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azamat
3 years ago
8

A steel ball rolls with constant velocity on a tabletop 1.95 m high. It rolls off and hits the ground 0.5 m away from the edge o

f the table. How fast was the ball rolling?
Physics
1 answer:
Alex777 [14]3 years ago
5 0

Answer:

0.79 m/s

Explanation:

First of all, we analyze the vertical motion of the ball. It is a free fall motion, so its vertical displacement is given by

s=ut+\frac{1}{2}at^2

where

s = 1.95 m is the displacement

u = 0 is the initial vertical velocity

t is the time

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t, we find the time it takes for the ball to reach the ground:

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.95)}{9.8}}=0.63 s

Now we can analyze the horizontal motion: this is a uniform motion with constant speed, so the horizontal distance covered by the ball is

d=v_x t

where

d = 0.5 m is the horizontal distance covered

t = 0.63 s is the time

Solving for vx, we find the horizontal velocity of the ball:

v_x = \frac{d}{t}=\frac{0.5}{0.63}=0.79 m/s

And this velocity is constant during the motion, so the ball was moving at 0.79 m/s when it rolls off the table.

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Hey there!

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3 years ago
Help me! Btw he’s playing golf.
fiasKO [112]

Answer:

Muscular energy

Explanation:

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3 years ago
Ccording to coulomb's law, which pair of charged particles has the lowest potential energy? according to coulomb's law, which pa
Sladkaya [172]

Coulombs law says that the force between any two charges depends on the amount of charges and distance between them. This force is directly proportional to the magnitude of the two charges and inversely proportional to the distance between them.

F=k\frac{|q_1| |q_2|}{r^2}

where q_1\hspace{1mm}and\hspace{1mm}q_2 are charges, r is the distance between them and k is the coulomb constant.

case 1:

q_1=-e\\ q_2=+3e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||3e|}{(100pm)^2}=3ke^2\times10^8

case 2

q_1=-e\\ q_2=+2e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||2e|}{(100pm)^2}=2ke^2\times10^8

case 3:

q_1=-e\\ q_2=+e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||e|}{(200pm)^2}=0.25ke^2\times10^8

Comparing the 3 cases:

The maximum potential force according to coulombs law is between -1 charge and +3 charge separated by a distance of 100 pm.

3 0
3 years ago
Read 2 more answers
A wave traveling in the positive x-direction with a frequency of 50.0 Hz is shown in the figure below. Find the following values
Klio2033 [76]

Answer:

Explanation:

a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.

b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or \lambda=20.0cm

c. The period, or T, of a wave is found in the equation

f=\frac{1}{T} were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:

50.0=\frac{1}{T} so

T=\frac{1}{50.0} and

T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)

d. The speed of the wave is found in the equation

f=\frac{v}{\lambda} and since we already have the frequency and we solved for the wavelength already, filling in:

50.0=\frac{v}{20.0} and

v = 50.0(20.0) so

v = 1.00 × 10³ m/s

And there you go!

5 0
3 years ago
An ice cream maker has a refrigeration unit which can remove heat at 120 Js'. Liquid ice
Rom4ik [11]

Answer:

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C is 45,360 J

Explanation:

The given parameters for the refrigeration unit and the ice cream are;

The power of the refrigeration unit = 120 J/s

The mass of the liquid ice cream, m = 0.6 kg

The initial temperature of the liquid ice cream, T₁ = 20°C

The freezing point temperature of the ice cream, T₂ = -16°C

The specific heat capacity of the ice cream, c = 2,100 J/kg⁻¹·°C⁻¹

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, ΔQ, is given as follows;

ΔQ = m × c × ΔT

Where;

ΔT = T₁ - T₂

∴ ΔQ = m × c × (T₁ - T₂)

Therefore, by substituting the known values, we have;

ΔQ = 0.6 × 2,100 × (20 - (-16)) = 45,360

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C = ΔQ = 45,360 J.

8 0
3 years ago
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