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4 years ago

When an object of weight w is suspended from the center of a massless string?

Physics

2 answers:

pickupchik [31]4 years ago

6 0

<span>An object suspended from its own center will not have any motion or rotation unless acted on by an outside force. In the question the idea of a massless string is introduced. The mass of the string is pointless. In this trivial example the weight w did not need to be defined as "w" as there is only one object considered.</span>

igomit [66]4 years ago

4 0

I attached a picture of the diagram associated with this question.

Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ

The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ

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A red laser beam goes from crown glass with refraction index n=1.3 to air (n=1) with an incident angle of 0.23 radians. What is

Klio2033 [76]

Answer:

θ = 10.28º

Explanation:

To find the angle of refraction use the equation of refraction

n₁ sin θ₁ = n₂ sin θ₂

where index 1 is for incident light and index 2 is for refracted light.

sin θ₂ = n₁ / n₂ sin θ

let's calculate

sin = 1 / 1.3 sin 0.23

sin = 0.175

θ= 0.17528 rad

let's reduce to degrees

θ = 0.17528 rad (180ª / pi rad)

θ = 10.28º

6 0

3 years ago

The CIA investigated hypnosis as a possible tool for interrogating prisoners. Why did they decide it was unsuitable for that

alukav5142 [94]

Answer:

D. Hypnosis can make the subjects talk, but they talk only about their childhoods.

Explanation:

7 0

3 years ago

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Which of these rotational quantities is analogous to mass in a linear system?

Rom4ik [11]

Answer:

Explanation:

i think it is rotational inertia

because analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in transnational motion.

hope it's right & helps !!!!!!!!!

5 0

3 years ago

Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω ,

Degger [83]

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41 , RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω

Using law of Ohm

V = I * R

I = Vc / Rc = 364 V / 473 Ω

I = 0.76 A

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -) √ 150.74² 77.0²

RL = 473 (+ / -) (129.58)

RL₁ = 34.41 , RL₂ = 602.58

The higher value have the less angular frequency

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

6 0

4 years ago

All Physicsts over here plz help in these questions!!!!!!!!!

mrs_skeptik [129]

Hello!

First one we can use that PE=mgh so we have

4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m

Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have

g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2

Hope this helps. Any questions please ask. Thank you.

6 0

4 years ago

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