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nikdorinn [45]
3 years ago
11

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

= 30~\rm m/s^2. When he releases the ball from chin height without giving it a push, how will the ball's behavior differ from its behavior on Earth? Ignore friction and air resistance. (Select all that apply.)a.It will smash his face.b.It will stop well short of his face.c.It will take less time to return to the point from which it was released.d.Its mass will be greater.e.It will take more time to return to the point from which it was released
Physics
1 answer:
Ne4ueva [31]3 years ago
7 0

Answer:

C

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

                                 t = v / a

                                 t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

                                 t = v / g

                                 t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

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tangare [24]

Answer:

Friction force is equal to the coefficient of friction times the normal force.

Explanation:

It does not really matter if the force is side or not. You need to draw a free body diagram, add weight, normal force, friction force and active force. Then see the horizontal and vertical components to calculate what is the normal force.

5 0
4 years ago
Use what you know about mass and how you use it to calculate force in the following situation. If each washer has a mass of 4. 9
mina [271]

Considering the change of units and the definition of weight, the mass of two washers would be 0.0098 kg and if 4 washers are attached to the string, then the applied force on the car will be 0.196 N.

  • <u><em>Mass of two washers</em></u>

If each washer has a mass of 4.9 g, then the mass of two washers will be calculated simply by multiplying the mass of 1 washer by the amount you have of them, in this case 2:

4.9 g×2= 9.8 g

Being 1 kg = 1000 g, you apply the following rule of three to make the change of units: if 1000 g equals 1 kg, 9.8 g equals how many kilograms?

mass=\frac{9.8 gramsx1 kg}{1000g}

Solving:

<u><em>mass= 0.0098 kg</em></u>

Finally, the mass of two washers would be 0.0098 kg.

  • <u><em>Applied force</em></u>

On the other side, mass is the amount of matter that a body contains and weight is the action exerted by the force of gravity on the body.

The mass of an object will always be the same, no matter where it is located. Instead, the weight of the object will vary according to the force of gravity acting on it.

Weight is calculated as the product of the object's mass and the value of the gravitational acceleration:

W= m×g

In this case, you know:  

  • m= 4.9 g×4= 19.6 g= 0.0196 kg → the mass of 4 washers in kilograms
  • g= 10 m/s²

Replacing:

W= 0.0196 kg× 10 m/s²

Solving:

<u><em>W= 0.196 N</em></u>

Finally, if 4 washers are attached to the string, then the applied force on the car will be 0.196 N.

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7 0
3 years ago
Read 2 more answers
Mass is most closely related to an objects weight? Size? Shape or temperature?
Elenna [48]

Weight is the most closely related object to mass.

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3 years ago
Two loudspeakers placed 6.0 m apart are driven in phase by an audio oscillator, whose frequency range is 1595 to 2158 Hz. A poin
denpristay [2]

Answer:

The frequency produced by the oscillator, for which destructive interference occurs at point P, in the range of frequencies for the oscillator = 2033 Hz

Explanation:

For destructive interference, the difference in path length of the waves from each loudspeaker is related to the wavelength through.

(m + ½)λ = |d₁ - d₂|

where m can take on positive whole number values 0,1,2,3...

Point P is 4.7 m from A and 3.6 m from B

d₁ = 4.7 m

d₂ = 3.6 m

|d₁ - d₂| = 4.7 - 3.6 = 1.1 m

(m + ½)λ = |d₁ - d₂| = 1.1

(m + ½)λ = 1.1

where m could be any whole number from 0,1,2...

And the relationship between velocity of a wave, v, its frequency, f, and the wavelength, λ, is given as

v = fλ

The frequency range of the audio oscillator is frequency range is 1595 to 2158 Hz.

We can find a wavelength for the sound within this range, so as to obtain the exact frequency.

The options include 2001, 2033, 2127, 2095, or 2064 Hz.

Taking just 1 frequency in that range, f = 2033 Hz.

v = fλ

λ = (v/f) = (344/2033) = 0.169 m

Inserting in the destructive interference equation

(m + ½)λ = 1.1

If λ = 0.169 m

(m + ½) = (1.1/0.169) = 6.5

m + ½ = 6.5

Hence, it is evident that m = 6 for this question.

And the corresponding frequency at this level of destructive interference is 2033 Hz

Hope this Helps!!!

8 0
4 years ago
which of the following statements is correct when describing the forces at work on a bridge a. wires are in tension B. large pos
stiv31 [10]

Answer:

All the statements are true

Explanation:

A) Wires are in tension.

It is true because the loads exerted on the bridge are taken by each of the cables of the bridge. these cables transmit the force to the pillars or columns of the bridge. (check the image 01)

B) Large posts are in compression.

It is true each of the loads of the cables is tension, are transmitted to the columns of the bridge, in this way the balance will be maintained (check the image 01)

C) The bridge deck is both in tension and compression.

Yes it is true, the superior part of the deck will be in compression and the inferior part will be in tension, this can be seen in the image 02

8 0
3 years ago
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