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3 years ago

8

What is the momentum of an 85 kg runner traveling at 8 m/s?

1 answer:

dolphi86 [110]3 years ago

5 0

As per the equation: P(momentum)= (mass x velocity) The answer is 680 kg*m/s.

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A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantit

Bess [88]

Answer:

C. 85%

Explanation:

A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat?

A. 15%

B. 30%

C. 85%

D. 100%

work done by the system will be

W=PdV

p=pressure

dV=change in volume

3tam will be changed to N/m^2

3*1.01*10^5

W=3.03*10^5*(1.5-1)

convert 0.5L to m^3

5*10^-4

W=3.03*10^5*5*10^-4

W=152J

therefore

to find the percentage used

152/1000*100

15%

100%-15%

85% uf the fuel's energy was lost to friction and heat

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3 years ago

All organisms require a set of instructions that specify its traits. The instructions that are responsible for all the inherited

tino4ka555 [31]

DNA
DNA provides the set of instructions that tell the organism how to form

3 0

3 years ago

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4. Heat is added to an ideal gas and the gas expands. In such a process the temperature

Bumek [7]

Answer:

is high as 100 degrees c

Explanation:

due to high heat gas expands fast than normal

4 0

3 years ago

Which of these best defines weather? (3 points) a Atmospheric conditions over 30-year period Ob Day-to-day condition of the atmo

prisoha [69]

Answer:

b Day-to-day condition of the atmosphere

Explanation:

Weather is short term, Climate is long term

3 0

2 years ago

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The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the

Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

$s=ut+0.5at^2$

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, $a=-10.0 m/s^2$

$\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m$

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is $s=44.0 m$

The acceleration is same, $a=-10.0 m/s^2$

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

$v^2-u^2=2as$

$0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s$

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.

7 0

3 years ago