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IgorC [24]
3 years ago
7

Charlie throws a ball up into the air. While the ball is airborne, which is the greatest force acting on the ball to slow it dow

n? applied force friction force gravitational force normal force
Physics
2 answers:
Naya [18.7K]3 years ago
5 0

Answer:

Gravitational force

Explanation:

When ball is thrown upwards then there will be two forces on the ball opposite to the motion of ball

1. Gravitational force (mg)

2. Air friction or air drag

Both forces are opposite to the motion of the ball and it will cause the decrease in the speed of the ball.

Now here due to less value of drag coefficient on the ball we will say that ball will decrease its speed more due to gravitational force compare to friction force of air.

so here correct answer would be

Gravitational force

oksian1 [2.3K]3 years ago
3 0
I believe it is the gravitational force for gravity controls the speed of the object hurdling towards the ground.
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Alona [7]

Answer:

Power=720[watt]

Explanation:

We need to remember the definition of mechanical work which is equal to the product of the force applied by the distance traveled.

In this problem, we have to find the power which is defined as the work divided into the time in which such work is performed. This way if we have the displacement and the time, this will be the speed with which this work is done.

Power= W/T\\T=time [s]\\W=work [J]\\Power = F*V\\where\\V=velocity [m/s]\\Power=1800 * 0.4 = 720 [watt]

4 0
3 years ago
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Answer:

2\:\mathrm{m/s^2}

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3 years ago
A piano has a mass of 185 kg, and the coefficient of friction between it and the floor is 0.39. What is the maximum force of fri
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What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?
liq [111]

Answer:

46.9 C

Explanation:

The heat released by the gold bar is equal to the heat absorbed by the water:

m_g C_g (T_g-T_f)=m_w C_w (T_f-T_w)

where:

m_g = 3.0 kg is the mass of the gold bar

C_g=129 J/kg C is the specific heat of gold

T_g=99 C is the initial temperature of the gold bar

m_w = 0.22 kg is the mass of the water

C_w=4186 J/kg C is the specific heat of water

T_w=25 C is the initial temperature of the water

T_f is the final temperature of both gold and water at equilibrium

We can re-arrange the formula and solve for T_f, so we find:

m_g C_g T_g -m_g C_g T_f = m_w C_w T_f - m_w C_w T_w\\m_g C_g T_g +m_w C_w T_w= m_w C_w T_f +m_g C_g T_f \\T_f=\frac{m_g C_g T_g +m_w C_w T_w}{m_w C_w + m_g C_g}=\\=\frac{(3.0)(129)(99)+(0.22)(4186)(25)}{(0.22)(4186)+(3.0)(129)}=\frac{38313+23023}{921+387}=\frac{61336}{1308}=46.9 C

5 0
3 years ago
What would happen if a group of force of 45 N compete against a group of 50 N
larisa [96]
The 50N group of force would be greater
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