Please help! the first person to answer this correctly will get a brainlist

A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula

Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

$V_{ab}$ = i R

$V_{ab}$ = (0.1832) (65)

$V_{ab}$ = 11.91 Volts

(c)

Power dissipated in the resister R is given as

$P_{R}$ = i²R

$P_{R}$ = (0.1832)²(65)

$P_{R}$ = 2.18 Watt

Power dissipated in the internal resistance is given as

$P_{r}$ = i²r

$P_{r}$ = (0.1832)²(0.5)

$P_{r}$ = 0.0168 Watt

5 0

3 years ago

A pure musical torte causes a thin wooden

victus00 [196]

Hello! Your answer would be D. Interference

This is because interference causes vibration!!!

4 0

3 years ago

I'm not really sure how to go about creating the equation, can anyone help me?

AlexFokin [52]

The displacement vector (SI units) is
$\vec{r} =At\hat{i}+A[t^{3}-6t^{2}]\hat{j}$

The speed is a scalar quantity. Its magnitude is
$v= \sqrt{A^{2}t^{2}+A^{2}(t^{3}-6t^{2})^{2}} \\ v=A \sqrt{t^{2}+t^{6}-12t^{5}+36t^{4}} \\ v=At \sqrt{t^{4}-12t^{3}+36t^{2}+1}$

Answer: At√(t⁴ - 12t³ + 36t² + 1)

3 0

3 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago

A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it

Darina [25.2K]

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given: </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.

<u>Solution </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next

Ef=Ei+W (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form

(Kf + mc^2) = (KJ+ mc^2) (2)

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by

W = FΔr (4)

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut

W = F Δr= (190N)(6 m) = 1140 J

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W' (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate

1/2mvf'^2=1/2mvi'^2+W'

vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

7 0

3 years ago