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3 years ago

5

The microwaves produced inside a microwave have a wavelength of 12.0m and a frequency of 25Hz. At what speed do the microwaves t

ravel in units of m/s ?

1 answer:

inn [45]3 years ago

8 0

Answer: 300 m/s

Explanation:

The equation for finding wave speed is the wavelength times the frequency which would be

12.0mx 25hz=300m/s

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An ammeter is connected in series with a resistor of unknown resistance R and a DC power supply of known emf e. A student uses t

rosijanka [135]

Answer:

B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Explanation:

Here,power has been calculated using current I and total EMF \ε . So,P=EMF*current= ε I will represent total power dissipated in resistor and ammeter.

Now, this total power P has been dissipated in both resistor and ammeter. So, power dissipated in resistor must be less than P as some power is also dissipated in ammeter because it has non-zero resistance.

So, the answer is B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Note that option A,C and E are ruled out as they state power dissipated by resistor is greater than or equal to P which is false.

Also,option D is ruled out as ammeter is connected in series.

8 0

4 years ago

When a mass of 29 g is attached to a certain spring, it makes 20 complete vibrations in 3.1 s. What is the spring constant of th

never [62]

Answer:

The spring constant of the spring is 47.62 N/m

Explanation:

Given that,

Mass that is attached with the spring, m = 29 g = 0.029 kg

The spring makes 20 complete vibrations in 3.1 s. We need to find the spring constant of the spring. We know that the number of oscillations per unit time is called frequency of an object. So,

$f=\dfrac{20}{3.1}$

f = 6.45 Hz

The frequency of oscillator is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

k is the spring constant

$k=4\pi^2f^2m$

$k=4\pi^2\times (6.45)^2\times 0.029$

k = 47.62 N/m

So, the spring constant of the spring is 47.62 N/m. Hence, this is the required solution.

4 0

3 years ago

At which point does the image appear? <br> W <br> X <br> Y <br> Z

ahrayia [7]

Z I think...........

3 0

3 years ago

Read 2 more answers

The 5-lb collar is released from rest at A and travels along the frictionless guide. Determine the speed of the collar when it s

Sholpan [36]

Answer:

Explanation:

Stiffness of spring k equal to 4 lb/ft.

Unstretched length of the spring L is equal to 0.5 feet.

Weight of the collar W is 15lb

Radius of curvature of curve guide is 1 feet

length of vertical rod is 1.5 feet

Initial speed of collar when released from rest at A is 0 feet per seconds

use the energy conservation equation

$P_A +K_A=P_B+K_B$

Estimate the potential energy , component as position B as below

$P_A=Wh_1+\frac{1}{2} ks^2_1\\\\=5\times(1.5+1)+\frac{1}{2} \times 4 \times(1.5+1-0.5)^2\\\\=20.5lb.ft$

Estimate the kinetic energy , component as position A as below

$K_A=\frac{1}{2} \frac{W}{g} V^2_1\\\\=\frac{1}{2} \frac{5}{32.2} \times0^2\\\\=0lb.ft$

Estimate the kinetic energy , component as position B as below

$K_A=\frac{1}{2} \frac{W}{g} V^2_2\\\\=\frac{1}{2} \frac{5}{32.2} \times V^2_2\\\\=00777V^2_2$

Substitute 20.5lb- ft for $P_A$

0.5lb-ft for $P_B$

0lb -ft for $K_A$

$0.0777V_2^2 for K_B$

$20.5+0=0.5+0.777V^2_2\\\\V^2_2=257.6\\\\V_2=16.05$

= 16.05ft/sec

7 0

3 years ago

????????????????what’s the answer

kompoz [17]

Answer:

I don't know what do u. mean

6 0

3 years ago