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Mamont248 [21]
3 years ago
14

. What is the relationship between potential energy, kinetic energy, and speed as the skater moves down and up the U-shaped ramp

?
Physics
2 answers:
kykrilka [37]3 years ago
6 0
Top of the U ramp: potential energy is the highest, while kinetic energy is the lowest

Bottom of the U ramp(aka the curve part): potential energy is the lowest and the kinetic energy is the highest

THEREFORE, PE and KE have an INVERSE RELATIONSHIP.
vladimir1956 [14]3 years ago
4 0

Top of the U ramp: potential energy is the highest, while kinetic energy is the lowest

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Pendulum clocks generally run fast in winter and slow in summer
Harman [31]
If the question is true or false then the answer is true
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3 years ago
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A point charge Q moves on the x-axis in the positive direction with a speed of A point P is on the y-axis at The magnetic field
oee [108]

Answer: q = -52.5 μC

Explanation:

The complete question is given thus;

A point charge Q moves on the x-axis in the positive direction with a speed of 280 m/s. A point P is on the y-axis at y=+70mm. The magnetic field produced at the point P, as the charge moves through the origin, is equal to -0.30uTk. What is the charge Q? (uo=4pi x 10^-7 T m/A).

SOLVING:

from the given parameters we can solve this problem.

Given that the

Speed = 280 m/s

y = 70mm

B =  -30 * 10⁻⁶T

Using the equation for magnetic field we have;

Β = μqv*r / 4πr²

making q (charge) the subject of formula we have that;

q  = B * 4 *πr² / μqv*r

substituting the values gives us

q = (-0.3*10⁻⁶Tk * 4π * 0.07²) / (4π*10⁻⁷ * 280 ) = - [14.7 * 10⁻¹⁰k / 2.8 * 10⁻⁵ k ]

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3 years ago
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viva [34]

Answer:

21.67 rad/s²

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Explanation:

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\omega_i = Initial angular velocity = 78 rad/s

\alpha = Angular acceleration

\theta = Angle of rotation

t = Time taken

r = Radius = 0.13

I = Moment of inertia = 1.25 kgm²

From equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875

Frictional force is given by

F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N

The magnitude of the force of friction applied by the brake shoe is 208.36538 N

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3 years ago
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nataly862011 [7]

Answer:

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2. Transverse wave: These waves are defined as the waves in which the particles of the medium travel perpendicularly to the direction of the wave. This does not require a medium to travel. These can travel in vacuum also. For Example: Light waves.

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