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guajiro [1.7K]
3 years ago
10

A student climbs to the top of the gym and drops a baseball to the ground below. A physics student standing nearby has a motion

sensor and determines that the baseball hit the ground with a velocity of 17 m/s. if you ignore air resistance, how long was the baseball falling through the air?​
Physics
1 answer:
sladkih [1.3K]3 years ago
6 0

Answer:

The baseball was falling during 1.733 seconds.

Explanation:

The baseball experimented a free fall, which is a particular case of uniformly accelerated motion, in which object is accelerated by gravity. In this case, we need to determine the time spent by the ball before hitting the ground (t), measured in seconds, which is determined by the following kinematic formula:

t = \frac{v-v_{o}}{g} (1)

Where:

v_{o}, v - Initial and final speeds of the baseball, measured in meters per second.

g - Gravitational acceleration, measured in meters per square second.

If we know that v_{o} = 0\,\frac{m}{s}, v = 17\,\frac{m}{s} and g = 9.807\,\frac{m}{s^{2}}, then the time spent by the baseball is:

t = \frac{17\,\frac{m}{s}-0\,\frac{m}{s}  }{9.807\,\frac{m}{s^{2}} }

t = 1.733\,s

The baseball was falling during 1.733 seconds.

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Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.

Mathematically also potential energy is represented as

Potential energy= mgh

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At the bottom of the swing,the height is zero, hence the potential energy is also zero.

The kinetic energy is represented mathematically as

Kinetic energy= 1/2 mv^2

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3 years ago
A long solenoid has a radius of 4.0 cm and has 800 turns/m. If the current in the solenoid is increasing at the rate of 3.0 A/s,
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Answer:

Explanation:

Given that,

Radius of solenoid R = 4cm = 0.04m

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Induced electric field?

At r = 2.2cm=0.022m

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The magnetic field inside a solenoid is give as

B = µo•N•I

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From gauss law

∮E•dA =qenc/εo

We can find the tangential component of the electric field from Faraday’s law

∮E•dl = −dΦB/dt

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E∮dl = −d/dt •(πr²B)

2πrE = −πr²dB/dt

2πrE = −πr² d/dt(µo•N•I)

2πrE = −πr² × µo•N•dI/dt

Divide both sides by 2πr

E =- ½ r•µo•N•dI/dt

Now, substituting the given data

E = -½ × 0.022 × 4π ×10^-7 × 800 × 3

E = —3.32 × 10^-5 V/m

E = —33.2 µV/m

The magnitude of the electric field at a point 2.2 cm from the solenoid axis is 33.2 µV/m

where the negative sign denotes counter-clockwise electric field when looking along the direction of the solenoid’s magnetic field.

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Answer:

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