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valentinak56 [21]
4 years ago
12

If a ball is accelerating down through the air with no horizontal motion, what must be true about the net forces acting on the b

all?(1 point) The net force on the ball is directed upward. The net force on the ball is zero. The gravitational force is greater than the drag. The drag is greater than the gravitational force.
Physics
1 answer:
disa [49]4 years ago
8 0

Answer:

The net force on the ball is zero.

Explanation:

Newton's first law of motion states that an object will remain in the same state of motion or at rest unless a resultant force acts on it, and if the net force on an object is zero, then that means that the object will keep moving at the same acceleration or the object stays stationary.

So, the truth about the net forces acting on the ball is that the net force on the ball is zero.

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Question number 6 :<br><br> Is my answer right?? Plzz help
Dennis_Churaev [7]
No.  You have the units wrong.

Example with numbers: 

When you multiply  (1/9) by 2, you get 2/9 .

With units:

When you multiply  (m/s) by (kg), you get (kg-m/s).
3 0
4 years ago
#4<br> HELP PLEASE!<br> Find Gravitational Potential Energy! <br> THANK YOU!
Scilla [17]

Answer:

\Delta GPE=-3.92\ 10^9\ J

Explanation:

<u>Gravitational Potential Energy</u>

It's the capacity of an object to do work due to its relative height from a fixed reference point.

It's computed as

GPE=m.g.h

Where m is the mass of the object, h is its height and g is the acceleration of gravity, g=9.8 m/sec^2

The mass of water is given as

m=8,000,000\ kg

The height above the rocks is h=50 m. Let's compute the GPE

GPE_1=(8,000,000\ kg)(9.8\ m/s^2)(50\ m)

GPE_1=3,920,000,000\ Joule

It should be expressed in scientific notation

GPE_1=3.92\ 10^9\ J

The GPE at the bottom, where h=0

GPE_2=0

The change of gravitational potential energy is:

\Delta GPE=GPE_2-GPE_1

\boxed{\Delta GPE=-3.92\ 10^9\ J}

8 0
3 years ago
If a cheetah can maintain a constant speed of 25 m/s, it will cover 25 meters every second. At this rate, how far will it travel
Ksju [112]
10 seconds 25m*10s= 250m
60 seconds 25m*60s=1500m
5 0
4 years ago
a uniform resistance wire is stretched till its length becomes 4 times what happens to the resistance​
Rzqust [24]
The answer is in the box


Step by step

Yegehgehd
3 0
3 years ago
Find the reaction supports at Ta and TB as shown in the loaded beam.
koban [17]

ANSWER

\begin{gathered} T_A=17.5N \\ T_B=12.5N \end{gathered}

EXPLANATION

First, we have to make a sketch of the direction of the moments of the forces about 12m from the left in the diagram:

The sum of upward forces must be equal to the sum of downward forces. This implies that:

\begin{gathered} T_A+T_B=20+10 \\ T_A+T_B=30N \end{gathered}

Also, the sum of clockwise moments must be equal to the counter-clockwise moments:

\begin{gathered} (20\cdot8)+(T_B\cdot4)=(T_A\cdot12) \\ 160+4T_B=12T_A \\ \Rightarrow12T_A-4T_B=160 \end{gathered}

From the first equation, make TA the subject of the formula:

T_A=30-T_B

Substitute that into the second equation:

\begin{gathered} 12(30-T_B)-4T_B=160 \\ 360-12T_B-4T_B=160 \\ -16T_B=160-360=-200 \\ T_B=\frac{-200}{-16} \\ T_B=12.5N \end{gathered}

Substitute that into the equation for TA:

\begin{gathered} T_A=30-12.5 \\ T_A=17.5N \end{gathered}

Therefore, the reaction supports at TA and TB are:

\begin{gathered} T_A=17.5N \\ T_B=12.5N \end{gathered}

5 0
2 years ago
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