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aleksandr82 [10.1K]

3 years ago

10

A ball of mass is released from rest at a height of 30 how fast is it going when it hits the ground

1 answer:

elena55 [62]3 years ago

7 0

Answer:

If the height is in metres, the speed is 24.25m/s

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An 30-turn coil has square loops measuring 0.341 m along a side and a resistance of 3.61 Ω. It is placed in a magnetic field tha

Verdich [7]

Answer:

Thus, the induced current in the coil at $t =1.73s$ is 9.98 A.

Explanation:

Faraday's law says

$\varepsilon = N \frac{d \Phi_B}{dt}$

where $N$ is the number of turns and $\Phi_B$ is the magnetic flux through the square coil:

Now,

$N = 30$

$\theta = 37.5^o$

$A = (0.341m)^2= 0.11623m^2$

$B = 1.45t^3$ ;

therefore,

$\varepsilon = N \frac{d \Phi_B}{dt} = N\frac{d ( BA\:cos(\theta))}{dt} = 30*\frac{d ( (1.45t^2)(0.1163)\:cos(37.5^o))}{dt}$

$=30*(0.1163)\:cos(37.5^o)*1.45*\dfrac{d ( t^3)}{dt} = 12.04t^2$

$\boxed{\varepsilon = 12.04t^2}$

is the emf induced in the coil.

Now, the loop is connected to $R = 3.61\Omega$ resistance; therefore, at $t = 1.73s$

$\varepsilon = RI = 12.04t^2$

$RI = 12.04(1.73)^2$

$RI = 36.03$

$I = \dfrac{36.03}{3.61\Omega }$

$\boxed{I = 9.98A }$

Thus, the current in the coil at $t =1.73s$ is 9.98 A.

8 0

4 years ago

A child on a playground swings through a total of 32 degrees. If the displacement is equal on each side of the equilibrium posit

Black_prince [1.1K]

Answer: A = y/cos32

That is the amplitude A in terms of the displacement y.

Explanation: Since the displacement in the question In the question is the same in both direction, it is a Simple Harmonic Motion problem. In s.h.m the amplitude of displacement A is related to the displacement itself y by this simple equation

y = A* cos(theta)

So, A = y/cos(theta)

A = y/cos32.

If the magnitude of the displacement y is given, you just substitute in.

4 0

3 years ago

eimsori [14]

B. Unbalanced force

3 0

3 years ago

I need to choose a theme for my physics assignment My experiment is finding g

Kobotan [32]

<h3>Question:</h3>

How to find g (acceleration due to gravity)

<h3>Solution:</h3>

We know,

Acceleration due to gravity (g)

$= \frac{GM}{ {R}^{2} }$

where, G = Gravitational constant

$= 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\$

M = Mass of the earth

$= 6 \times {10}^{24} \: kg$

R = Radius of the earth

$= 6.4 \times {10}^{6} m$

Putting these values of G, M and R in the above formula, we get

$g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2}$

So, the value of acceleration due to gravity is

$9.8m/s ^{2}$

Hope it helps.

Do comment if you have any query.

5 0

3 years ago

This chart shows characteristics of three different types of waves.

prohojiy [21]

Answer:

1. Primary or P waves are push and pull waves

2. Secondary, S or Shear Waves are also called transverse wave

3. L or surface waves reach the earth's surface after P and S waves

7 0

3 years ago