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Vanyuwa [196]
2 years ago
4

A group of students perform the same "Conservation of Mechanical Energy" experiment that you performed in lab by allowing a soli

d sphere and then a solid cylinder to roll down the ramp. Both objects were released from the same position at the top of the ramp. If the speed vsphere of the solid sphere at the bottom of the ramp was 1.25 m/s, what would be the speed vcylinder at the bottom of the ramp?
Physics
1 answer:
Keith_Richards [23]2 years ago
3 0

Answer:

V_ {cylinder} = 1.11m / s

Explanation:

For this case we simply find the speeds for both

in the situations mentioned as well,

V_ {sphere} = \sqrt {\frac {10gh} {7}}

1.15 = \sqrt {\frac {10 (9.8) h} {7}}

Solving for h,

h = 0.0944m

V_ {cylinder} = \sqrt {\frac {4gh} {3}}

V_ {cylinder} = \sqrt {\frac {4 (9.8) (0.0944)} {3}}

V_ {cylinder} = 1.11m / s

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Rhea kicks a soccer ball at 13 km/h to Sean. After kicking the ball, the speed of the soccer ball from Rhea’s reference frame is
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Answer:  Sean is standing still, and Rhea is running toward Sean while   kicking the ball

Explanation: Your welcome :)

5 0
2 years ago
We observe that a moving charged particle experiences no magnetic force. From this we can definitely conclude that:_______
Leto [7]

Answer:

b. the particle must be moving parallel to the magnetic field.

Explanation:

The magnetic force on a moving charged particle is given by;

F = qvBsinθ

where;

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field

θ is the angle between the magnetic field and velocity of the moving particle.

When is the charge is stationary the magnetic force on the charge is zero.

Also when the charge is moving parallel to the magnetic field, the magnetic force is zero.

Therefore, when a moving charged particle experiences no magnetic force, we can definitely conclude that the particle must be moving parallel to the magnetic field.

b. the particle must be moving parallel to the magnetic field.

5 0
2 years ago
The diagram shows the electric field due to point charge Q. Which statements are correct? Check all that apply.
Bess [88]

Answer:

The <em>correct</em> statements  are:

  • <em>A. The electric field is nonuniform.</em>
  • <em>D. Charge Q is positive.</em>
  • <em>E. If charge A moves toward charge Q, it must be a negative charge</em>

Explanation:

The answer choices are:

  • A. The electric field is nonuniform.
  • B. The electric field is uniform.
  • C. Charge Q is negative.
  • D. Charge Q is positive.
  • E. If charge A moves toward charge Q, it must be a negative charge.
  • F. If charge A moves toward charge Q, it must be a positive charge.

<h2>Solution</h2>

The <em>electric field</em> is the electrostatic force per unit of charge,  

         \vec E=\dfrac{\vec F}{Q}

around around a charge, where another charge would experience the electrostatic force.

The electric field lines are shown in a diagram with arrows ditributed radially away from a positive charge and radially toward a negative charge.

Since the arrows are away from Q, Q is a positive charge: <em>statement D.</em>

Since the size of the arrows decreases as you move away  from Q the stregth of the field is not uniform: <em>statement A.</em>

Since the charge Q is positive, a negative charge would be attracted toward it: <em>statement E.</em>

4 0
2 years ago
You jump off a platform 134 m above the Neuse River. After you have free-fallen for the first 40 m, the bungee cord attached to
omeli [17]

Answer:

The acceleration at lowest point is 19.62 m/s^2

Explanation:

Conservation of energy is an concept in which it is stated that the energy of an isolated object remains the same. Energy changes from one form to another.

Lets Assume

Constant of string is K

By using the conservation of energy we will have the following equation

1/2 x 80^2 x K = m x 9.81 x 120

3200 K =  1177.2 m

K = 1177.2 m / 3200

K = 0.368 m

At the lowest point we will have

a = ( K x X - m x g ) / m

a = ( 0.368 m x 80 - m x 9.81 ) / m

a = 19.62 m / s^2

So, the acceleration at lowest point is 19.62 m/s^2

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Ahat [919]
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