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VMariaS [17]
3 years ago
13

Which astronomer spent 20 years plotting the positions of the planets

Physics
1 answer:
Harlamova29_29 [7]3 years ago
3 0

Tycho Brahe is the answer... i already investigate this :)


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A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di
Vikki [24]

Answer:

Mass of the cart = 146 kg

Explanation:

A cart is pulled by a force of 250 N at an angle of 35° above the horizontal.

The cart accelerates at 1.4 m/s² horizontally.

Horizontal force = Fcosθ = 250 cos35° = 204.79N

We have F = ma

Substituting

        204.79 = m x 1.4  

              m = 146.28 kg = 146 kg

Mass of the cart = 146 kg

3 0
3 years ago
A meteor moving 468 km per minute traveling in a south-to-north direction passed near Earth in 2013. Does this statement describ
Thepotemich [5.8K]

Answer:

this statement describes meteor's velocity,

because velocity is a vector quantity which has both magnitude as well as a specific direction and here the meteor's direction is specified in the statement hence we conclude that this statement describes meteor's velocity as well as speed too.

3 0
3 years ago
1. If I dig a 6FT hole how deep is that hole?
iren [92.7K]

Answer:

For the first one its about 25 feet

Explanation:

5 0
2 years ago
Read 2 more answers
When light encounters a barrier with slits cut it in it, the light will bend through the slits creating a pattern like that seen
lawyer [7]

Answer:

reflection

Explanation:

an example would be looking in the mirror

8 0
3 years ago
Read 2 more answers
A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

H_{max}= \dfrac{v^2 sin^2(\theta)}{g}

now,

H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}

H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}

now, equating both the equations

\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}

\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}

   v₂² = 31061.79

   v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0
3 years ago
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