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dalvyx [7]
3 years ago
14

What is the force felt by the 64-kg occupant of the car? Express your answer to two significant figures and include the appropri

ate units. Enter positive value if the direction of the force is in the direction of the initial velocity and negative value if the direction of the force is in the direction opposite to the initial velocity. F =
Physics
1 answer:
Yakvenalex [24]3 years ago
5 0

Answer:

The force is -1.67\times10^{5}\ N

Explanation:

Given that,

Mass of car = 64 kg

Suppose,  a 1400-kg car that stops from 34 km/h on a distance of 1.7 cm.

We need to calculate the acceleration

Using formula of acceleration

v^2-u^2=2as

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value into the formula

0^2-(34\times\dfrac{5}{18})^2=2\times a\times 1.7\times10^{-2}

a=\dfrac{(34\times\dfrac{5}{18})^2}{2\times1.7\times10^{-2}}

a=-2623.45\ m/s²

We need to calculate the force

Using formula of force

F=ma

F=64\times(-2623.45)

F=-1.67\times10^{5}\ N

Negative sign shows the direction of the force is in the direction opposite to the initial velocity.

Hence, The force is -1.67\times10^{5}\ N

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Answer:

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Hello everyone. This is a question about Dimensional Analysis and I came across this question but I am unable to wrap my head ar
omeli [17]

Answer:

2. [B] = [L]/[T] and [C] = [L]/[T]

Explanation:

I assume you mean this:

A = B² + 2B⁴/C²

Since you can't add numbers with different units (for example, you can't add seconds to meters), each term in the sum must have the same units as A.

B² = [L]²/[T]²

B = [L]/[T]

B⁴/C² = [L]²/[T]²

C²/B⁴ = [T]²/[L]²

C² = B⁴ [T]²/[L]²

C² = ([L]/[T])⁴ [T]²/[L]²

C² = [L]²/[T]²

C = [L]/[T]

Notice we ignore the 2 coefficient, which is unitless.

7 0
4 years ago
A 45.0-kg sample of ice is at 0.00°C. How much heat is needed to melt it? For water, Lf=334 kJ/kg and Lv=2257 kJ/kg 
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Heat required to change the phase of ice is given by

Q = m* L

here

m = mass of ice

L = latent heat of fusion

now we have

m = 45 kg

L = 334 KJ/kg

now by using above formula

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Q = 1.5 * 10^7 J

In KJ we can convert this as

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7 0
3 years ago
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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

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