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Kipish [7]
3 years ago
12

A bowling ball of mass m=1.7kg is launched from a spring compressed by a distance d=0.31m at an angle of theta=37 measured from

the horizontal. It is observed that the ball reaches a maximum height of h=4.9m, measured from the initial position of the ball. Let the gravitational potential energy be zero at the initial height of the bowling ball.
Part a) What is the spring constant k, in newtons per meter?

Part b) Calculate the speed of the ball, V0 in m/s, just after the launch.
Physics
1 answer:
vodomira [7]3 years ago
8 0

Answer:

k = 1 700.7 N/m

v0 = 9.8 m/s^2

Explanation:

Hello!

We can answer this question using conservation of energy.

The potential energy of the spring (PS) will transform to kinetic energy (KE) of the ball, and eventually, when the velocity of the ball is zero, all that energy will be potential gravitational (PG) energy.

When the kinetic energy of the ball is zero, that is, when it has reached its maximum heigh, all the potential energy of the spring will be equal to the potential energy of the gravitational field.

PS = (1/2) k x^2  <em>where x is the compresion or elongation of the spring</em>

PG = mgh

a)

Since energy must be conserved and we are neglecting any energy loss:

PS = PG

Solving for k

k = (2mgh)/(x^2) = ( 2 * 1.7 * 9.81 * 4.9 Nm)/(0.31^2 m^2)

k = 1 700.7 N/m

b)

Since the potential energy of the spring transfors to kinetic energy of the ball we have that:

PS = KE

that is:

(1/2) k x^2 = (1/2) m v0^2

Solving for v0

v0 = x √(k/m) = (0.31 m ) √( 1 700.7 N/m / 1.7kg)

v0 = 9.8 m/s^2

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3 years ago
Find the radius Rrigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7×
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r = 1.61 x 10^{11} m

Explanation:

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we can find the radius of the star from the equation below

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where area (A) = 4 x π x r^{2} (assuming it is a sphere)

therefore  the equation becomes

H = 4 x π x r^{2} x  ε x σ x T^{4}  

2.7 x 10^31  = 4 x π x r^{2} x  1 x 5.67 x 10^{-8}  x (11,000)^{4}

r = \sqrt{\frac{2.7 x 10^31}{4 x π x 1 x 5.67 x 10^{-8}  x (11,000)^{4}} }

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3 years ago
Kayla starts at -3, walks 5 blocks right and 3 blocks left. What is her displacement?
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Then if you start at position A, and you walk N blocks to the right, the new position is:

A + N

And if you start at position A, and you walk M blocks to the left, the new position is:

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In this case, we know that Kayla starts at -3 and she walks 5 blocks to the right.

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