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3 years ago

8

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity (in m/s) must a basket ball player l

eave the grownd to rist 1.24 m above the floor in an attempt to get the ball?

1 answer:

mamaluj [8]3 years ago

7 0

Answer:4.93 m/s

Explanation:

Given

height to reach is (h )1.24 m

here Let initial velocity is u

using equation of motion

$v^2-u^2=2ah$

here Final Velocity v=0

a=acceleration due to gravity

$0-u^2=2\left ( -g\right )h$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.81\times 1.24}$

$u=\sqrt{24.328}$

u=4.93 m/s

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Jlenok [28]

Answer:

Newton’s Three Laws of Motion has a great impact.

Explanation:

Newton’s Three Laws of Motion has a great impact on the bowling game for the 2 students. When the student one throw ball to the student 2, the ball decrease its speed due to the gravity and opposing air. If these forces are removed from the system the ball will continue its motion till another force is applied on it. When the force applied to the ball it produces acceleration in the direction to the applied force. If the ball touches the ground it bounce back with equal force which is a reaction of the ground.

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When I wave a charged golf tube at the front of the classroom with a frequency of two oscillations per second, I produce an elec

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To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.

From the definition we know that the wavelength is described under the equation,

$\lambda = \frac{c}{f}$

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c = Speed of light (vacuum)

f = frequency

Our values are,

$f = 2Hz$

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$\lambda = \frac{c}{f}$

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3 years ago

I NEED MAJOR HELP ON THIS AS WELL PLEASE SOMEONE HELP ME

yuradex [85]

Answer:

The total distance is 130.2 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the cart starts from rest, i.e. its initial speed is zero.

$v_{f} =v_{o} +(a*t)$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 3 [m/s²]

t = time = 8[s]

Vf = 0 + (3*8)

Vf = 24 [m/s]

With this velocity we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} +2*a*x$

where

x = distance traveled [m]

24² = 0 + (2*3*x)

x = 576/(6)

x = 96 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

$v_{f}= v_{i} -(a*t)$

Vf = final velocity [m/s]

Vi = initial velocity = 24 [m/s]

a = desacceleration = 1.6 [m/s²]

t = time = 15 [s]

Vf = 24 - (1.6*15)

Vf = 21.6 [m/s]

With this velocity, we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} -2*a*x$

where

x = distance traveled [m]

21.6² = 24² - (2*1.6*x)

x = 109.44/(3.2)

x = 34.2 [m]

Note: The negative sign in the equations is because the car is desaccelerating, it means its velocity is decreasing.

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3 years ago

A transformer has input voltage and current of 12 V and 4 A

Advocard [28]

Explanation:

It is given that,

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We need to find the number of turns in the primary side of the transformer. The current to the number of turns in the input and output is given by :

$\dfrac{N_s}{N_p}=\dfrac{I_p}{I_s}$

Substituting all the above values

So,

$N_p=\dfrac{N_sI_s}{I_p}\\\\N_p=\dfrac{1177\times 4}{0.8}\\\\N_p=5885$

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What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate

Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The values is $E =248.2 \ N/C$

Explanation:

From the question we are told that

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The distance from the center is $k = 0.1 \ mm = 1*10^{-4} \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{0.12}{2}$

=> $r = 0.06 \ m$

Generally electric field is mathematically represented as

$E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]$

substituting values

$E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]$

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3 years ago