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beks73 [17]
3 years ago
12

Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At w

hich points in time does a zero net force act on the ball? Ignore air resistance.
Physics
1 answer:
valentina_108 [34]3 years ago
6 0

Answer:before throwing and after catching the ball

Explanation:

When basketball is in the hand of player net force on it zero as holding force is canceled by gravity Force. During its entire motion gravitational force is acting on the ball which is acting downward. Even at highest point gravity is constantly acting downwards.

After catching the ball net force on it zero as holding force is canceled by gravity force and ball is continue to be in stationary motion.          

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Don't give links or anything besides an answer to this problem
alexdok [17]

Answer:

I think its the last one

Explanation:

The particles always move perpendicular to the direction of the wave.

7 0
3 years ago
Place the balloon in a bell jar. If available also add some shaving cream and fresh marshmallows. Ask the instructor for help if
irina [24]

Answer:

The balloon will collapse

Explanation:

When air is removed from the bell jar, the balloon will collapse if the internal pressure from the balloon does not balance the atmospheric pressure from the surroundings.

5 0
3 years ago
Fill in the appropriate values for each blank as it refers to ATOM 1. The number of protons present in ATOM 1 is _________.​
wlad13 [49]
3, protons are positive and there are 3 positive atoms visible
6 0
4 years ago
9. Una jeringa contiene cloro gaseoso, que ocupa un volumen de 95 mL a una presión de 0,96 atm. ¿Qué presión debemos ejercer en
masha68 [24]

Answer:

2.61 atm

Ley de Boyle

Explanation:

P_1 = Presión inicial = 0.96 atm

P_2 = Presión final

V_1 = Volumen inicial = 95 mL

V_2 = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}

La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.

4 0
3 years ago
A ball was rolling downhill at 2 m/s. After 5s, it was rolling at 90 m/s. What is its acceleration?
olga55 [171]

Answer:

17.6 m/s²

Explanation:

Given:

v_{f} = 90 m/s (final velocity)

v_{i} = 2 m/s (initial velocity)

Δt = 5s (change in time)

The formula for acceleration is:

a_{avg} = Δv / Δt

We can find Δv by doing

Δv = v_{f} - v_{i}

Replace the values

Δv = 90m/s - 2m/s

Δv= 88m/s

Using the equation from earlier, we can find the acceleration by dividing the average velocity by time.

a_{avg} = Δv / Δt

a_{avg} = \frac{88m/s}{5/s}

acceleration = 17.6 m/s^{2}

4 0
3 years ago
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