Ask question

Ask question

All categories

3 years ago

9

A loop circuit has a resistance of R1 and a current of 2.2 A. The current is reduced to 1.6 A when an additional 2.8 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

1 answer:

kvasek [131]3 years ago

7 0

Answer:

Value of $R_1$ is 7.466 ohm

Explanation:

Let emf of the battery is e

In first case when only $R_1$ is there current through battery is 2.2 A

So $2.2=\frac{e}{R_1}$

$e=2.2R_1$ ......eqn 1

In second case current is reduced to 1.6 A when additional resistance of 2.8 ohm is added in series

So $1.6=\frac{e}{R_1+2.8}$

$e=1.6R_1+4.48$ ....eqn 2

Fro eqn 1 and eqn 2

$2.2R_1=1.6R_1+4.48$

$0.6R_1=4.48$

$R_1=7.466ohm$

So value of $R_1$ is 7.466 ohm

You might be interested in

A 50.0 kg object is moving at 18.2 m/s when a 200 N force is applied opposite the direction of the objects motion, causing it to

Gekata [30.6K]

Answer:

t = 1.4[s]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

$P=m*v\\or\\P=F*t$

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 50 [kg]

v = velocity [m/s]

F = force = 200[N]

t = time = [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

$(m_{1}*v_{1})-F*t=(m_{1}*v_{2})$

where:

m₁ = mass of the object = 50 [kg]

v₁ = velocity of the object before the impulse = 18.2 [m/s]

v₂ = velocity of the object after the impulse = 12.6 [m/s]

$(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]$

3 0

3 years ago

lbvjy [14]

Answer:

Impulse = 322.5[kg*m/s], the answer is D

Explanation:

This method it is based on the principle of momentum and the amount of movement; and used to solve problems involving strength, mass, speed and time.

If units of the SI are used, the magnitude of the impulse of a force is expressed in N * s. however, when remembering the definition of the newton.

$N*S=(kg*m/s^{2} )*s = kg*m/s$

Now replacing the values on the following equation that express the definition of impulse

$Impulse = Force * Time\\\\Impulse = 215 * 1.5 = 322.5 [kg*m/s]$

4 0

3 years ago

Read 2 more answers

A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed

Tresset [83]

For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
-24.96 = (-0.65) M2
M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg

4 0

3 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

3 years ago

Climate models are computer simulations of all the complex interactions of the atmosphere, oceans, land surface, ice and biosphe

Nady [450]

Answer: False

Sorry if this is wrong I’m not positive this answer is right but you can still try it. :)

8 0

3 years ago