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Bumek [7]
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?
Physics
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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Student one used bowling ball A in a bowling game against Student 2, who used bowling ball B. Use Newton’s Three Laws of Motion
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.

8 0
3 years ago
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
borishaifa [10]

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}

Where,

c = Speed of light (vacuum)

f = frequency

Our values are,

f = 2Hz

c = 3*10^8km/s

Replacing we have,

\lambda = \frac{c}{f}

\lambda = \frac{3*10^8km/s}{2Hz}

\lambda = 1.5*10^8m

<em>Therefore the wavelength of this wave is 1.5*10^{8}m</em>

8 0
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.

Therefore the total distance is Xt = 34.2 + 96 = 130.2 [m].

5 0
3 years ago
A transformer has input voltage and current of 12 V and 4 A
Advocard [28]

Explanation:

It is given that,

Input voltage, V_i=12\ V

Input current, I_i=4\ A

Output current, I_o=0.8\ A

Number of turns in the secondary side of transformer, N_s=1177

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

So, the number of turns in primary side of the transformer is 5885.

4 0
3 years ago
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

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    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 } } ]

     E =248.2 \  N/C

4 0
3 years ago
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