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exis [7]
1 year ago
10

A simple pendulum consists of a point mass suspended by a weightless, rigid wire in a uniform gravitation field. Which of the fo

llowing statements are true when the system undergoes small oscillations?
Check all that apply.

A. The period is inversely proportional to the suspended mass.
B. The period is proportional to the square root of the length of the wire.
C. The period is independent of the suspended mass.
D. The period is proportional to the suspended mass.
E. The period is independent of the length of the wire.
F. The period is inversely proportional to the length of the wire.
Physics
1 answer:
jarptica [38.1K]1 year ago
4 0

Answer:

- The period is independent of the suspended mass.

- The period is proportional to the square root of the length of the wire.

Explanation:

A simple pendulum consists of a point mass suspended by a weightless, rigid wire in a uniform gravitation field. Which of the following statements are true when the system undergoes small oscillations?

Check all that apply.

A. The period is inversely proportional to the suspended mass.

B. The period is proportional to the square root of the length of the wire.

C. The period is independent of the suspended mass.

D. The period is proportional to the suspended mass.

E. The period is independent of the length of the wire.

F. The period is inversely proportional to the length of the wire.

Simple harmonic motion is periodic motion under the action of a restoring force that is directly proportional to the displacement from equilibrium

from the relation of period T

T=2\pi \sqrt{l/g}

from the above formula , it can be concluded that

C. The period is independent of the suspended mass

B.The period is proportional to the square root of the length of the wire.

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An ice rescue team pulls a stranded hiker off a frozen lake by throwing him a rope and pulling him horizontally across the essen
Anuta_ua [19.1K]

Answer:T=116.84 N

Explanation:

Given

Weight of hiker =1040 N

acceleration a=1.1 m/s^2

Force exerted by Rope is equal to Tension in the rope

F_{net}=T=ma_{net}

T=\frac{1040}{g}\times 1.1

T=116.84 N

8 0
2 years ago
Determine the magnitude and sense (direction) of the current in the 500-latex: \omega ω resistor when i = 30 ma.
VARVARA [1.3K]

Complete Question:

Check the circuit in the file attached to this solution

Answer:

Total current = 0.056 A(From left to right)

Explanation:

Let the current in loop 1 be I₁ and the current in loop 2 be I₂

Applying KVL to loop 1

30 - (I₁ - I₂)500 + I₂R + 15 = 0

45 - 500I₁ - 500I₂ + RI₂ = 0

I₁ = 30mA = 0.03 A

45 - 500(0.03) - 500I₂ + RI₂ = 0

30 -500I₂ + RI₂ = 0...............(1)

Applying kvl to loop 2

-RI₂ - 15 + 10 - 400I₁ = 0

-RI₂ = 5 + 400*0.03

RI₂ = -17 ................(2)

Put equation (2) into (1)

30 -500I₂ -17 = 0

-500I₂ = 13

I₂ = -13/500

I₂ = -0.026 A

The total current in the 500 ohms resistor = I₁ - I₂ = 0.03+0.026

Total current = 0.056 A

The current will flow from left to right

5 0
1 year ago
A diver explores a shallow reef off the coast of Belize. She initially swims d1 = 74.8 m north, makes a turn to the east and con
Nataly [62]

Answer:R=1607556m

θ=180degrees

Explanation:

d1=74.8m

d2=160.7km=160.7km*1000

d2=160700m

d3=80m

d4=198.1m

Using analytical method :

Rx=-(160700+75*cos(41.8))= -160755.9m

Ry= -(74.8+75sin(41.8))-198.1=73m

Magnitude, R:

R=√Rx+Ry

R=√160755.9^2+20^2=160755.916

R=160756m

Direction,θ:

θ=arctan(Rx/Ry)

θ=arctan(-73/160755.9)

θ=-7.9256*10^-6

Note that θ is in the second quadrant, so add 180

θ=180-7.9256*10^6=180degrees

8 0
2 years ago
If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol
xenn [34]

Answer:

Average density of Sun is 1.3927 \frac{g}{cm}.

Given:

Radius of Sun = 7.001 ×10^{5} km = 7.001 ×10^{10} cm

Mass of Sun = 2 × 10^{30} kg = 2 × 10^{33} g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Solution:

Density of Sun is given by,

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Volume of Sun = \frac{4}{3} \pi r^{3}

Volume of Sun = \frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}

Volume of Sun = 1.436 × 10^{33} cm^{3}

Density of Sun = \frac{ 2\times 10^{33} }{1.436 \times 10^{33} }

Density of Sun = 1.3927 \frac{g}{cm}

Thus, Average density of Sun is 1.3927 \frac{g}{cm}.

4 0
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A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
Dvinal [7]
Nobody will do that for 5 points loll
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