m.

the acceleration of gravity is 9.8 m/s 2 . calculate the value of the force constant k for the spring. answer in units of n/m
Physics
1 answer:
melamori03 [73]3 years ago
7 0
The force = 0.3 * 9.8 = 2.94
x = 0.13

k = 2.94/0.13
k = 22.6 N/m
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An astronaut is moving in space when a big explosion occurs about 50 meters behind him. How will the astronaut come to know abou
LUCKY_DIMON [66]

Answer:

The correct answer is B.

The astronaut will know due to the light from the explosion.

Explanation:

Sound and vibrations require a medium such as air to travel through. Space, there is no air. Only a vacuum. So sound and vibrations are unable to travel. Light requires no medium to travel. It can go through a vacuum.  

Therefore the Astronaut will see a bright flash of light as it travels from the explosion to outer space. It is also important to note that light can travel very far because nothing else interacts with its wave particles and as such, it cannot be impeded.

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7 0
3 years ago
A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal
Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, a=g=-9.8\ m/s^2(Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

S=ut+\frac{1}{2}at^2

Plug in the given values and solve for 'u'. This gives,

0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0
4 years ago
On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs
user100 [1]

Answer:

A) \Delta P =  14512.5 Pa = 14.512 kPa

B) F = 1632.65 N

Explanation:

Given details

outside air speed is given as v_2 = 150 m/s

since inside air is atmospheric , v_1 = 0 m/s

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\Delta P = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2

\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]

\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]

\Delta P =  14512.5 Pa = 14.512 kPa

b) force exerted on window

Area of window  = 25\times 45 = 1125 cm^2 = 0.1125 m^2

We know that force is given as

F = P\times A

F = 14512.5 \times 0.1125

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