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andrew-mc [135]
3 years ago
13

A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.5 A when an additional 1.6 Ω resistor is

added in series with R1. What is the value of R1? Assume the internal resistance of the source of emf is zero. Answer in units of Ω.
Physics
2 answers:
myrzilka [38]3 years ago
7 0

Answer:

4.8 Ω

Explanation:

From Ohm's Law,

Using,

I = E/(R+r)................. Equation 1

E = I(R+r)................. Equation 2

Where I = current, E = emf, R = external resistance, r = internal resistance

Given: I = 2 A, R = R1, r = 0 Ω

Substitute into equation 2

E = 2(R1)

E = 2R1.

When an additional 1.6 Ω  resistor is added in series,

E = 1.5(R1+1.6)

2R1 = 1.5R1+2.4

2R1-1.5R1 = 2.4

0.5R1 = 2.4

R1 = 2.4/0.5

R1 = 4.8 Ω

sasho [114]3 years ago
3 0

Answer:

R1 = 4.8Ω

Explanation:

The loop circuit has an initial voltage of  V = IR

I = 2 A , R1 = R

V = 2R1

with the current reduced to 1.5A with an additional 1.6Ω resistor

the total resistance of the circuit is 1.6 + R1

the voltage of the two scenarios has to be equal , since the same voltage flows through the circuit

therefore V = 2R1

from Ohms law V = IR

2R1= 1.5 (1.6 + R1)

2R1 = 2.4 + 1.5R1

collecting like terms

2R1 - 1.5R1 = 2.4

0.5R1 = 2.4

R1 = \frac{2.4}{0.5}

R1 = 4.8Ω

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3 years ago
You have an incident ray from a medium with a n1=1, through a medium with a n2 =3.325. If the incident angle is equal to 0.478 r
sesenic [268]

Answer:

0.139 rad

Explanation:

We use Snell's law n_1sin\theta_1=n_2sin\theta_2, where if n_1 is the <em>refractive index</em> of the medium containing the <em>incident ray</em>, \theta_1 would be the <em>incident angle</em>, and if n_2 is the <em>refractive index</em> of the medium containing the <em>refracted ray</em>, \theta_2 would be the <em>refraction angle</em>, which we want, so we do:

sin\theta_2=\frac{n_1}{n_2}sin\theta_1

And finally:

\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)

We then insert our values:

\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)=Arcsin(\frac{1}{3.325}sin(0.478rad))=arcsin(0.13834714686&#10;)=0.139 rad

6 0
3 years ago
Torque can cause the angular momentum vector to rotate in UCM. This motion is called ___________.
emmainna [20.7K]

Torque can cause the angular momentum vector to rotate in UCM. This motion is called _Conservation of Angular momentum__________.

Answer:

Conservation of Angular momentum

Explanation:

The motion of an object in a circular path at constant speed is known as uniform circular motion (UCM). An object in UCM is constantly changing direction, and since velocity is a vector and has direction, you could say that an object undergoing UCM has a constantly changing velocity, even if its speed remains constant.

The law of conservation of angular momentum states that when no external torque acts on an object, no change of angular momentum will occur.

Key Points

When an object is spinning in a closed system and no external torques are applied to it, it will have no change in angular momentum.

The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation.

If the net torque is zero, then angular momentum is constant or conserved.

Angular Momentum

The conserved quantity we are investigating is called angular momentum. The symbol for angular momentum is the letter L. Just as linear momentum is conserved when there is no net external forces, angular momentum is constant or conserved when the net torque is zero. We can see this by considering Newton’s 2nd law for rotational motion:

τ→=dL→dt, where  

τ is the torque. For the situation in which the net torque is zero,  

dL→dt=0.

If the change in angular momentum ΔL is zero, then the angular momentum is constant; therefore,

⇒

L  =constant

L=constant (when net τ=0).

This is an expression for the law of conservation of angular momentum.

Example and Implications

An example of conservation of angular momentum is seen in an ice skater executing a spin,  The net torque on her is very close to zero,

because (1) there is relatively little friction between her skates and the ice, and (2) the friction is exerted very close to the pivot point.

Conservation of angular momentum is one of the key conservation laws in physics, along with the conservation laws for energy and (linear) momentum. These laws are applicable even in microscopic domains where quantum mechanics governs; they exist due to inherent symmetries present in nature.

7 0
3 years ago
Which part of the electromagnetic spectrum is nearest to radio waves?
Sav [38]

Answer:

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Explanation:

8 0
3 years ago
Read 2 more answers
A rubber-wheeled 50 kg cart rolls down a 35 degree concrete incline. Coefficient of rolling friction between rubber and concrete
lapo4ka [179]

Answer:

(a) 5.62 m/s²

(b) 5.46 m/s²

Explanation:

The given values are:

mass,

m = 50 kg

angle

= 35°

(a)

If friction neglected,

⇒ F_x=mgSin \theta=ma

⇒ a_x=gSin \theta

       =9.8 \ Sin35^{\circ}

       =5.62 \ m/s^2

(b)

If friction present,

⇒ F_x=mgCos \theta

⇒ F_x=mgSin \theta-\mu_r mgCos \theta

⇒ a=gSin \theta-\mu_rgCos \theta

      =9.8 \ Sin35^{\circ}-0.02\times 9.8 \ Cos30^{\circ}

      =5.46 \ m/s^2

8 0
3 years ago
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