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svet-max [94.6K]
2 years ago
14

A 10 kg mass rests on a table. What acceleration will be generated when a force of 20 N is applied and encounters a frictional f

orce of 15 N?
Physics
1 answer:
11Alexandr11 [23.1K]2 years ago
6 0
The frictional force is oposing the force that we apply to this mass. Therefore the overall force is F_{total}=F_{applied}-F_{friction}=20-15 =5 (Newton)
The second principle of physics tells us that the acceleration of a mass is direct proportional to the applied force, the mass of that object being the proportionality constant.
F=m*a
Hence a= \frac{F}{m}=5/10 =0.5 (m/s^2)
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Ram has power of 550 watt. What does it mean?
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2 years ago
a child hits a ball with a force of 350 N. (a) If the ball and bat are in contact for 0.12 is, what impulse does the ball receiv
Lina20 [59]

Explanation:

Given that,

Force with which a child hits a ball is 350 N

Time of contact is 0.12 s

We need to find the impulse received by the ball. The impulse delivered is given by :

J=F\times t\\\\J=350\times 0.12\\\\J=42\ N-m

So, the impulse is 42 N-m..

We know that he change in momentum is also equal to the impulse delivered.

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7 0
2 years ago
A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta
BartSMP [9]

Answer:

35%

Explanation:

Assuming no external torques present during the collision between the record and the turntable, total angular momentum must be conserved.

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L = I* ω

So, as initial angular momentum and final angular momentum must be the same, we have:

Li = Lf

⇒ I₁ * ω₁ = I₂ * ω₂ (1)

where I₁ is the rotational inertia of the turntable, and I₂, is the combined rotational inertia of the turntable and the record:

I₂ = I₁ + 0.54 I₁ = 1.54 I₁

We can solve (1) for the new common angular speed, as follows:

ω₂ = ω₁ / 1.54 (2)

The initial rotational kinetic energy is given by definition for the following equation:

Kroti = 1/2 * I₁ * ω₁² (3)

The final rotational kinetic  energy takes into account the new rotational inertia and the common final angular speed:

Krotf = 1/2* I₂ * ω₂² = 1/2* 1.54 I₁* (ω₁/1.54)² (4)

Dividing both sides in (3) and (4), we get:

Krotf/Kroti = 1/1.54 = 0.65

This means that the final rotational kinetic energy, has reduced to 0.65 of the initial value, or that has lost 35% of the initial kinetic energy.

8 0
2 years ago
A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one
forsale [732]

Answer:

C. Both reach the bottom at the same time.

Explanation:

For a rolling object down an inclined plane , the acceleration is given below

a = g sinθ / (1 + k² / r² )

θ is angle of inclination , k is radius of gyration , r is radius of the cylinder

For cylindrical object

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acceleration =  g sinθ  /( 1 + 1/2 )

= 2 g sinθ / 3  

Since it does not depend upon either mass or radius , acceleration of both the cylinder will be equal . Hence they will reach the bottom simultaneously.

6 0
2 years ago
Technician a says that using a pressure transducer and lab scope is a similar process to using a vacuum gauge. technician b says
Phantasy [73]

Answer: Both Technician A and B

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And also, the pressure transducer can be used to tie any issues to individual cylinders if paired with a second trace consisting of the ignition pattern. Therefore, both Technician A and B are correct.

7 0
2 years ago
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