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natima [27]
3 years ago
6

Kaushik had three pendulum bobs made of steel of the same length but different masses of 10 g, 50 g and 100 g respectively. When

he performed an experiment with them to find the time period, which of these statements would he consider to be correct? a. The time period is maximum with 10 g bob. b. The time period is maximum with 50 g bob. c. Time period remains the same in all. d. None of these is true.
Physics
1 answer:
Alexandra [31]3 years ago
8 0

Answer:

c. Time period remains the same in all.

Explanation:

In order to answer this question, we need to analyze the parameters, upon which the time period of a pendulum depends. We know that the time of a pendulum is given by the following formula:

T = 2π√(L/g)

where,

T = Time period

L = Length of pendulum

g = acceleration due to gravity

The formula clearly shows that the time period of the pendulum depends only upon the length of pendulum and value of g. And the time period of a pendulum does not depend upon the mass of the bob. Hence, the time period for each of the three pendulums will remain same. So, the correct option will be:

<u>c. Time period remains the same in all.</u>

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If an object accelerates from rest, with a constant of 8 m/s2, what will its velocity be after 35s?
AlladinOne [14]

Answer:

<em>Its speed will be 280 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the speed of an object changes by an equal amount in every equal period of time.

If a is the constant acceleration, vo the initial speed, vf the final speed, and t the time, vf can be calculated as:

v_f=v_o+at

The object accelerates from rest (vo=0) at a constant acceleration of a=8\ m/s^2. The final speed at t=35 seconds is:

v_f=0+8*35

v_f=280\ m/s

Its speed will be 280 m/s

5 0
3 years ago
A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block
MariettaO [177]

Answer:

v₀ = 0.5058 m/s

Explanation:

From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m

Now, the potential energy of the block at x = 0.08 m is ½kx²

where;

k is the spring constant given by; k = ω²m

ω is the angular velocity of the oscillation

m is the mass of the block.

Thus, potential energy of the spring at the bottle(x = 0.08 m) is;

U = ½ω²m(0.08m)²

Also, potential energy of the spring at the bottle(x = 0.05 m) is;

U = ½ω²m(0.05m)²

and the kinetic energy of the block at x = 0.05 m is;

K = ½mv₀²

Thus;

½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²

Inspecting this, ½m will cancel out to give;

ω²(0.08)² = ω²(0.05)² + v₀²

Making v₀ the subject, we have;

v₀ = ω√((0.08)² - (0.05)²)

So,

v₀ = 8.1√((0.08)² - (0.05)²)

v₀ = 0.5058 m/s

7 0
3 years ago
A 12 volt battery in a motor vehicle is capable of supplying the starter motor with 150 A. It is noticed that the terminal volta
qwelly [4]

Answer:0.0133 \Omega

Explanation:

Given

Voltage=12 V

Current(I)=150 A

V_{terminal}=10 V

r_{internal}=\frac{\Delta V}{I}

r_{internal}=\frac{12-10}{150}

r_{internal}=\frac{2}{150}=0.0133 \Omega

4 0
3 years ago
A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho
creativ13 [48]

Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano (m) = 190 kg

Inclined angle (\theta) = 18 degree

Considering gravity, g = 9.8 ms^-^2

And

Using, sin(18) =0.30 and cos(18)=0.95

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

a.

When the man pushes it parallel to the incline.

Balancing the forces as  \sum F=0 .

⇒ F+mgsin(\theta) =0

⇒ F=-mgsin(\theta)

⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass (m) and angle (\theta) .

⇒ F=190\times 9.8\times sin(18)

⇒ F=575.38 N

b.

When the force is parallel to the floor.

⇒ Fcos(\theta)=mgsin(\theta)

⇒ F=\frac{mgsin(\theta)}{cos(\theta)}

⇒ Plugging the values.

⇒ F=\frac{190\times 9.8\times sin(18)}{cos(18)}

⇒ F=605 N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

6 0
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