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SpyIntel [72]
2 years ago
7

Knowledge of projectile motion might help you aim a basketball when you throw it so it is more likely to go into the basket. tru

e or false
please need a answer​
Physics
1 answer:
8_murik_8 [283]2 years ago
5 0

Answer: True.

Explanation:

You would be able to visualize the basketballs height going up and when it sinks down into the hoop.

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The scientist to first introduce the concept of inertia was
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The first scientist to introduce the concept of inertia was Galileo
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3 years ago
A rock is dropped off a cliff and falls the first half of the distance to the ground in 2.0 seconds. how long will it take to fa
son4ous [18]

Answer:

The answer is 0.83 seconds.

Explanation:

The formula of free fall is following:

h=1/2*g*t^2

Where g=9.8 m/s^2 and t=2 seconds, the rock takes:

h=1/2*9.8*2^2=19.6

19.6 meters. This is the half distance of the cliff. The whole distance is 39.2 meters. So it takes:

39.2=1/2*9.8*t^2\\t^2=8\\t=2.83

2.83 second to fall down completely. The rock takes the second half of the cliff in 0.83 seconds

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3 years ago
What are 5 wave interactions with matter ?
vichka [17]
An echo
Refraction
Diffraction
Transmission
reflection
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3 years ago
True or false: Everything in the universe works with harmony because of balanced energy
Eduardwww [97]

Answer:

Explanation:

True

7 0
2 years ago
Read 2 more answers
Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot
Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² =  2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ =  m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁  = 4.427 - 2v₂  ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂  = 8.854

v₂ = 8.854 / 3

v₂  = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁  = - 1.477 m/s

v₁  = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

mgh_{max} = \frac{1}{2}mv_{max}^2

(a) The maximum height achieved by the first ball (v₁  = 1.477 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max}  = 0.11 \ m

(b) The maximum height achieved by the second ball (v₂  = 2.95 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max}  = 0.44 \ m

6 0
3 years ago
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