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gregori [183]
3 years ago
11

A tennis ball with a speed of 11.3 m/s is moving perpendicular to a wall. after striking the wall, the ball rebounds in the oppo

site direction with a speed of 7.2207 m/s. if the ball is in contact with the wall for 0.00803 s, what is the average acceleration of the ball while it is in contact with the wall? take "toward the wall" to be the positive direction. answer in units of m/s 2 .
Physics
1 answer:
Arlecino [84]3 years ago
4 0

To calculate the average acceleration of the ball we use the formula,

a= \frac{v_{f} -v_{i}}{t}

Here, v_{f} is the final velocity of the ball and v_{i} is the initial velocity of the ball t is the time in contact with the wall.

Given v_{i} = 11.3 m/s towards the wall and v_{f} = 7.2207 m/s

away from the wall and t=0.00803 s.

Substituting these values in above formula , we get

a=\frac{(-7.2207 m/s)-11.3 m/s}{0.00803 s}

Herev_{f} is negative because ball is moving away from the wall.

a= - 2306 .4 \ m/s^{2}

Therefore, average acceleration of the ball is - 2306 .4 \ m/s^{2} (away from the wall).

 

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What happens in a tug of war if the net forces are balanced and why?
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Answer:

Balanced forces are responsible for unchanging motion. Balanced forces are forces where the effect of one force is cancelled out by another. A tug of war, where each team is pulling equally on the rope, is an example of balanced forces. The forces exerted on the rope are equal in size and opposite in direction.

Explanation:

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3 years ago
Why do you often freeze after a shower? <br> Help mee
V125BC [204]
When water evaporates of your body it cools you down, hence why after a shower when the water droplets begin to evaporate you feel cold
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3 years ago
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t
AVprozaik [17]

Answer:

v=115 m/s

or

v=414 km/h

Explanation:

Given data

A_{area}=0.140m^{2}\\  p_{air}=1.21 kg/m^{3}\\  m_{mass}=80kg

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

mg=1/2*C*p_{air}*v^{2}*A_{area}\\   v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

v=(115*3.6)km/h\\v=414km/h

5 0
3 years ago
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
Hunter works to fix wires and paneling. Hunter is a(n)
Serjik [45]

Answer:

Hunter is a electrician

7 0
3 years ago
Read 2 more answers
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