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2 years ago

PLZ WHAT I DOO I BROKE A WINDOW WHAT I DO

1 answer:

3 0

Hmmmm, you either tell your parent, try to fix it, or blame it on the dog with a baseball bat that always comes on your yard and tries to eat you.

o(*￣▽￣*)ブ

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The shape of the earth is an _______ ________

Nastasia [14]

The shape of the earth is an <span>oblate spheroid.</span>

3 0

3 years ago

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

emmainna [20.7K]

Answer:

$5,2$

Explanation:

From the question we are told that:

Speed of light $C=3.0×10^8 m/s.$

Generally the equation for Average Speed is mathematically given by

$V_{avg}=\frac{d}{t}$

Where

d=Distance between the Earth and the sun

$d=1.5*10^11m$

Therefore

$t=\frac{d}{V_{avg}}$

$t=\frac{1.5*10^11m}{3.0×10^8 m/s.}$

$t=5*10^2s$

Since m and n is given in the form of

$m*10^n$

Therefore

$m=5 & n=2$

$5,2$

3 0

2 years ago

What is the pressure at 5000 km below the surface of the earth

attashe74 [19]

Answer:

The centre of the earth is harder to study than the centre of the sun." Temperatures in the lower mantle the reach around 3,000-3,500 degrees Celsius and the barometer reads about 125 gigapascals, about one and a quarter million times atmospheric pressure.

Explanation:

7 0

2 years ago

A 1.2 kg hammer slams down on a nail at 5.0 m/s and bounces off at 1.0 m/s. If the impact lasts 1.0 ms, what average force is ex

Delvig [45]

Answer:

Explanation:

Impulse results in a change of momentum

FΔt = mΔV

F = mΔV/Δt

The impulse acting on the hammer will equal the impulse acting on the nail

If we assume upward is the positive direction

F = m(vf - vi)/t

F = 1.2(1.0 - (-1.5)) / 0.001

F = 3000 N

7 0

2 years ago

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

2 years ago