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2 years ago

5

A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference,

this resistor will dissipate:

1 answer:

Stels [109]2 years ago

7 0

Answer:

<h2>0.056 W</h2>

Explanation:

$Power = IV$

From ohms law we know that

$V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}$

Given data

P1 = 0.5 Watt

P2 = ?

V1= 3 Volts

V2= 1 Volt

Thus we can solve for the power dissipated as follows

$P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}$

$\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W$

<em>The resistor will dissipate 0.056 Watt</em>

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

2 years ago

Noah drops a rock with a density of 1.73 g/cm3 into a pond. Will the rock float or sink?

bearhunter [10]

Answer:

It will sink

Explanation:

An object in the water can float only if its density is lower than the density of the water.

In fact, for an object completely immersed in water, there are two forces acting on it:

- Its weight, $W=mg=\rho_o V g$ , downward, where $\rho_o$ is the density of the object, V its volume and g the gravitational acceleration

- The buoyant force, $B=\rho_w V g$ , upwards, there $\rho_w$ is the density of the water

We see that when the density of an object is larger than the density of the water, $\rho_o > \rho_w$ , the weight is greater than the buoyant force, $W>B$ , so the object sinks.

In this case, the rock has a density of 1.73 g/cm3, while water has a density of 1.0 g/cm^3, so the rock will sink.

5 0

2 years ago

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You are in a spacecraft moving at a constant velocity. The front thruster rocket fires incorrectly, causing the craft to slow do

Alchen [17]

Answer:

It continue to move forward at a constant velocity which will be slower than before the front thruster was fired.

Explanation:

Before the front thruster was fired, the spacecraft was already moving at a particular velocity.

After the malfunction, the front thruster is fired and then the force exerted by that front thruster slows the spacecraft down, as we are told.

By using the rear thruster to exert a force equal to that from the front thrusters, a force equal in magnitude to that of the front thrusters is added, cancelling out the effect of the front thrusters. Because the spacecraft is already moving at a slower speed at this point compared to the beginning, it continues to move at that speed.

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2 years ago

A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft>s. if the attached cord is pulled

Leya [2.2K]

Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer: 44.25 ft-lb

6 0

2 years ago

Nitrogen (n2) gas within a piston–cylinder assembly undergoes a compression from p1 = 20 bar, v1 = 0.5 m3 to a state where v2 =

Bingel [31]

Part a)

As we know that

$P_1V_1^{1.35} = P_2V_2^{1.35}$

here we know that

P1 = 20 bar

V1 = 0.5 m^3

V2 = 2.75 m^3

from above equation

$20* 0.5^{1.35} = P * (2.75)^{1.35}$

$P = 2 bar$

so final state pressure will be 2 bar

Part b)

now in order to find the work done

$W = \int PdV$

$W = \int \frac{c}{V^{1.35}}dV$

$W = c\frac{V^{-0.35}}{-0.35}$

$W = \frac{P_1V_1 - P_2V_2}{0.35}$

$W = \frac{20* 0.5 - 2 * 2.75}{0.35}* 10^5 = 12.86 * 10^5 J$

3 0

2 years ago

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