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3 years ago

If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?

Physics

1 answer:

Lena [83]3 years ago

5 0

Answer:2 volts

Explanation:

Power=current x voltage

6=3 x voltage

Divide both sides by 3

6/3=(3 x voltage)/3

2=voltage

Voltage=2volts

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Un carro parte del reposo y acelera uniformemente a 4.0 m / s^2 durante 5.0 s. A continuación, mantiene la velocidad que alcanzó

Arada [10]

Answer:

Average speed of the car = 16.53 m.s

Explanation:

For first 5 seconds:

As the car starts from rest, so the initial speed of the car, u=0

Acceleration, a= 4 m/s^2

So, the final speed (speed at time, t=5 second)

v=u+at

v=0 + 4 x 5 = 20 m/s

Distance covered,

$s=ut+\frac 1 2 a t^2 \\\\s=0+\frac 1 2 \times 4 \times 5^2$

s=50 m

As the car maintained the constant speed of 20 m/s for the next 10 seconds.

Distance traveled in 10 seconds, d=20 x 10 =200 m

For the last 4 seconds:

The car slows down at the rate of 2.0 m / s ^ 2 from the speed of 20 m/s.

As the car slows down, so the acceleration will be negative, i.e a=-2 m/s^2

$s=ut+\frac 12 at^2 \\\\s=20\times 4 +\frac 12 (-2)4^2$

s=80-16=64 m.

Considering all three cases, the total distance covered=

=50+200+64=314 m

Total time= 5+10+4=19 s

So, the averate peed= Totat distance covered / total time taken

=314/19=16.53 m/s

6 0

3 years ago

. How does a simple cell work????

Delvig [45]

A simple cell can be made by connecting two different metals in contact with an electrolyte. A number of cells can be connected in series to make a battery , which has a higher voltage than a single cell. In non-rechargeable cells, eg alkaline cells, a voltage is produced until one of the reactants is used up

5 0

3 years ago

A 250 g air-track glider is attached to a spring with springconstant 4.0 N/m. Th damping constant due to air resistance is0.015

vaieri [72.5K]

Answer:

33.33 seconds

Explanation:

$N=\dfrac{1}{e}N_0$

$N_0$ = Initial length pulled = 20 cm

b = Damping constant = 0.015 kg/s

k = Spring constant = 4 N/m

m = Mass of glider = 250 g

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.25}{4}}\\\Rightarrow T=1.57079632679\ s$

Using exponential decay formula

$N=N_0e^{\dfrac{-bt}{m}}$

Final amplitude = Initial times decay

$\dfrac{1}{e}0.2=0.2e^{\dfrac{-0.015t}{2\times 0.25}}\\\Rightarrow 0.2=0.2e^{\frac{-0.015t}{2\cdot \:0.25}+1}\\\Rightarrow e^{\frac{-0.015t}{2\cdot \:0.25}+1}=1\\\Rightarrow \ln \left(e^{\frac{-0.015t}{2\cdot \:0.25}+1}\right)=\ln \left(1\right)\\\Rightarrow \left(\frac{-0.015t}{2\cdot \:0.25}+1\right)\ln \left(e\right)=\ln \left(1\right)\\\Rightarrow \frac{-0.015t}{2\cdot \:0.25}+1=\ln \left(1\right)\\\Rightarrow -\frac{0.015t}{0.5}=-1\\\Rightarrow -0.000225t=-0.0075\\\Rightarrow t=33.33\ s$

The time taken is 33.33 seconds

7 0

4 years ago

A 50-turn circular coil with radius 4 cm is placed in magnetic field of 6000 G. An amount of 800 mA current passes through the c

satela [25.4K]

Explanation:

n=50,r=0.02mn=50,r=0.02m,

I=5AandB=0.20TI=5AandB=0.20T

τismaxiμmwhensinθ=90∘τismaxiμmwhensinθ=90∘

τmax=niabsin90∘=mbτmax=niabsin90∘=mb

=50×5×3.14×4×10−4×2×10−1=50×5×3.14×4×10-4×2×10-1

=6.28×10−2Nm=6.28×10-2Nm

Given τ=12×τmaxτ=12×τmax

⇒sinsinθ=12⇒sinsinθ=12 or sinθ=30∘sinθ=30∘

=∠betweenareavar→randmag≠ticfield=∠betweenareavar→randmag≠ticfield.

So angle between magnetic field and the plane of the coil

=90∘−30∘=60∘=90∘-30∘=60∘.

<h3>HOPE IT HELPS </h3>

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4 0

3 years ago

10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr

12345 [234]

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

$F_{g0}=G \frac{mM_{E0}}{r^{2}}$

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=2M_{E0}$

so:

$F_{gf}=G \frac{2mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=2$

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=3M_{E0}$

so:

$F_{gf}=G \frac{3mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=3$

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

$M_{Ef}=\frac{M_{E0}}{4}$

so:

$F_{gf}=G \frac{mM_{E0}}{4r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=\frac{1}{4}$

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

4 0

3 years ago