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2 years ago

A roller coaster with a potential energy of 235,200 J sits at the top of a 30 m high hill. What is the mass of the roller

Physics

2 answers:

xxTIMURxx [149]2 years ago

5 0

Answer:

800 kg

Explanation:

I just did it :)

igomit [66]2 years ago

4 0

Answer:

the correct answer is A. 800 kg

Explanation:

i got the answer right on the quiz, i hope this helps!

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Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity

Anna11 [10]

Answer:

the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

the distance that A slides relative to B is $S = \frac{v_o^2M}{2 \mu g (M+m)}$

Explanation:

From the diagram below;

acceleration of A relative to B is : $a = - ( \mu g + \frac{ \mu mg}{M})$

where

v = u + at

$0 = v_o + ( - \mu g - \frac{\mu m g }{M})t$

Making t the subject of the formula; we have:

$t = \frac{v_o M}{(\mu g )(M+m)}$

$v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S\\\\$

$S = \frac{v_o^2M}{2 \mu g (M+m)}$ which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

$v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}\\\\$

$v = \frac{mv_o}{m+M}$

Thus, the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

4 0

3 years ago

A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?

Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

$H_m = \frac{u^2sin^2 \theta}{2g}$

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

$H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta = sin^{-1}(0.2123)\\\\\theta = 12.26^0$