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spin [16.1K]
2 years ago
15

HURRY The compressed spring of a dart gun has potential energy of 50 J. If the spring constant is 200 N/m, what is the displacem

ent of the spring?
A. 0.5 m
B. 0.2 m
C. 0.7 m
D. 0.4 m
Physics
2 answers:
Dovator [93]2 years ago
8 0

the answer on edg is C, 0.7

Crank2 years ago
7 0
Spring potential energy:
E = 0.5 * k * x²

k spring constant
x spring compression

x = √(2 * E / k) = 0.7


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Answer:a b c

Explanation: I’m not sure tho

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3 years ago
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A 60 kg runner accelerates during a race at 3m/s2. what force is exerted on the earth with each step
marusya05 [52]

Answer:

180 Newton(N)

Explanation:

force =mass *acceleration

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3 years ago
Calculate the magnitude of the total impulse applied to the car to bring it to rest
n200080 [17]

Answer:

Total impulse = mv = Initial momentum of the car

Explanation:

Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.

The final velocity of the car is 0 m/s as it is brought to rest.

Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.

The magnitude of impulse is the absolute value of the change in momentum.

|J|=|p_f-p_i|

Momentum of an object is equal to the product of its mass and velocity.

So, the initial momentum of the car is given as:

p_i=mv

The final momentum of the car is given as:

p_f=m(0)=0

Therefore, the impulse is given as:

|J|=|p_f-p_i|=|0-mv|=|-mv|=mv

Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.

5 0
3 years ago
An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f
sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

f=\frac{v_s}{2L}         (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

f=\frac{343m/s}{2(0.47228m)}=362.36Hz      

Closed tube:

f'=\frac{v_s}{4L'}

L': length of the closed tube = 0.702821m

f'=\frac{343m/s}{4(0.702821m)}=122.00Hz

Next, you use the following formula for the beat frequency:

f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0
3 years ago
Hooke’s law describes the linear relationship between stress and strain through Young’s modulus. Given two materials under the s
stiks02 [169]

Answer:

The material with higher modulus will stretch less than

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E= stress/stain

Hooks law states that provided the elastic limit is not exceeded the extension e of a spring is directly proportional to the load or force attached

F=ke

Where k is the constant which gives the measure of the spring under tension

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3 years ago
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