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2 years ago

Answer the attached question.No Spam!⚔️⚔️⚔️

Physics

1 answer:

RoseWind [281]2 years ago

6 0

<h3>Question:5</h3>

The given data can be written in following way in coordinates axes.

(0,0), (2,2), (4,4), (6,4), (8,4), (10,6),(12,4), (14, 2).

a) Average velocity for first 4 seconds

Average velocity = Total Displacement/ Time taken

= (4-0)/(4 - 0)

=4/4

= 1/1

b) Average velocity for 4 to 8 seconds

Average velocity = Total Displacement/ Time taken

= (4 - 4)/(8-4)

= 0/4

c) Average velocity for last 6 seconds

Last 6 seconds = from 8 to 14 seconds

Average velocity = Total Displacement/Time taken

= (2 - 4)/(14 - 8)

= -2/6

<h3>Question 6:</h3>

The given data can be written in following way in coordinates axes.

(0,0), (10,20), (20,20), (30,20), (40,0)

a) State the kind of motion from Os to 10s and from 30s to 40s

It is obvious from the graph that the velocity between Os to 10s has been increased from 0 m/s to 20m/s. Hence there is a uniform Acceleration in the body.

b) What is the velocity of the body after 10s and 40s ?

It is clear from the graph and table as well that,

<h3>Velocity after 10s is 20m/s</h3>

and

<h3>Velocity after 40s is 0 m/s</h3>

c) Calculate the distance covered by the body between 10s to 30s.

Distance covered by the body between 10s to 30s will be given by area of rectangle ABCD

Area of rectangle ABCD= (30 - 10) × (20 - O)

=20×20

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3 years ago

Two blocks A and B with mA = 2.2 kg and mB = 0.84 kg are connected by a string of negligible mass. They rest on a frictionless h

madreJ [45]

Answer:

(a) a = 1.875 m/s²

(b)T = 1.575 N

(c)T increase

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the blocks on the horizontal surface and the y-axis in the direction perpendicular to it.

Forces acting on the block A

WA: Weight of of the A block : In vertical direction downaward (-y)

NA : Normal force of the A block :In vertical direction upaward (+y)

F= 5.7 N In in the direction parallel to the movement of the blocks (+x)

T : tension of the string: In in the direction (-x)

Forces acting on the block B

WB: Weight of the B block: In vertical direction downaward (-y)

NB : Normal force of the B block :In vertical direction upaward (+y)

T : tension of the string: In in the direction (+x)

Data

mA = 2.2 kg

mB = 0.84 kg

(a) Magnitude of the acceleration of the blocks

Newton's second law to B block:

∑Fx = m*a

T = (0.84)*a Equation (1)

Newton's second law to A block:

∑Fx = m*a

5.7 - T = (2.2)*a We replace T of the Equation (1)

5.7 - (0.84)*a = (2.2)*a

5.7 = (2.2)*a + (0.84)*a Equation (2)

5.7 =(3.04)*a

a =5.7 / (3.04)

a = 1.875 m/s²

(b)Tension (in N) in the string connecting the two blocks

We replace data in the Equation (1)

T = (0.84)*a

T = (0.84)*(1.875)

T = 1.575 N

(c) How will the tension in the string be affected if mA is decreased?

We observed Equation (2)

5.7 = (2.2)*a + (0.84)*a

5.7 = (mA)*a + (0.84)*a

5.7 =a*( mA+ 0.84)

if mA decrease , then, the acceleration increase and T increase

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Answer the attached question.No Spam!⚔️⚔️⚔️​

Answer the attached question.No Spam!⚔️⚔️⚔️