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3 years ago

a microwave uses a 14 a of electricity if 120 v is run through it what is the resistance of the microwave

1 answer:

5 0

Resistance = voltage / current.

That's. 120v / 14A = 8.57 ohms.

By the way, voltage doesn't "run through" anything. Current does. That would be the 14 Amps.

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When Cesium metal is illuminated with light of wavelength 300 nm, the photoelectrons emitted have a maximum kinetic energy of 2.

Allisa [31]

Answer:

i have no clue im sorry

Explanation:

8 0

3 years ago

A 50-cm-long spring is suspended from the ceiling. A 330g mass is connected to the end and held at rest with the spring unstretc

Anvisha [2.4K]

Explanation:

Given that,

Length of the spring, l = 50 cm

Mass, m = 330 g = 0.33 kg

(A) The mass is released and falls, stretching the spring by 28 cm before coming to rest at its lowest point. On applying second law of Newton at 14 cm below the lowest point we get :

$kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.14}\\\\k=23.1\ N/m$

(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.

(C) The frequency of the oscillation is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{23.1}{0.33}} \\\\f=1.33\ Hz$

5 0

3 years ago

A 9.0-V battery is connected to a resistor so that there is a 0.50-A current through the resistor.For how long should the batter

salantis [7]

Answer:

35.6 s

Explanation:

The power through the resistor is given by:

$P=VI$

where V=9.0 V is the voltage and I=0.50 A is the current. Substituting into the formula, we find

$P=(9.0 V)(0.5 A)=4.5 W$

The power is also equal to:

$P=\frac{W}{t}$

where W is the work done while t is the time taken. Since we know the work done, W=160 J, we can re-arrange the equation to find the time taken:

$t=\frac{W}{P}=\frac{160 J}{4.5 W}=35.6 s$

7 0

3 years ago

A 6.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en

Jlenok [28]

Answer:

$F=52.1301\ N$

Explanation:

Given:

mass attached to the rope, $m=6\ kg$

given time, $t=4.4\ s$

function of height of the mass connected via a rope:

$y=2.8\ t+0.61\ t^3$

differentiate the above eq. with respect to time t gives us the velocity in vertical direction:

$v_y=\frac{d}{dt} y$

$v_y=2.8+1.83\ t^2$

put the value of given time:

$v_y=2.8+1.83\times 4.4^2$

$v_y=38.2288\ m.s^{-1}$

As we know form the Newton'a second law of motion that the rate of change in momentum is proportional to the applied force.

$F=\frac{\Delta (m.v_y)}{t}$

$F=\frac{6\times 38.2288}{4.4}$

$F=52.1301\ N$

6 0

3 years ago

f the arm is 7.75 m long and pivots about one end, at what angular speed (in rpm) should it spin so that the acceleration of the

777dan777 [17]

The angular speed of the lander must be 3.92 rpm

Explanation:

The (centripetal) acceleration of the lander is given by:

$a=\omega^2 r$

where

$\omega$ is the angular speed

r is the length of the arm

In this problem, we have

r = 7.75 m

$a=1.31 m/s^2$ (we are told that the acceleration of the lander must be the same as the acceleration due to gravity at the surface of Europa, which is $1.31 m/s^2$