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shutvik [7]
2 years ago
11

A basketball has a mass of 1 kg and is traveling 15 m/s. How fast would a 5 kg bowling ball have to travel to have the same mome

ntum?
A) 3 m/s

B) 15 m/s

C) 1 m/s

D) 5 m/s
Physics
2 answers:
DiKsa [7]2 years ago
8 0
A) 3 m/s. Hope this helps.
Andrei [34K]2 years ago
5 0

Answer:

A 3m/s

Explanation:

Took the test (k12)

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Given 4.8 moles of a gas at 37 degrees Celsius and at 792 torr, what is the volume of the gas? (The ideal gas constant is 0.0821
Anastaziya [24]

Answer:

1.17 x 10^2 L

Explanation:

We can find the volume of the gas by using the ideal gas law:

pV=nRT

where we have:

p=792 Torr \cdot \frac{1 atm}{760 Torr} = 1.04 atm is the pressure

V is the volume

n = 4.8 mol is the number of moles

R = 0.0821 L · atm/mol · K is the ideal gas constant

T=37^{\circ}+273 =310 K is the temperature

Solving the equation for V, we find the volume

V=\frac{nRT}{p}=\frac{(4.8 mol)(0.0821 L atm/mol K)(310 K)}{1.04 atm}=117.5 L = 1.17\cdot 10^2 L

4 0
3 years ago
At what point of thermometric scale does Kelvin scale reading coincide with Fahrenheit scale
amid [387]

Answer:

<em>At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.</em>

Explanation:

We first need to find a relation between the Kelvin scale and the Fahranheit scale. We'll use the Celsius scale to relate them.

The Kelvin and Celsius scales are related by the formula:

K = °C + 273.15

Solving for °C:

°C = K - 273.15

Besides, the Kelvin and Celsius scales are related by:

°C = 5 ⁄ 9(°F - 32)

Now we find a temperature, say X, where both scales coincide. Equating both formulas:

X - 273.15=5 ⁄ 9(X - 32)

Multiply by 9:

9X - 2,458.35 = 5X - 160

Simplifying:

4X = 2,458.35 - 160=2,298.35

Solving:

X =2,298.35 / 4 = 574.59

At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.

3 0
3 years ago
A student uses 60 Newtons of force to climb 18 meters in 6 seconds. Calculate her Power.
puteri [66]

Answer:

15

Explanation:

P=W/T

T=6sec

W=?

F=60N

S=18m

W=F X S. .s indicate displacement

W=60x18

W=108

So p=108 j/6sec

P=15watt

5 0
2 years ago
2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her he
Ratling [72]

Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, v_s = d/t₂

∴ v_s = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, v_s ≈ 5.79 m/s.

5 0
2 years ago
A block of mass 5 kg is descending with a constant velocity on an inclined plane at 30°. What is the value of the frictional for
sattari [20]

Answer:28.9N

Explanation:

F=UR R=mg R=5×10 R=50N

U=tan© U=Tan30 U=1/√(3)

F=UR F=1/√(3) ×50

F=28.9N

4 0
2 years ago
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