All categories

3 years ago

The rectangular coils in a 345-turn generator are 12 cm by 12 cm. Part A

Physics

1 answer:

Likurg_2 [28]3 years ago

4 0

Answer:

118.166 volt

Explanation:

We have given number of turns N =345

Sides of the rectangular coils is 12 cm =0.12 m

So area A =0.12×0.12=0.0144 $m^2$

Magnetic field B =0.45 T

Angular speed =505 rpm

Speed in rad/sec $\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 505}{60}=52.8566rad/sec$

The emf is given by $E=NAB\omega SIN\omega t$

For maximum emf sinwt =1

So $E=345\times 0.0144\times 0.45\times 52.8566=118.166volt$

You might be interested in

The fluid inside the hydraulic jack has a pressure of 30,000 Pa. If the surface of

aleksley [76]

Explanation:

p = F /A

F = P×A

F = 30,000 Pa / 0.1 m²

F = 300,000 N

5 0

2 years ago

A particle moves along the x axis so that its velocity at time t is given by v(t)=

vodomira [7]

Answer:

a = 0.7267 , acceleration is positive therefore the speed is increasing

Explanation:

The definition of acceleration is

a = dv / dt

they give us the function of speed

v = - (t-1) sin (t² / 2)

a = - sin (t²/2) - (t-1) cos (t²/2) 2t / 2

a = - sin (t²/2) - t (t-1) cos (t²/2)

the acceleration for t = 4 s

a = - sin (4²/2) - 4 (4-1) cos (4²/2)

a = -sin 8 - 12 cos 8

remember that the angles are in radians

a = 0.7267

the problem does not indicate the units, but to be correct they must be m/s²

We see that the acceleration is positive therefore the speed is increasing

6 0

3 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

Read 2 more answers

Movement of a stationary object (2words)

Whitepunk [10]

Induced motion or movement

6 0

3 years ago

Identify the 4 components of an atom and what the charge is on each one ?<br><br> Please help

luda_lava [24]

Answer:

Structure Of The Atom: Our current model of the atom can be broken down into three constituents parts – protons, neutron, and electrons. Each of these parts has an associated charge, with protons carrying a positive charge, electrons having a negative charge, and neutrons possessing no net charge.

6 0

2 years ago

Read 2 more answers