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3 years ago

An electric current of 0.25 A passes through a circuit that has a resistance of

Physics

2 answers:

Katen [24]3 years ago

8 0

Answer:

The voltage is V = 37.5 [V]

Explanation:

To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance.

V = I*R

where:

V = voltage [Volt]

I = current = 0.25[amp]

R = resistance = 150 [ohm]

V = 0.25*150 = 37.5 [V]

adoni [48]3 years ago

7 0

Answer: 37.5 V

Explanation: Just finished the quiz on a p e x :)

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A cannonball with a mass of 1.0 kilogram is fired horizontally from a 500.-kilogram cannon, initially at rest, on a horizontal,

vfiekz [6]

The impulse J is equal to the magnitude of the force applied to the cannonball times the time it is applied:
$J=F \Delta t$
But the impulse is also equal to the change in momentum of the cannonball:
$J=\Delta p$
If we put the two equations together, we find
$F \Delta t= \Delta p$
And since we know the magnitude of the average force and the time, we can calculate the change in momentum:
$\Delta p= F \Delta t=(8.0 \cdot 10^3 N)(1.0 \cdot 10^{-1} s)=800 kg m/s$

7 0

3 years ago

The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level β = 10 log(I/ $I_{0}$ ) ......................a

put here value (I/ $I_{0}$ ) = 10^−12 W/m² and β = 115

115 = 10 log(I/10^−12)

I = 0.316228 W/m²

and we know power = intensity × 4π r² ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0

3 years ago

Cotton balls could be used to represent clouds because of their white color and appearance of texture. Which of these other thin

iris [78.8K]

I just did it the answer is..... B.cobwebs

8 0

3 years ago

Read 2 more answers

An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be

Inessa05 [86]

Explanation:

The given data is as follows.

Inner wheel Radius = 20 m,

Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

Angular velocity of automobile = w = $\frac{V}{R}$

where, V = linear velocity of automobile m/min,

R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

u = linear velocity of inner wheel

and, u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

w =

= $\frac{u}{(R - d)}$

u = $V \times \frac{(R - d)}{R}$

= $V \times (1 - \frac{d}{R})$

If radius of wheel is r it will cover distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus, u = $2\pi rn$ = $V \times (1 - \frac{d}{R})$

Now, rotation per minute of inner wheel is calculated as follows.

n = $\frac{V}{2 \pi r \times (1 - \frac{d}{R})}$

= $\frac{V}{2 \pi r \times (1 - \frac{0.75}{20})}$ (since 2d = 1.5m given, d = 0.75m),

= $\frac{V}{r} \times 0.1532$

So, rotation per minute of outer wheel; n' =

= $\frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}$

= $\frac{V}{r} \times 0.1651$

5 0

3 years ago

A car moves along a curved road of diameter 2 km. If the maximum velocity for safe driving on this path is 30 m/s, at what angle

Leto [7]

The maximum velocity in a banked road, ignoring friction, is given by;

v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.

Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°

Therefore, the road has been banked at 5.24°.

4 0

3 years ago