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lorasvet [3.4K]

2 years ago

10

A ball of mass m is dropped from rest from a height h and collides elastically with the floor, rebounding to its original height

.
What is the magnitude of the average force applied by the floor on the ball during the time the ball is in contact with the floor?

(A) Zero
(B) mg
(C) 2mg
(D) 4mg
(E) It cannot be determined without knowing the length of time that the ball is in contact with the floor.

2 answers:

grin007 [14]2 years ago

5 0

Answer:

option (E)

Explanation:

mass = m

height = h

initial velocity u = 0

The average force applied by the floor on the ball is the rate of change of momentum of the ball.

So, we need the time of contact of ball and floor, so it is not given in the question.

Thus, option (E)

Jet001 [13]2 years ago

3 0

Answer:

Explanation:

Given

mass m collides elastically with floor and jump to its original height I .e.velocity of rebound is same as the initial velocity

Impulse imparted by steel ball to the ground is given by the change in momentum of steel ball

$J=\Delta P$

and impulse is product of average force and time

so we need time to calculate the average force. So, option e is correct

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If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w

marshall27 [118]

Answer:

$P(3600)=593.247W$

Explanation:

First, let's find the voltage through the resistor using ohm's law:

$V=IR=20*8=160V$

AC power as function of time can be calculated as:

$P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)$ (1)

Where:

$\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency$

Because of the problem doesn't give us additional information, let's assume:

$\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi$

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

$P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W$

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147.5 ml of ethyl alcohol (coefficient of volume expansion = 1120 x 10-6K-1) at a temperature of 273.1 K is measured into a 150.

marishachu [46]

Answer:

288.2 K

Explanation:

$\beta$ = Volumetric expansion coefficient = $1120\times 10^{-6}\ /K$

$T_i$ = Initial temperature = 273.1 K

$T_f$ = Final temperature

$v_0$ = Original volume = 150 mL

Change in volume is given by

$\Delta v=\beta v_0(T_f-T_i)\\\Rightarrow T_f=\frac{\Delta v}{\beta v_0}+T_i\\\Rightarrow T_f=\frac{150-147.5}{1120\times 10^{-6}\times 147.5}+273.1\\\Rightarrow T_f=288.2\ K$

The temperature of the ethyl alcohol should be 288.2 K to reach 150 mL

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Annette [7]

The car tyre contains air initially at a pressure of 195 kPa after travelling several km the temperature of the air inside a car tyre rises from 30 to 70°C if the tyre is rigid and does not expand then the new pressure inside the tyre would be 220.74 kPa.

<h3>What is pressure?</h3>

The total applied force per unit of area is known as the pressure.

The pressure depends both on externally applied force as well the area on which it is applied.

The mathematical expression for the pressure

Pressure = Force /Area

the pressure is expressed by the unit pascal or N /m²

By using the Charles law for gases which states that the volume of the gas remains constant then the pressure of the gas is directly proportional to the temperature.

As given in the problem the tyre is rigid and does not expand this means the volume of the tyre remains constant.

The mathematical expression for Charles's law is as follows

P₁/P₂ = T₁/T₂

First, we have to change the temperature from degree Celcius to the kelvin temperature scale

K = 273 + C

where k is the temperature in kelvin and the C is degrees of Celcius

Initially, the temperature was 30° C

T₁ = 273 + 30

T₁ = 303 K

Then after travelling the temperature of the air inside a car tyre rises from 30 to 70°C

T₂= 273+ 70

T₂ =343 K

The car tyre contains air initially at a pressure of 195 KPa

P₁ = 195 kPa

Lets us take the final pressure of the air would be P₂

By substituting the values in the formula

P₁/P₂ = T₁/T₂

195/P₂ = 303/343

P₂ = 220.74 kPa

Thus, the new pressure inside the tyre would be 220.74 kPa.

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