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3 years ago

If the moon were 2 times closer to earth than it is now, the gravitational force between earth and the moon would be

1 answer:

7 0

So much brighter and the moon would be so much darker than it is now because the moon is further away from the moon than it is now...

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The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to

LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power ,P = F .v

F=force

v=velocity

v= 100 km/h

$v=\dfrac{1000}{3600}\times 100\ m/s$

v=27.77 m/s

75 x 1000 = F x 27.77

$F= \dfrac{75000}{27.77}\ N$

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

3 0

3 years ago

A horizontal force in used to pull a 5 kilogram cart at a constant speed of 5 meters per second across the floor as shown in the

ddd [48]

A cart is pulled by horizontal force such that it moves with constant velocity

So here since velocity is constant we can say that its acceleration will be ZERO

now here for zero acceleration we can say

$F_{net} = 0$

So here we will have

$F_{net} = F_{ap} - F_f = 0$

here we know that

$F_f = 10 N$

so we will have

$F_{ap} - 10 = 0$

$F_{ap} = 10 N$

so here applied force on handle will be

5 0

3 years ago

Just this last one!!

Pepsi [2]

Answer:

$47 \ \frac{m}{s}$

Explanation:

s = displacement (m)

u = initial velocity $(\frac{m}{s})$

v = final velocity $(\frac{m}{s})$

a = acceleration $(\frac{m}{s^{2} })$

t = time (s)

s = 235

a = -4.7

v = 0

v² = u² + 2as

(0)² = u² + 2(-4.7)(235)

u² - 2209 = 0

u² = 2209

u = 47

4 0

1 year ago

Read 2 more answers

H e l p a h o m i e o u t t h a n k s

finlep [7]

Uh it’s D i’m pretty sure

4 0

2 years ago

Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 2.6 m long, its mass is 0.5

Irina-Kira [14]

Answer:

15.13 m/s

Explanation:

The wave speed of the stretched rope can be calculated using the following formula

$v = \sqrt{\frac{F_T}{\mu}}$

where $F_T = 44N$ is the tension on the rope and $\mu = m/L = 0.5 / 2.6 = 0.1923 kg/m$ is the density of the rope per unit length

$v = \sqrt{\frac{44}{0.1923}} = \sqrt{228.8} = 15.13 m/s$

6 0

3 years ago