Which use of iron is due to its chemical properties?

A uniform electric field of magnitude 110 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.

Marizza181 [45]

Answer:

1.7×10^5 ms-1

Explanation:

From

qE= qvB

q= charge on the electron

E = electric field

v= velocity

B= magnetic field

E= vB

v= E/B= 110×10^3/0.6

v= 1.7×10^5 ms-1

3 0

3 years ago

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Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

Ne4ueva [31]

Answer:

C

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

t = v / a

t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

t = v / g

t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

7 0

2 years ago

An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of

Gnoma [55]

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength = $\frac{2L}{3}$

The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength

Distance from center = 0.25 × wavelength

Distance = $0.25 x \frac{2L}{3}$

Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm

Distance = $0.25 x \frac{2 \times 39}{3}$

Distance from the center to the nearest anti - node = 6.5 cm

Hence, the nearest distance to the anti - node from the center = 6.5 cm

So, option C is correct.

7 0

3 years ago

A 174 pound Jimmy Cheek is riding on a 54 ft diameter Ferris Wheel. The normal force on Jimmy Cheek is 146 pounds when Jimmy is

Veronika [31]

To solve this problem we will apply the concepts related to the balance of Forces, the centripetal Force and Newton's second law.

I will also attach a free body diagram that allows a better understanding of the problem.

For there to be a balance between weight and normal strength, these two must be equivalent to the centripetal Force, therefore

$F_c = W-N$

$m\omega^2r = W-N$

Here,

m = Net mass

$\omega$ = Angular velocity

r = Radius

W = Weight

N = Normal Force

$m\omega^2r = 174-146$

The net mass is equivalent to

$F = mg \rightarrow m = \frac{F}{g}$

Then,

$m = \frac{174lb}{32.17ft/s^2}$

Replacing we have then,

$(\frac{174lb}{32.17ft/s^2})\omega^2 (54ft) =174lb-146lb$

Solving to find the angular velocity we have,

$\omega = 0.309rad/s$

Therefore the angular velocity is 0.309rad/s

6 0

2 years ago

In physics 28.97 + 45.876 adds up to what?

myrzilka [38]

$28.97 + 45.876$

equals 74.846

6 0

3 years ago

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