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3 years ago

A _______ is a rhythmic movement that carries energy through space or matter.

1 answer:

7 0

The answer is wave.

A wave can be defined as a rhythmic flow that moves over a medium from one place to the different area.

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A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =

castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

$emf=NBA\omega sin(\omega t)$

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

$emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V$

(b) the maximum rate of change of the magnetic flux is given by:

$\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}$

(c) $emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V$

(d) The torque is given by:

$\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm$

3 0

3 years ago

Pls help me with this problem!!

Olenka [21]

Answer:

v = 19.6 m/s.

Explanation:

Given that,

The radius of the circle, r = 5 m

The time period of the ball, T = 1.6s

We need to find the ball's tangential velocity.

The formula for the tangential velocity is given by :

$v=\dfrac{2\pi r}{T}$

Putting all the values in the above formula

$v=\dfrac{2\pi \times 5}{1.6}\\\\v=19.6\ m/s$

So, the tangential velocity of the ball is 19.6 m/s. Hence, the correct option is (c).

7 0

3 years ago

To win the game, a place kicker must kick a

masya89 [10]

Answer:

0.57 m

Explanation:

First of all, we need to calculate the time it takes for the ball to cover the horizontal distance between the starting position and the crossbar. This can be done by analzying the horizontal motion only. In fact, the horizontal velocity is constant and it is

$v_x = u cos \theta = (15)(cos 51.7^{\circ})=9.30 m/s$

And the distance to cover is

d = 19 m

So the time taken is

$t=\frac{d}{v_x}=\frac{19}{9.30}=2.04 s$

Now we want to find how high the ball is at that time. The initial vertical velocity is

$u_y = u sin \theta = (15)(sin 51.7^{\circ})=11.77 m/s$

So the vertical position of the ball at time t is

$y(t) = u_y t - \frac{1}{2}gt^2$

where g = 9.8 m/s^2 is the acceleration of gravity. Substituting t = 2.04 s, we find

$y=(11.77)(2.04)-\frac{1}{2}(9.8)(2.04)^2=3.62 m$

The crossbar height is 3.05 m, so the difference is

$\Delta h = 3.62 - 3.05 =0.57 m$

So the ball passes 0.57 m above the crossbar.

8 0

3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

4 years ago

Suppose one night the radius of the earth doubled but its mass stayed the same. What would be an approximate new value for the f

Alex17521 [72]

Answer:

g' = g/4

Explanation:

The value of the free-fall acceleration at the surface of the earth, can be obtained applying Newton's 2nd law, assuming that the only force acting on an object at the surface of the earth, is the one produced by the mass of the Earth, i.e. gravity.
This force can be expressed according the Newton's Universal Law of Gravitation , as follows:

$F_{g} = G*\frac{m_{x} *m_{E} }{r_{E}^{2} } (1)$

From Newton's 2nd Law, we have:
$F = m* a (2)$
Since the left sides in (1) and (2) are equal each other, both right sides must be equal each other also.
Simplifying the mass m, we can write the acceleration a in (2) as the acceleration due to gravity, g, as follows:

$g = G*\frac{m_{E} }{r_{E}^{2} } (3)$

Since G is an universal constant, and the mass mE remains constant, if we double the radius of the Earth, the new value for the acceleration due to gravity (let's call it g'), is as follows:

$g' = G*\frac{m_{E} }{(2r_{E})^{2} } =G*\frac{m_{E} }{4*r_{E}^{2} } = g*\frac{1}{4} =\frac{g}{4} (4)$

4 0

3 years ago