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tigry1 [53]
3 years ago
15

A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed of 10 m/s. If air

resistance is negligible, the distance from the edge of the base of the cliff to where the rock hits the level ground below the cliff is most nearly
a. 20 m

b. 5 m

c. 10 m

d. 40 m
Physics
1 answer:
Gwar [14]3 years ago
8 0

Answer:

option A

Explanation:

given,

height of the cliff = 20 m

initial speed = 10 m/s

distance from the edge of the base of the cliff  = ?

now,

time taken to cover vertical distance

initial vertical speed = 0 m/s

using equation of motion

s = u t + \dfrac{1}{2}gt^2

s =\dfrac{1}{2}gt^2

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 20}{9.8}}

  t = 2 s

in same time the ball will travel horizontal distance

horizontal velocity of ball = 10 m/s

distance = s x t

d = 10 x 2

d = 20 m

Hence, the correct answer is option A

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An atom of a certain element has 36 protons, 36 electrons, and a mass number of 84. At room temperature, this element is a very
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Answer:

12 neutrons

Explanation:

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A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The
S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

\sum F_x = 0

f-N_w = 0

N_w = f

The forces in the horizontal direction would be,

\sum F_y = 0

N_f -W =0

N_f = W

The sum of Torques at equilibrium,

\sum \tau = 0

Wdcos\theta - N_wLsin\theta = 0

WdCos\theta = fLSin\theta

f = \frac{Wd}{Ltan\theta}

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

f_{max} = \mu W=\frac{Wd}{Ltan\theta}

\theta = tan^{-1} (\frac{d}{\mu L})

Replacing,

\theta = tan^{-1} (\frac{0.9}{0.42*2})

\theta = 46.975\°

Therefore the minimum angle that the person can reach is 46.9°

8 0
3 years ago
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