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Rus_ich [418]
3 years ago
7

A small rock is thrown straight up with initial speed V0 from

Physics
1 answer:
bagirrra123 [75]3 years ago
8 0

Answer:

12 feet

Explanation:

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A flat, rectangular coil consisting of 60 turns measures 23.0 cmcm by 34.0 cmcm . It is in a uniform, 1.40-TT, magnetic field, w
Travka [436]

Answer:

(a)   ΔФ = -0.109W

(b)  emf = 28.43V

(c)   Iin = emf/R

Explanation:

(a) In order to calculate the magnetic flux you use the following formula:

\Delta\Phi_B=\Phi-\Phi_o=BAcos(90\°)-BAcos(0\°)   (1)

B: magnitude of the magnetic field = 1.40T

A: area of the rectangular coil = (0.23m)(0.34m)=0.078m^2

Where it has been taken into account that at the beginning the normal vector to the cross sectional area of the coil, and the magnetic field vector are parallel. When the coil is rotated the vectors are perpendicular.

Then, you obtain:

\Delta\Phi_B=(1.40T)(0.078m^2)=-0.109W

The change in the magnetic flux is -0.109 W

(b) During the rotation of the coil the emf induced is given by:

emf=-N\frac{\Delta \Phi}{\Delta t}         (2)

N: turns of the coil = 60

ΔФ: change in the magnetic flux = 0.109W

Δt: lapse time of the rotation = 0.230s

You replace the values of the parameters in the equation (2):

emf=-(60)(\frac{-0.109W}{0.230s})=28.43V

The induced emf is 28.43V

(c) The induced current in the coil is given by:

I_{in}=\frac{emf}{R}      (3)

R: resistance of the coil     (it is necessary to have this value)

emf :induced emf  = 28.43V

7 0
4 years ago
If a 70kg swimmer pushes off a pool wall with a force of 250N. What is her acceleration?
Alenkasestr [34]

Newton taught us:                              Force = (mass) x (acceleration)

Divide each side by (mass) :              Acceleration = (force) / (mass) .

The only problem here is:  This formula applies when the "Force" is the
only force on the object.  When the objects in these school problems are
falling out of airplanes, shot from guns, or being hit by baseball bats, we
routinely ignore the force of air resistance against the object.  We're
comfortable with that, maybe because it's become a habit.  But now,
we're not so comfortable about ignoring the force of water resistance.

All I can tell you is that if you DO ignore the water resistance, that is,
if the water were not there, her acceleration would be

                    (250 newtons) / (70 kg) = 3.57 m/s² = about  0.36 g .

But what is it really, in the water ? 

If you've spent any substantial amount of time anywhere near competitive
swimmers, then you know that it depends on their position coming off the
wall, what they do with their knees and knuckles, how straight they hold
their body, how deep the texture of their swim-cap is, and how well they've
shaved their legs.

6 0
4 years ago
Read 2 more answers
A constant torque of 26.6 N · m is applied to a grindstone for which the moment of inertia is 0.162 kg · m2 . Find the angular s
ozzi

Answer:

156.1 rad/s = 24.8 rev/s

Explanation:

Torque = Momentum of inertial × radial acceleration = Iα

τ = 26.6 N.m

I = 0.162 kg.m²

26.6 = 0.162 × α

α = 164.2 rad/s²

Using equations of motion,

θ = 11.8 rev = 11.8 × 2π = 74.2 rad

w₀ = 0 rad/s (since the grindstone starts from rest)

w = ?

α = 164.2 rad/s²

w² = w₀² + 2(α)(θ)

w² = 0² + (2×164.2)(74.2)

w = 156.1 rad/s = 24.8 rev/s

Hope this Helps!!!

7 0
4 years ago
Read 2 more answers
Suppose that the clay balls model the growth of a planetesimal at various stages during its accretion. Choose the planetesimal t
motikmotik

Answer:2

Explanation:

8 0
4 years ago
Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not
musickatia [10]

Answer:

Part A: t = v_0/a_0

Part B: t = v_0/a_0

Part C: v_0^2/a_0

Explanation:

Part A:

We will use the following kinematics equation:

v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}

Part B:

We will use the same kinematics equation:

v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}

Part C:

The total time takes is 2t.

So the train moves a distance of

x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}

And the car moves a distance in Part A and in Part B:

d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}

So the total distance that the car traveled is d = \frac{v_0^2}{a_0}

The difference between the train and the car is

x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}

8 0
3 years ago
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