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2 years ago

Please help asap!!!!!!!!!!!

Physics

1 answer:

Tomtit [17]2 years ago

8 0

The answer would be 20000

The answer in standard form/scientific notation would be 2 x 10^4

(The exponent is 4 because that's how many digits after the 2 there is)

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a student standing in an elevator at rest notices that her pendulum has a frequency of oscillation of 0.5 Hz. a. what is the len

EastWind [94]

Answer:

L = 0.99 m = 99 cm

Explanation:

The period is the reciprocal of the frequency.

T = 1/0.5 = 2.0 s

T = 2π√(L/g)

L = g(T/2π)²

L = 9.8(2.0/2π)² = 0.99 m

If the system accelerates upward, it will cause the apparent gravity to increase. This will require a longer pendulum to keep the same period, or shorten the period if the length remains the same. This shows up in the equation where the product of gravity and the square of the period must remain constant for the length to remain constant.

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3 years ago

What is the source of geothermal energy? User: What is the source of geothermal energy?

bazaltina [42]

The answer is; C

In particular points in the earth’s surface, underground water is naturally heated to steam that can be harness for geothermal energy. The steam that ejects from the ground with high kinetic energy can be used to turn turbines that generate electricity. The underground water is usually heated by the hot rocks beneath that are subjected to the immense heat of magma or the enormous pressure of overlying crust.

8 0

3 years ago

Read 2 more answers

A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60∘ away from vertical increases fro

bixtya [17]

Answer:

The induced emf is $|\epsilon|=0.0261 V$

Explanation:

From the question we are told that

The number of turn is $N = 100$

The diameter of the coil is $d = 2.0 cm = 2.0 *10^{-2} m$

The uniform magnetic at initial is $B_i = 0.50 T$

The uniform magnetic at initial is $B_f = 1.50 T$

The time taken is $t = 0.60s$

The angle the magnetic field makes with vertical is $\theta = 60^o$

Generally induced emf is mathematically represented as

$\epsilon = -N \frac{d \o}{dt}$

where $d \o$ is the change magnetic flux

Magnetic flux is mathematically represented as

$\O = \= B \cdot \= A$

$= BA cos \theta$

Substituting this above

$\epsilon = -N \frac{d (BA cos \theta)}{dt}$

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

Where B is the magnetic field and A is the area which is mathematically evaluated as

$A = \frac{\pi d^2}{4}$

Substituting values

$A = \frac{\pi (2.0 *10^{-2})^2}{4}$

$A= 3.142*10^{-4}m^2$

From the equation of emf

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

dB = $B_2 -B_1$

$|\epsilon| = N A \frac{ (B_2 -B_1 cos \theta)}{dt}$

substituting values

$|\epsilon| = 100(3.142*10^{-4}) \frac{ (1.50 -0.50 cos(60))}{0.60}$

$|\epsilon|=0.0261 V$

6 0

3 years ago

Fig.4.1

Nutka1998 [239]

Answer:

Explanation:

7 0

4 years ago

a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string. The string is hung over a pulley as shown in t

svet-max [94.6K]

The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately

<h3>What are we to consider in equilibrium ?</h3>

Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.

Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.

The acceleration at which they will both move will be;

a = ( $m_{1}$ - $m_{2}$ ) / ( $m_{1}$ + $m_{2}$ )

a = (5.75 - 3.53) / (5.75 + 3.53)

a = 2.22 / 9.28

a = 0.24 m/s²

Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m

We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.

v² = u² + 2as

since u =0

v² = 2 × 0.24 × 2.47

v² = 1.1856

v = √1.19

v = 1.0888 m/s

Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately

Learn more about Equilibrium here: brainly.com/question/517289

#SPJ1

5 0

2 years ago