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3 years ago

Plz help, and show work

1 answer:

7 0

1.
Given:
S=-35m(downward), a=-9.8m/s2, u=0(start from rest)

Solution:
Formula: S=ut+(1/2)(a)(t^2)
-35m=(0)t+(1/2)(-9.8m/s)(t^2)
t^2= 7.142857
t= 2.67s

Answer: t=2.67s

2.
Given:
u=0(start from rest), t=20s, v=2m/s

Solution:
Formula: v=u+at
2m/s=0+a(20s)
a=0.1m/s^2

Answer: a=0.1m/s^2

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3 years ago

A car with a mass of 1.50 X 10^3 kg starts from rest and accelerates to a speed of 18.0 m/s in 12.0 s. Assume that the force of

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The average power is 20.3 kW

Explanation:

First of all, we calculate the work done on the car: the work-energy theorem states that the work done on the car is equal to the change in kinetic energy of the car, so we have

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

W is the work done

$K_i, K_f$ are the initial and final kinetic energy of the car

$m=1.50\cdot 10^3 kg = 1500 kg$ is the mass of the car

u = 0 is the initial velocity

v = 18.0 m/s is the final velocity

Substituting,

$W=\frac{1}{2}(1500)(18)^2=2.43\cdot 10^5 J$

Now we can find the average power developed by the car's engine, which is given by

$P=\frac{W}{t}$

where

$W=2.43\cdot 10^5 J$ is the work done

t = 12.0 s is the time taken

Substituting,

$P=\frac{2.43\cdot 10^5 J}{12.0}=20,250 W = 20.3 kW$

Learn more about power:

brainly.com/question/7956557

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4 years ago

Which of the following is another term for a planned economy

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Answer:

Another term for planned economy is a socialist system.

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The Andromeda galaxy is moving toward us at 3.01x10^5 m/s. By how much will the frequency of the 6.17x10^14 Hz hydrogen line be

just olya [345]

Answer:

f - f '= 6.18*10^11 Hz

Explanation:

The change in frequency is given by:

$\frac{1}{f'}=\frac{1}{f}\frac{\sqrt{1+v/c}}{\sqrt{1-v/c}}$

f': observed frequency

f: source frequency = 6.17*10^14 Hz

v: speed of the source = 3.01*10^55 m/s

c: speed of light = 3*10^8 m/s

By replacing all these values you obtain:

$\frac{1}{f'}=\frac{1}{6.17x10^{14} Hz}\frac{\sqrt{1+(3.01*10^5m/s)(3*10^8m/s)}}{\sqrt{1-(3.01*10^5m/s)(3*10^8m/s)}}=1.62*10^{-15}\\\\f'=6.16*10^{14}Hz$

hence, the change in frequency f-f' will be:

f - f '= 6.18*10^11 Hz

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