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Vikentia [17]
3 years ago
8

How does a car’s average speed differ from its instantaneous speed?

Physics
2 answers:
polet [3.4K]3 years ago
8 0

Answer:

Option (A) is correct.

Explanation:

The speed of a car is defined as the ratio of the distance traveled by the car to the time taken by the car.

speed = distance / time

The average speed of the car is defined as the ratio of the total distance traveled by the car to the total time taken by the car.

Average speed = total distance traveled / total time taken

Instantaneous speed is defined as the speed of the car at any instant.

attashe74 [19]3 years ago
7 0

Answer: Instantaneous speed is measured at each particular instant, and average speed is the sum of the instantaneous speeds.

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bogdanovich [222]
<span>First sum applied the Newton's second law motion: F = ma Force = mass* acceleration This motion define force as the product of mass times Acceleration (vs.Velocity). Since acceleration is the change in velocity divided by time, force=(mass*velocity)/time such that, (mass*velocity)/time=momentum/time Therefore we get mass*velocity=momentum Momentum=mass*velocity Elephant mass=6300 kg; velocity=0.11 m/s Momentum=6300*0.11 P=693 kg (m/s) Dolphin mass=50 kg; velocity=10.4 m/s Momentum=50*10.4 P=520 kg (m/s) The elephant has more momentum(P) because it is large.</span>
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3 years ago
Use the equation PiVi PfVf. Assume that Pi 101 kPa and Vi 10.0 L. If Pf 43.0 kPa, what is Vf
iren2701 [21]
Solving for vf gives you PiVi/Pf. Now plug in 101kPa*10L/43kPa = 23.48L. Using significant figures i would round to 23.5L
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4 years ago
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umka2103 [35]
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7 0
3 years ago
A ball is hurled straight up at a speed of 15 m/s, leaving the hand of the thrower 2.00 m above the ground. Compute the times an
SVETLANKA909090 [29]

Answer:

5.37 m/s

0.98 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 10+15^2}\\\Rightarrow v=5.37\ m/s

Velocity of the ball when it passes an observer sitting at a window is 5.37 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{5.37-15}{-9.81}\\\Rightarrow t=0.98\ s

Time taken by the ball to pass the observer sitting at a window is 0.98 seconds

3 0
3 years ago
An ideal gas is brought through an isothermal compression process. The 3.00 mol 3.00 mol of gas goes from an initial volume of 2
ozzi

Answer:

The answers can be found by considering the isothermal expansion equation as well as the ideal gas equation from where we have

The temperature T = 602.64K and the final pressure P = 110.24MPa

Explanation:

Numbeer of moles of gas = 3.00 mol  

initial volume = 230.8×10−6 m3  

final volume = 133.4×10−6 m3 .

released energy =  8240 J

Temperature = Constant = T

Pressure =  p_{f} =unknown

From the relation the combined ideal gas law, PV = nRT

Where R = 8.314 4621.JK−1mol−1

we have The release energy from compression P1V1 -P2V2

-qrev = -nRTln(\frac{V_{2} }{V_{1} }) = 8240J

n = 3

Hence -nRTln(\frac{V_{2} }{V_{1} }) =  3×8.314 462×ln(\frac{133.4}{230.8}) × T=  -8240 J

or -13.67×T = -8240J, thus T = -8240/-13.67 = 602.64K

The Final pressure is given by

PV = n×R×T from where we have V = final volume thus

P = (n×R×T)/V = (3×8.134×602.64)÷(133.4×10^{-6}) = 110237041.1 N/m^{2} = 110.237MPa

8 0
3 years ago
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