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Lena [83]
3 years ago
13

the end of the tracks, 8.8 m lower vertically, is a horizontally situated spring with constant 5 × 105 N/m. The acceleration of

gravity is 9.8 m/s 2 . Ignore friction. How much is the spring compressed in stopping the ore car? Answer in units of m.
Physics
1 answer:
devlian [24]3 years ago
6 0

Answer

Assuming the mass of the car, m = 43000 kg

initial speed u = 0

vertical distance moved, h = 8.8 m

spring constant k = 5 x  10⁵ N / m

acceleration of gravity = 9.8 m/s²

From law of conservation of energy ,

Gravitational potential energy at starting position =potential energy of the spring at maximum compression

                m g h =\dfrac{1}{2}k x^2

                x = \sqrt{\dfrac{2mgh}{k}}

                x = \sqrt{\dfrac{2\times 43000\times 9.8\times 8.8}{5\times 10^5}}

                    x = 14.83 m

If the mass of the car is equal to 43000 Kg the spring is compressed to 14.83 m

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borishaifa [10]

To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.

From the definition we know that the wavelength is described under the equation,

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f = frequency

Our values are,

f = 2Hz

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\lambda = 1.5*10^8m

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8 0
3 years ago
an object with a mass of 2.4 kg has a force of 12.6 N applied to it. What is the resulting acceleration of the object? WITH PROO
Scorpion4ik [409]
Newton’s Law: F = MA
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2 years ago
In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0
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Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims
777dan777 [17]

Answer:

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From work energy theorem,

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After cutoff kinetic energy is converted into potential energy,

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Put value of KE

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Answer:

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