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Andreyy89
3 years ago
14

A boy on rollerskates is travelling along at 8 m/s. He has a mass of 60 kg and is carrying his

Physics
1 answer:
Brrunno [24]3 years ago
5 0

Answer:

6m/s

Explanation:

the original momentum = mass x velocity = 8x (60+10) = 560

momentum after = mass x velocity of the school bag + mass x velocity of the boy = 10x20 + 60x A

200+60A = 560

A=6

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A 4kg box accelerated from rest by a force across the floor at a rate of 2m/s^2 for 7 seconds. What is the net work done on the
AlladinOne [14]

Answer and work is shown in the image attached.

please mark me brainliest :)

7 0
3 years ago
Read 2 more answers
Part II # 1 A mass on a string of unknown length oscillates as a pendulum with a period of 4 sec. What is the period if: (Parts
Mrac [35]

Answer:

C. 1) No Change (4 sec) 2) 5.7 sec 3) 2.8 sec 4) No Change (4 sec)

Explanation:

Given that:

Period (T) = 4 s

1) If the mass is doubled.

The period of a pendulum is given by the formula:

T=2\pi\sqrt{\frac{L}{g} } where L is the length and g is the acceleration due to gravity.

From the formula, the period does not depend on the mass of the spring therefore if the mass is doubled the period does not change.

2) The string length is doubled

Given that:

T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }

if the length is doubled, the new spring length is 2L. Therefore the new period (T1) is given as:

T_1=2\pi\sqrt{\frac{2L}{g} }=\sqrt{2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{2}*4=5.7\ sec

3) The string length is halved

Given that:

T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }

if the length is halved, the new spring length is L/2. Therefore the new period (T1) is given as:

T_1=2\pi\sqrt{\frac{L}{2g} }=\sqrt{1/2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{1/2}*4=2.8\ sec

4) The amplitude is halved

From the formula, the period does not depend on the amplitude therefore if the amplitude is halved the period does not change.

6 0
3 years ago
Mr. Llama walked from his house to the bus stop. The bus stop is 2 miles from his house. He returned back to his house from the
Hoochie [10]

Answer:

Displacement of Mr. Llama: Option D. 0 miles.

Explanation:

The magnitude of the displacement of an object is equal to the distance between its final position and its initial position. In other words, as long as the initial and final positions of the object stay unchanged, the path that this object took will not affect its displacement.

For Mr. Llama:

  • Final position: Mr. Llama's house;
  • Initial position: Mr. Llama's house.

The distance between the final and initial position of Mr. Llama is equal to zero. As a result, the magnitude of Mr. Llama's displacement in the entire process will also be equal to zero.

7 0
3 years ago
Five 6 ohm resistors are connected in parallel. Which of the following is their effective resistance?
Ganezh [65]

1/Rt = 1/R1 + 1/R2 + 1/R3+ 1/R4+ 1/R5

1/Rt = 1/6+ 1/6+1/6+1/6+1/6

1/Rt = 5/6

Rt = 6/5

Rt = 1.2 ohm

so B is the answer

3 0
3 years ago
how much current is in a circuit that includes a 9-volt battery and a bulb with a resistance of 3 ohms?
ExtremeBDS [4]
3 Amper 9 divided by 3
8 0
3 years ago
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