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CaHeK987 [17]
4 years ago
11

For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this

is simple harmonic motion (in the vertical direction) with amplitude 5 cm and frequency 2 Hz. There is a box on the floor with mass m = 1 kg.
(a) Assuming the box remains in contact with the floor throughout, find the maximum and minimum values of the normal force exerted on it by the floor over an oscillation cycle.
(b) How large would the amplitude of the oscillations have to become for the box to lose contact with the floor, assuming the frequency remains constant? (Hint: what is the value of the normal force at the moment the box loses contact with the floor?)
Physics
1 answer:
ikadub [295]4 years ago
7 0

Answer:

no idea

Explanation:

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How would you get your friend to change his/her mind from using PED's?​
kvasek [131]

Answer:

Watch funny videos and start eating a lot of candy

6 0
3 years ago
A slingshot fires a pebble from the top of a building at a speed of 17.5 m/s. The building is 31.5 m tall. Ignoring air resistan
Jlenok [28]

Answer:

The speed with which the pebble strikes the ground is 30.4m/s

Explanation:

Initial velocity (u) = 17.5m/s, height (h) = 31.5m, acceleration due to gravity (g) = 9.8m/s^2

V^2 = u^2 + 2gh = 17.5^2 + 2×9.8×31.5 = 306.25 + 617.4 = 923.65

V^2 = 923.65

V = √923.65 = 30.4m/s

6 0
3 years ago
In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor.
USPshnik [31]
V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )  
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At  α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t² 
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
7 0
3 years ago
When the atmosphere is not quite clear, one may sometimes see colored circles concentric with the Sun or the Moon. These are gen
stealth61 [152]

Answer:

D) diffraction

Explanation:

Corona is an optical phenomenon produced by the diffraction of sunlight or moonlight, as light moves through water droplets in the atmosphere.

This phenomenon produces one or more diffuse concentric rings of light around the Sun or Moon, usually seen as colored circles.

Therefore, the explanation for these phenomena of colored concentric circles, sometimes seen with the Sun or the Moon involves diffraction.

5 0
3 years ago
A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.
miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

5 0
1 year ago
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