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sergiy2304 [10]

1 year ago

6

(a) State and explain which of the arrangements would have the greater extension of spring(s). (b) Explain if there are any chan

ges observed in the extension of the springs if the same set-up is done on the moon.

2 answers:

Lostsunrise [7]1 year ago

8 0

Answer:

2................................

Explanation:

swat321 year ago

3 0

Answer:
arrangement 2

Explanation:
arrangement 1's spring would broke idek

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Is there a change in speed from when a tennis ball first starts bouncing to when it is near the end of the bounce? How can you t

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Explanation:

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1 year ago

The driver of a car traveling on the highway suddenly slams on the brakes because of a slowdown in traffic ahead. A) If the car’

ladessa [460]

Answer:

Acceleration is -30000 mi/h²

Distance travelled in the 3 seconds of deceleration is 261.888 feet

Explanation:

t = Time taken for the car to slow down = 3 s =

u = Initial velocity = 75 mi/h

v = Final velocity = 50 mi/h

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{50-75}{\frac{3}{3600}}\\\Rightarrow a=-30000\ mi/h^2$

Acceleration is -30000 mi/h²

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{50^2-75^2}{2\times -30000}\\\Rightarrow s=0.0496\ mi$

Converting to feet

1 mile = 5280 feet

0.0496 mile = 0.0496×5280 = 261.888 feet

Distance travelled in the 3 seconds of deceleration is 261.888 feet

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2 years ago

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2 years ago

Read 2 more answers

A missile is launched upward with a speed that is half the escape speed. What height (in radii of Earth) will it reach?

timofeeve [1]

Answer:

The height (h) will be: $\frac{3}{4}R =h$

Explanation:

The scape speed equation is given by:

$v_{scape}=\sqrt{\frac{2GM}{R}}$

Now, the speed of the missile is

$v_{missile}=\frac{1}{2}v_{scape}$

$v_{scape}=\frac{1}{2}\sqrt{\frac{2GM}{R}}$

Using the conservation of energy, we can find the maximu height of the missile.

$E_{i}=E_{f}$

$\frac{1}{2}mv_{scape}^{2}-mgR =-mgh$

$\frac{1}{2}\frac{2GM}{4R}-gR =-gh$

$\frac{GM}{4R}-gR =gh$

Let's recall that g = GM/R², using the equivalence principle. When R is the radius of the earth and M is the mass of the earth.

$\frac{1}{4}gR-gR =-gh$

$\frac{1}{4}R-R =-h$

Therefore the height (h) will be:

$\frac{3}{4}R =h$

I hope it helps you!

4 0

1 year ago