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prohojiy [21]
2 years ago
9

A child approaches a stop sign and applies the brakes to slow to a stop. If the bicycle and child were originally traveling at 9

.00 m/s and the breaks cause a deceleration of magnitude 1.20 m/s^2, how long will it take the child to stop?
Physics
1 answer:
user100 [1]2 years ago
6 0

Answer:

7.5secs

Explanation:

Using the equation of motion

v = u + at

v is the final velocity = 0m/s

u is the initial velocity = 9m/s

a is the deceleration = -1.2m/s²

t is the required time

Substitute;

0 = 9 + 1.2(t)

-9 = -1.2t

t = -9/-1.2

t = 7.5secs

hence it takes the child 7.5secs to stop

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A thermally insulated vessel containing a gas whose molar mass is equal to M and the ratio of specific heats cP /cV = γ moves wi
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Answer:

∆T = Mv^2Y/2Cp

Explanation:

Formula for Kinetic energy of the vessel = 1/2mv^2

Increase in internal energy Δu = nCVΔT

where n is the number of moles of the gas in vessel.

When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas

We say

1/2mv^2 = ∆u

1/2mv^2 = nCv∆T

Since n = m/M

1/2mv^2 = mCv∆T/M

Making ∆T subject of the formula we have

∆T = Mv^2/2Cv

Multiple the RHS by Cp/Cp

∆T = Mv^2/2Cv *Cp/Cp

Since Y = Cp/CV

∆T = Mv^2Y/2Cp k

Since CV = R/Y - 1

We could also have

∆T = Mv^2(Y - 1)/2R k

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3 years ago
a wooden block has a mass of 1.2 kg, a specific heat of 710, and is at a temperature of 25*C. what is the block's final temperat
mash [69]

The final temperature of the block is 27.5^{\circ} C

Explanation:

The amount of thermal energy Q supplied to a substance is related to the increase in temperature of the substance, \Delta T, according to the equation

Q=mC_s \Delta T

where:

m is the mass of the substance

C_s is the specific heat capacity of the substance

In this problem, we have:

m = 1.2 kg is the mass of the block

Q = 2,130 J is the amount of energy supplied to the block

C_s = 710 J/kg^{\circ}C is the specific heat capacity of the block

Solving for \Delta T, we find the increase in temperature:

\Delta T = \frac{Q}{m C_s}=\frac{2130}{(1.2)(710)}=2.5^{\circ}C

And since the initial temperature was

T_i = 25^{\circ}C

The final temperature will be

T_f = T_i + \Delta T = 25+2.5=27.5^{\circ} C

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

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