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rodikova [14]
3 years ago
15

How did people studying lunar eclipses learn that Earth is round?

Physics
2 answers:
qwelly [4]3 years ago
4 0

Answer:

Scientists have studied eclipses since ancient times. Aristotle observed that the Earth's shadow has a circular shape as it moves across the moon. He posited that this must mean the Earth was round. Another Greek astronomer named Aristarchus used a lunar eclipse to estimate the distance of the Moon and Sun from Earth

bearhunter [10]3 years ago
3 0
What the person said above me is exactly correct
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A rat is pushed with a massive force. The mass of the rat is 2.3 kg and the acceleration of the rat is 9.5 m/s^2. What was the f
Phoenix [80]

Answer:

Force = 21.85N

Explanation:

Given the following data;

Mass = 2.3kg

Acceleration = 9.5m/s²

To find the force;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

F = ma

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

Substituting into the equation, we have

Force = 2.3 * 9.5

Force = 21.85N

Therefore, the force applied to the rat is 21.85 Newton.

8 0
3 years ago
What is the force on an object that goes from 35 m/s to 85 m/s in 20 seconds and has a mass of 148 kg
Sever21 [200]
F=ma
F = 148×(85-35)÷20
F = 148×(50÷20)
F = 148×2.5
F = 370N
3 0
3 years ago
A Smart Car, which has a mass of 1000 kg, is going 20 m/s. When it hits the barrier, it stops with a time of 0.5 seconds. What i
antiseptic1488 [7]

Answer:

The change in momentum = -20000 kg m/s.

Explanation:

Mass m = 1000 kg

speed v₁ = 20 m/s

speed v₂ = 0 m/s

We know that,

The change in momentum

ΔP = m (Δv)

ΔP = m (v₂ - v₁)

     = 1000 (0 - 20)

     = 1000 (-20)

     = -20000 kg m/s

Thus, the change in momentum = -20000 kg m/s.

Note: negative sign indicates that the velocity is reducing when it hits the barrier.

4 0
3 years ago
Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0
3 years ago
A spacecraft is moving past the earth at a constant speed of 0.60 times the speed of light. The astronaut measures the time inte
Afina-wow [57]

Answer:

the time interval that an earth observer measures is 4 seconds

Explanation:

Given the data in the question;

speed of the spacecraft as it moves past the is 0.6 times the speed of light

we know that speed of light c = 3 × 10⁸ m/s

so speed of spacecraft v = 0.6 × c = 0.6c

time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds

Now, from time dilation;

t = Δt₀ / √( 1 - ( v² / c² ) )

t = Δt₀ / √( 1 - ( v/c )² )

we substitute

t = 3.2 / √( 1 - ( 0.6c / c )² )

t = 3.2 / √( 1 - ( 0.6 )² )

t = 3.2 / √( 1 - 0.36 )

t = 3.2 / √0.64

t = 3.2 / 0.8

t = 4 seconds

Therefore, the time interval that an earth observer measures is 4 seconds

6 0
3 years ago
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