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alukav5142 [94]
3 years ago
10

What is a substance?

Physics
1 answer:
kogti [31]3 years ago
7 0

Answer:

a particular kind of matter with uniform properties.

Explanation:

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Which statement correctly describes the gravitational potential energy of the pendulum based on this diagram?
MrRa [10]
I think it's 'C' but I won't know for sure until you let me see the diagram.
8 0
4 years ago
Net force is the sum of all the forces acting on an object. If a spring balance pulls on a body with a force of 10 N, and fricti
julsineya [31]

Answer:

5N

Explanation:

(25 N - 20 N = 5 N)

3 0
4 years ago
A sailboat weighing 980 lb with its occupants is running downwind at 8 mi/h when its spinnaker is raised to increase its speed.
Effectus [21]

Answer:

78.498N

Explanation:

The Net force provided by the spinnaker can be obtained from Newton's second law of motion as follows;

F=\frac{m(v-u)}{t}................(1)

where m is the mass, v is the final velocity, u is the initial velocity and t is the time interval for which the force acted.

Given;

m =980lb

v = 12mi/h

u =8mi/hr

t = 10s.

It is important to convert all quantities to their SI units where necessary, so we do that as follows;

1lb = 0.45kg,

hence 980lb = 980 x 0.45kg = 441kg.

1mile = 1609.34m

1hour = 3600s,

therefore;

8mi/h=\frac{8*1609.34m}{3600s}=3.58m/s

12mi/h=\frac{12*1609.34m}{3600s}=5.36m/s

Substituting all values into equation (1), we obtain the following;

F=\frac{441(5.36-3.58)}{10}\\F=\frac{441*1.78}{10}\\F=\frac{784.98}{10}\\F=78.498N

4 0
3 years ago
A batter hits two baseballs with the same force. One hits the ground near third base. The other is a home run out of the park. W
Basile [38]
I think the answer is "<span>The ball that went out of the park shows more work because the distance was greater."</span>
3 0
3 years ago
A ball droped from a building. How fast is it traveling after falling 3.55s
algol [13]

Answer:

d = 61.75 m

Explanation:

Given that,

A ball droped from a building.

We need to find how fast is it traveling after falling 3.55 s.

As it is dropped, its initial velocity is equal to 0.

Let d is the distance it covers after falling 3.55 s.

We can use second equation of motion to find d.

d=ut+\dfrac{1}{2}at^2

Here, u = 0 and a =g

d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times (3.55)^2\\\\d=61.75\ m

So, it will cover 61.75 m after falling 3.55 seconds.

4 0
3 years ago
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