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2 years ago

.A 0.6 kg basketball is thrown straight up at a velocity of 4 m/s. What is the initial kinetic energy of the basketball?

Physics

1 answer:

motikmotik2 years ago

7 0

Answer:

the initial kinetic energy of the basket ball is 4.8 J

Explanation:

The computation of the initial kinetic energy of the basket ball is shown below:

As we know that

Initial kinetic energy is

$= \frac{1}{2}mv^2 \\\\= \frac{1}{2}\times 0.6 \times (4m/s)^2\\\\= \frac{1}{2}\times 0.6 \times 16m/s^2\\\\= 4.8 J$

Hence, the initial kinetic energy of the basket ball is 4.8 J

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Describe the energy transformations that occur from the time a skydiver jumps out of a plane until landing on the ground.

Kisachek [45]

When the Skydiver jump out a plane, his Potential Energy is being converted or transform into Kinetic energy due to gravity. Hope this helps

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2 years ago

Read 2 more answers

In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad

Crank

Answer:

a) Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c) Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

A₁ v₁ = A₂ v₂

v₂ = v₁ A₁ / A₂

We replace

ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

Q = A v

Q = A₁ v₁

Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

A₁ = π r₁

A₂ = π r₂

We replace

Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

r₁ = d₁ / 2 = 1.00 / 2

r₁ = 0.500 10⁻² m

r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

Q = 1.57 10⁻² √(12 10³/2550)

Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

Q = 1.57 10⁻² √(2 12 10³ / (850 3)

Q = 4.8 10⁻² m³ / s

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2 years ago

A physics student walks 100 meters in 80 seconds. The student stops for 30 seconds, and then walks 200 meters farther in 90 seco

Elan Coil [88]

Answer:

B i think is the answer

Explanation:

i feel like it is B because if you put them together and the answer is 1.5 so it is B

8 0

2 years ago

A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by

Delicious77 [7]

Answer:

W = -510.98J

Explanation:

Force = 43N, 61° SW

Displacement = 12m, 22° NE

Work done is given as:

W = F*d*cosA

where A = angle between force and displacement.

Angle between force and displacement, A = 61 + 90 + 22 = 172°

W = 43 * 12 * cos172

W = -510.98J

The negative sign shows that the work done is in the opposite direction of the force applied to it.

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2 years ago

A spacecraft of the Trade Federation flies past the planet Coruscant at a speed of 0.610 c. A scientist on Coruscant measures th

mamaluj [8]

Answer:

the length of the now stationary spacecraft = 89.65m

Explanation:

In contraction equation, Length contraction L is the shortening of the measured length of an object moving relative to the observer’s frame.

Thus, it has a formula;

L = L_o(√(1 - (v²/c²))

Where in this question;

L = 71m and v = 0.610 c

Thus;

71 = L_o (√(1 - ((0.61c)²/c²))

c² will cancel out to give;

71 = L_o (√(1 - 0.61²)

71 = 0.792L_o

L_o = 71/0.792

L_o = 89.65m

6 0

2 years ago