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3 years ago

A charge of 80 coulombs passes through a circuit in 5 seconds. What is the current through the circuit?

Physics

1 answer:

UkoKoshka [18]3 years ago

4 0

Answer:

I = 16amp

Explanation:

Charge coulomb ( Q ) = It

Where I =current in ampere

t = time = 5 seconds

80 = I × 5

I = 80/5

I = 16amp

The current through the circuit will be I = 16amp

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A man fires a silver bullet of mass 2g with a velocity of200m/sec into wall. What is the temperature changeof the bullet? Note:

Sphinxa [80]

F=nmv

where;

n=no. of bullets = 1

m=mass of bullets=2g *10^-3

V=velocity of bullets200m/sec

F=1

loss in Kinetic energy=gain in heat energy

1/2MV^2=MS∆t

let M council M

=1/2V^2=S∆t

M=2g

K.E=MV^2/2

=(2*10^-3)(200)^2/2

2 councils 2

2*10^-3*4*10/2

K.E=40Js

H=mv∆t

(40/4.2)

40Js=40/4.2=mc∆t

40/4.2=2*0.03*∆t

=158.73°C

7 0

3 years ago

Please help Need good grade

ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the range

Using formula of range

$R=\dfrac{v^2\sin(2\theta)}{g}$ ...(I)

The range for the other angle is

$R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$ ....(II)

Here, distance and speed are same

On comparing both range

$\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$

$\sin(2\theta)=\sin(2\times(\alpha-\theta))$

$\sin120=\sin2(\alpha-60)$

$120=2\alpha-120$

$\alpha=\dfrac{120+120}{2}$

$\alpha=120^{\circ}$

Hence, The other angle is 120°

6 0

3 years ago

Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission

bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law P = σ A e T⁴

Wien displacement law λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

P₁ = σ A T₁⁴

T = 12000K

P₂ = σ A T₂⁴

P₂ / P₁ = T₂⁴ / T₁⁴

P₂ / P₁ = (12000/3000)⁴

P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

λ T = 2,898 10-3

λ₁ = 2.89810-3 / T

λ₁ = 2,898 10-3 / 3000

λ₁ = 0.966 10-6 m

λ₁ = 966 nm

T = 12000K

λ₂ = 2,898 10-3 / 12000

λ₂ = 0.2415 10-6 m

λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0

3 years ago

Peter Piper picked a peck of pickled peppers. If Peter Piper picked a peck of pickled peppers, how many pickled peppers did Pete

maria [59]

Peter picked 8 quarts of peppers.

8 0

2 years ago

Read 2 more answers

Two blocks of ice, one four times as heavy as the other, are at rest on a frozen lake. A person pushes each block the same dista

qaws [65]

Answer:b

Explanation:

Given

mass of heavy object is 4m

mass of lighter object is m

A person pushes each block with same force F

According to Work Energy theorem Change in kinetic energy of object is equal to Work done by all the object

As launching velocity is same for both the object so heavier mass must possess greater kinetic energy . For same force heavier mass must be pushed 4 times farther than the light block .

$\Delta (K.E.)_H=\frac{1}{2}(4m)v^2$

$\Delta (K.E.)_L=\frac{1}{2}(m)v^2$

$\Delta K.E.=F\times d$

So the correct option is b

4 0

3 years ago