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3 years ago

While operating at 120 volts, an electric toaster has a resistance of 15 ohms. The power used by the toaster is

Physics

1 answer:

Darya [45]3 years ago

8 0

Answer:

960 Watt

Explanation:

From the question,

Electric power = Voltage squared/Resistance

P = V²/R ..................... Equation 1

Where P = power, V = Voltage, R = Resistance

Given: V = 120 volts, R = 15 ohms.

Substitute these values into equation 1

P = 120²/15

P = 14400/15

P = 960 Watt

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A car is driving at 99 km/h, calculate the distance it travels in 70 minutes.

Lapatulllka [165]

Answer:

The distance the car travels is 115500 m in S.I units

Explanation:

Distance d = vt where v = speed of the car and t = time taken to travel

Now v = 99 km/h. We now convert it to S.I units. So

v = 99 km/h = 99 × 1000 m/(1 × 3600 s)

v = 99000 m/3600 s

v = 27.5 m/s

The speed of the car is 27.5 m/s in S.I units

We now convert the time t = 70 minutes to seconds by multiplying it by 60.

So, t = 70 min = 70 × 60 s = 4200 s

The time taken to travel is 4200 s in S.I units

Now the distance, d = vt

d = 27.5 m/s × 4200 s

d = 115500 m

So, the distance the car travels is 115500 m in S.I units

8 0

3 years ago

Sphere is fired downwards into a medium with an initial speed of 27 m/s. If it experiences a

anygoal [31]

Answer:

33 m/s

first i do 27+6=33 +s2=

8 0

3 years ago

What is the angle of the 2nd order bright fringe produced by two slits that are 8.25x10-5m apart if the wavelength of the incide

dexar [7]

In order to calculate the angle, we can use the formula below for a constructive interference (the interference is constructive because the fringe is bright):

$d\sin\theta=m\lambda$

Where d is the distance between the slits, m is the order of the interference and lambda is the wavelength.

So, using d = 8.25 * 10^-5, m = 2 and lambda = 4.5 * 10^-7, we have:

$\begin{gathered} 8.25\cdot10^{-5}\cdot\sin\theta=2\cdot4.5\cdot10^{-7}\\ \\ \sin\theta=\frac{9\cdot10^{-7}}{8.25\cdot10^{-5}}\\ \\ \sin\theta=1.091\cdot10^{-2}\\ \\ \theta=0.625° \end{gathered}$

Therefore the correct option is the second one.

5 0

1 year ago

17. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106

sleet_krkn [62]

Answer:

Explanation:

Distance between plates d = 2 x 10⁻³m

Potential diff applied = 5 x 10³ V

Electric field = Potential diff applied / d

= 5 x 10³ / 2 x 10⁻³

= 2.5 x 10⁶ V/m

This is less than breakdown strength for air 3.0×10⁶ V/m

b ) Let the plates be at a separation of d .so

5 x 10³ / d = 3.0×10⁶ ( break down voltage )

d = 5 x 10³ / 3.0×10⁶

= 1.67 x 10⁻³ m

= 1.67 mm.

5 0

4 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

6 0

4 years ago