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lys-0071 [83]
2 years ago
9

A tennis ball bounces on the floor three times. If each time it loses 22.0% of its energy due to heating, how high does it rise

after the third bounce, provided we released it 2.3 m from the floor?
Physics
1 answer:
lesya692 [45]2 years ago
6 0

Answer:

H = 109.14 cm

Explanation:

given,                                                            

Assume ,                                                            

Total energy be equal to 1 unit                                

Balance of energy after first collision = 0.78 x 1 unit

                                                             = 0.78 unit

Balance after second collision = 0.78 ^2 unit

                                                   = 0.6084 unit

Balance after third collision = 0.78 ^3 unit

                                              = 0.475 unit

height achieved by the third collision will be equal to energy remained                                        

H be the height achieved after 3 collision

0.475 ( m g h) = m g H                  

H = 0.475 x h                                    

H = 0.475 x 2.3 m                          

H = 1.0914 m                      

H = 109.14 cm                      

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andreyandreev [35.5K]

The resultant static friction force is equal to 20 N to the left.

Why?

I'm assuming that you forgot to write the question of the exercise, so,  I will try to complete it:

"A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-n force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?"

So, if we are going to calculate the resulting static friction force, it means that there is no movement, we must remember that the friction coefficient will give us the maximum force before the crate starts to move.

We can calculate the static friction force by using the following formula:

Fr=F(appliedforce)

Since the crate is not moving (static), the static friction force acting on the crate will be equal to the applied force.

Calculating we have:

Fr=F(appliedforce)

Fr=20N

Hence, the static friction force is equal to 20 N to the left (since the applied force is acting to the right)

So,

FrictionForce=AppliedForce

Since the static friction force is equal to the applied force, the crate does not start to move.

Have a nice day!

8 0
2 years ago
Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and
Anna11 [10]

Kepler’s third law formula: T^2=4pi^2*r^3/(GM)

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M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

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=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/( 86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The closest answer is 1.99 × 10^30

(it may vary a little with rounding – the difference is less than 1%)

5 0
2 years ago
Read 2 more answers
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lesantik [10]

Answer:

h=14.2857\,m

Explanation:

Given:

radius of capillary, r=10^{-6}\,m

angle of contact, \theta=0^{\circ}

density of water, \rho=1000\,kg.m^{-3}

surface tension of water, T=0.07 \,N.m^{-1}

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We have the equation for the height of meniscus as:

h=\frac{2T.cos\, \theta}{\rho.g.r}

h=\frac{2\times 0.07\times cos\,0^{\circ}}{1000\times 9.8\times 10^{-6}}

h=14.2857\,m

No, the capillary action alone cannot be the mechanism of water transportation to the top of the trees. Transpiration also creates a suction pressure in the xylem complementary to the ascent of sap and cohesion of water being the other causes of movement of water up in the plants.

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Stella [2.4K]

Answer:

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\Delta t = \frac{m\cdot [c_{p,w}\cdot (T_{2}-T_{1})+h_{fg}]}{\dot Q}

\Delta t = \frac{(2.25\times 10^{5}\,kg)\left[\left(4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (100\,^{\textdegree}C - 10\,^{\textdegree}C)+2256.5\,\frac{kJ}{kg} \right]}{200000\,kW}

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8 0
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faltersainse [42]

Answer:

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4 0
2 years ago
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