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4 years ago

An electric device uses 650 watts of power. If the voltage is 120 volts, what is the resistance?

Physics

2 answers:

grigory [225]4 years ago

8 0

Answer: The correct answer is 22.15 ohm.

Explanation:

The expression for the power in terms of resistance and potential is as follows;

$P=\frac{V^{2} }{R}$

Here, V is the potential, P is the power and R is the resistance.

It is given in the problem that an electric device uses 650 watts of power. The voltage of this device is 120 V.

Calculate the resistance of the electric device by rearranging the above expression.

$R=\frac{V^{2} }{P}$

Put V= 120 V and P= 650 W.

$R=\frac{(120)^{2} }{650}$

$R=22.15 ohm$

Therefore, the value of the resistance is 22.15 ohm.

PSYCHO15rus [73]4 years ago

5 0

There are several handy-dandy formulas for electrical power.
Here they are:

-- Power = (voltage) x (current)
-- Power = (current²) x (resistance)
-- Power = (voltage²) x (resistance)

We just have to pick the one that's most convenient.

We know the power and the voltage, and we have to find the resistance.
So we would be smart to use a formula that has power, voltage, and
resistance in it. The last formula on my list is exactly what we want.

-- Power = (voltage²) / (resistance)

   650 w = (120 v)² / (resistance)

   (650 w) x (resistance) = (120 v)²

   resistance = (120 v)² / (650 w) = 22.2 ohms .

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The driver of a car traveling at 23.1 m/s applies the brakes and undergoes a constant

Vsevolod [243]

Answer:

The tires make 125 revolutions before the car stops

Explanation:

Circular and Linear Motion

A tire rotates around a fixed point and the tire when in contact with the ground, drives a vehicle in a linear path. This is an example of a relationship between both types of movements that can be taking place simultaneously.

The car is moving with an initial speed of $v_o=23.1\ m/s$ and then breaks at $a=-1.03\ m/s^2$ until it stops. We can compute the time take to stop by using

$\displaystyle v_f=v_o+a.t$

Solving for t

$\displaystyle t=\frac{v_f-v_o}{a}$

Putting in numbers

$\displaystyle t=\frac{0-23.1}{-1.03}$

$\displaystyle t=22.427\ sec$

Now, let's transfer this information to the circular motion. We know the tangent speed is

$\displaystyle v_t=w.r$

Being w the angular speed and r the radius of the circle, in this case, the tires. The tangent speed is the same as the speed of motion of the car. It gives us the initial angular speed

$\displaystyle w_o=\frac{v_t}{r}$

$\displaystyle w_o=\frac{23.1}{0.33}=70\ rad/s$

When the circular motion is not uniform, i.e. there is angular acceleration $\alpha$ , the angular speed is a function of time

$\displaystyle w=w_o+\alpha t$

We can compute the angular acceleration knowing the final angular speed is zero when the car stops.

$\displaystyle \alpha=\frac{w-w_o}{t}=\frac{0-70}{22.427}$

$\displaystyle \alpha=-3.121\ rad/s^2$

The rotation angle is also a function of time as shown

$\displaystyle \theta=w_o\ t+\frac{\alpha t^2}{2}$

Using the given and computed values

$\displaystyle =70(22.427)-\frac{3.121(22.427)^2}{2}$

$\displaystyle \theta =784.95\ rad$

Knowing each revolution is $2\pi$ radians, the number of revolutions is

$\displaystyle n=\frac{\theta }{2\pi}=125\ rev$

The tires make 125 revolutions before the car stops

8 0

4 years ago

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

nata0808 [166]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

7 0

4 years ago

alculate the kinetic energies of (a) a 2.00×103-kg automobile moving at 100.0 km/h; (b) an 80.0-kg runner sprinting at 10.0 m/s;

zzz [600]

Answer:

(a) 7.72×10⁵ J

(b) 4000 J

(c) 1.82×10⁻¹⁶ J

Explanation:

Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass, v = velocity

(a)

For a moving automobile,

Ek = 1/2mv².

Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s

Substitute into equation 1

Ek = 1/2(2.0×10³)(27.78²)

Ek = 7.72×10⁵ J

(b)

For a sprinting runner,

Given: m = 80 kg, v = 10 m/s

Substitute into equation 1 above,

Ek = 1/2(80)(10²)

Ek = 40(100)

Ek = 4000 J

(c)

For a moving electron,

Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s

Substitute into equation 1 above,

Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²

Ek = 1.82×10⁻¹⁶ J

8 0

3 years ago

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ

vitfil [10]

Answer:

Explanation:

Rest mass of Kaon $M_{0K}$ = 494 MeV/c²

Rest mass of proton $M_{0P}$ = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy $E_{0K}$ = 494c² MeV

for the proton, rest energy $E_{0P}$ = 938c² MeV

Recall that the rest energy, and the total energy are related by..

$E$ = γ $E_{0}$

which can be written in this case as

$E_{K}$ = γ $E_{0K}$ ...... equ 1

where $E$ = total energy of the kaon, and

$E_{0}$ = rest energy of the kaon

γ = relativistic factor = $\frac{1}{\sqrt{1 - \beta ^{2} } }$

where $\beta = \frac{v}{c}$

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

$E_{K}$ = $E_{0P}$ ......equ 2

where $E_{K}$ is the total energy of the kaon, and

$E_{0P}$ is the rest energy of the proton.

From $E_{K}$ = $E_{0P}$ = 938c²

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = $\frac{1}{\sqrt{1 - \beta ^{2} } }$ = 1.89

1.89 $\sqrt{1 - \beta ^{2} }$ = 1

squaring both sides, we get

3.57( 1 - $\beta^{2}$ ) = 1

3.57 - 3.57 $\beta^{2}$ = 1

2.57 = 3.57 $\beta^{2}$

$\beta^{2}$ = 2.57/3.57 = 0.72

$\beta = \sqrt{0.72}$ = 0.85

but, $\beta = \frac{v}{c}$

v/c = 0.85

v = <em>0.85c </em>

7 0

3 years ago

Max ran 100 feet in 10 seconds.

Ivanshal [37]

Answer:

for max :

100 feet in 10 secs

for molly :

60 feet in 5 secs = 120 feet in 10 secs

so, molly ran farther in the same time interval i.e. covered 120 feet where as Max covered 100 feet

Explanation:

brainliest plz

8 0

3 years ago

Read 2 more answers