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tamaranim1 [39]
3 years ago
10

PLZ ANSWER CORRECT! IF I GET THIS CORRECT I GET MY HUNTING PERMIT!

Physics
1 answer:
Cerrena [4.2K]3 years ago
4 0
A. Greater than 1 mile
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What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT
sergiy2304 [10]

Answer:

The minimum frequency of the coil is 7.1 Hz

Explanation:

Given;

number of turns, N = 200 turns

cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

magnitude of magnetic field strength, B = 30 x 10⁻³ T

maximum value of the induced emf, E = 8 V

Maximum induced emf is given as;

E = NBAω

where

ω is angular velocity (ω = 2πf)

E = NBA2πf

where;

f is the minimum frequency, measured in hertz (Hz)

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³  x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz

8 0
3 years ago
PLEASE HELP!!! WILL MARK THE BRAINLIEST! HIGH SCHOOL PHYSICAL SCIENCE List the independent, dependent, controlled variables of t
serious [3.7K]

Answer:

Explanation:

You didn't last any of the variables. You have to list the variables to tell which are which.

3 0
3 years ago
Read 2 more answers
A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s
ICE Princess25 [194]

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

5 0
3 years ago
A billiard ball moving at 0.5 m/s strikes another identical billiard ball, which is at rest, in an elastic head-on collision on
NikAS [45]

Answer: assuming that the billiard balls are of identical weight the impacted billiard ball will move forward at around 0.5m/s (not considering energy conservation). The ball impacting the 2nd one would stop because most of its Kinetic energy would have been transferred into the not moving ball.

Explanation: hope this helps!

4 0
3 years ago
A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50
PSYCHO15rus [73]

Answer:

(a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Explanation:

Given that,

Length = 50.00 cm

Time period = 2.50 s

Time period of pendulum is defined as the time for one complete cycle.

The period depends on the length of the pendulum.

Using formula of time period

T=2\pi\sqrt{\dfrac{I}{mgh}}

Where, I = moment of inertia

We need to calculate the value of d

Using parallel theorem of moment of inertia

I=I_{cm}+md^2

For a meter stick mass m , the rotational inertia about it's center of mass

I_{cm}-\dfrac{mL^2}{12}

Where, L = 1 m

Put the value into the formula of time period

T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}

T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}

T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})

Multiplying both sides by d

tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]

(\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0

Put the value of T, L and g into the formula

4.028d^2-6.25d+0.336=0

d = 0.056\ m, 1.496\ m

The value of d is 0.056 cm and 1.496 cm.

(b). Given that,

L = 50-5 = 45 cm

We need to calculate the time period

Using formula of period

T=2\pi\sqrt{\dfrac{l}{g}}

Put the value into the formula

T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}

T=1.35\ sec

Hence, (a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

7 0
3 years ago
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