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sammy [17]
3 years ago
13

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 base of the cliff to where the rock hits the level ground below the cliff is most nearly
a.5m
b.10m
c.20m
d.40m
e.200m
Physics
1 answer:
scZoUnD [109]3 years ago
5 0

Answer:

c.20

Explanation:

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Natali [406]

Answer:

How to find the maximum height of a projectile?

if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...

if α = 45°, then the equation may be written as: ...

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.

Explanation:

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3 years ago
Section 1: Experimental Overview
Luden [163]

Answer:

1-state what the lab is about, that is, what scientific concept (theory, principle, procedure, etc.) you are supposed to be learning about by doing the lab. You should do this briefly, in a sentence or two. If you are having trouble writing the opening sentence of the report, you can try something like: "This laboratory experiment focuses on X…"; "This lab is designed to help students learn about, observe, or investigate, X…." Or begin with a definition of the scientific concept: "X is a theory that…."

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Explanation:

7 0
3 years ago
An astronaut holds a rock 100m above the surface of Planet X. The rock is then thrown upward with a speed of 15m/s, as shown in
Harlamova29_29 [7]

The acceleration due to gravity of the planet X is 1 m/s².

The given parameters;

  • height above the ground, h = 100 m
  • initial velocity of the rock, u = 15 m/s
  • time of motion of the rock, t = 10 s

The acceleration due to gravity is calculated as follows;

h = ut - \frac{1}{2} gt^2\\\\100 = 15(10) - (0.5\times 10^2)g\\\\100 = 150 - 50g\\\\50g = 150-100\\\\50g = 50\\\\g = 1 \ m/s^2

Thus, the acceleration due to gravity of the planet X is 1 m/s²

Learn more here: brainly.com/question/24564606

7 0
2 years ago
A spring with spring constant 15.5 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.
Aloiza [94]

Answer:

A) 138.8g

B)73.97 cm/s

Explanation:

K = 15.5 Kn/m

A = 7 cm

N = 37 oscillations

tn = 20 seconds

A) In harmonic motion, we know that;

ω² = k/m and m = k/ω²

Also, angular frequency (ω) = 2π/T

Now, T is the time it takes to complete one oscillation.

So from the question, we can calculate T as;

T = 22/37.

Thus ;

ω = 2π/(22/37) = 10.5672

So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g

B) In simple harmonic motion, velocity is given as;

v(t) = vmax Sin (ωt + Φ)

It is from the derivative of;

v(t) = -Aω Sin (ωt + Φ)

So comparing the two equations of v(t), we can see that ;

vmax = Aω

Vmax = 7 x 10.5672 = 73.97 cm/s

6 0
3 years ago
A train at rest emits a sound at a frequency of 1057 Hz. An observer in a car travels away from the sound source at a speed of 2
il63 [147K]

Answer:

993.52 Hz

Explanation:

The frequency of sound emitted by the stationery train is 1057 Hz.

The car travels away from the train at 20.6 m/s.

The frequency the observer hears is given by the formula:

f_o = \frac{v - v_o}{v}f

where v = velocity of sound = 343 m/s

vo = velocity of observer

f = frequency from source

This phenomenon is known as Doppler's effect.

Therefore:

f_o = \frac{343 - 20.6}{343} * 1057\\ \\f_o = 322.4 / 343 * 1057\\\\f_o = 993.52 Hz

The frequency heard by the observer is 993.52 Hz.

4 0
4 years ago
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