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Westkost [7]
2 years ago
5

Plz, help the rubber ball be dropped from the top of a ladder. It bounces on the same spot on the ground several times to a less

er height than the previous bounce before coming to rest in place.
In one or two sentences, explain how the experiment shows the direction that gravity pulls an object, specifically showing that it does not pull an object parallel to the ground.
Physics
1 answer:
Vladimir [108]2 years ago
6 0

Answer:

What do you mean?

Explanation:

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How to do it? Urgent ​
mixer [17]

a)

F= ma

a=v/t

F=5*(35/5)

F=35N

b)

a=F/m

a=(35-2)/5

a=33/5

a=6.6N

8 0
3 years ago
If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is
expeople1 [14]

Answer:

We can use  2 g H = v2^2 - v1^2    or

v2^2 = 2 g H + v1^2

Since 88 ft/sec = 60mph   we have 30 mph = 44 ft/sec

The object will return with the same speed that it had initially so the object

starts out with a downward speed of 44 ft/sec

Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2

v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2

v2 = 110 ft/sec

8 0
2 years ago
For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with d
Sedaia [141]

Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material (\sigma) = 200 M Pa

Diameter of the bar = 4.5 cm = 45 mm

We know that yield stress of the bar is given by the formula

                Yield Stress = \frac{Maximum load}{Area of the bar}

⇒                                \sigma = \frac{P_{max} }{A}  ---------------- (1)

⇒ Area of the bar (A) = \frac{\pi}{4} ×D^{2}

⇒                            A  = \frac{\pi}{4} × 45^{2}

⇒                            A = 1589.625 mm^{2}

Put all the values in equation (1) we get

⇒ P_{max} = 200 × 1589.625

⇒ P_{max} = 317925 N

In this bar the P_{max} is equal to the weight of the bar.

⇒ P_{max} = M_{max} × g

Where M_{max} is the maximum mass the bar can support.

⇒ M_{max} = \frac{P_{max} }{g}

Put all the values in the above formula we get

⇒ M_{max} = \frac{317925}{9.81}

⇒ M_{max} = 32408.26 Kg

There fore the maximum mass the bar can support without yielding = 32408.26 kg

3 0
3 years ago
I NEED THIS A SOON AS POSSIBLE
vivado [14]

Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.

6 0
2 years ago
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The name of C (S) + o2 (g) CO2 (g)
USPshnik [31]

Answer:

carbon + oxygen → carbon dioxide

6 0
2 years ago
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