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3 years ago

Suppose you drop a tennis ball from a height of 6 feet. After the ball hits the floor, it rebounds to 80% of

Physics

2 answers:

statuscvo [17]3 years ago

7 0

Answer: The height after the third bounce is 3.1 meters

Explanation: We know that after every bounce the ball rebounds to 80% (or 0.8 in decimal form) of its previous hight. So if the first height was 6m, after the first bounce, the height will be:

H1 = 6m*0.8 = 4.8m

after the second bounce we have:

H2 = 4.8m*0.8 = 3.84m

and in the third bounce, the height is:

H3 = 3.84m*0.8 = 3.072m

Now we need to round it to the nearest thenth, this is the first number after the decimal point.

We can see that the second number after the decimal point is a 7, so we should round up, and get H3 = 3.1m

Another way to model this situation is with the equation:

H(b) = 6m*0.8^b

Where b is the number of bounces that the ball did.

in this situation, the third bounce is when b= 3, and we have that:

H(3) = 6m*0,8^3 = 3.072m, which is the same result as before.

sasho [114]3 years ago

3 0

The formula to solve this is (.80)^3 X 6 and the answer would be 3.1 feet. That is how high the ball will rebound after its third bounce. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.

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One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

3 years ago

Which factors could be potential sources of error in the experiment? Check all that apply.

mote1985 [20]

Answer:

1. energy lost in the lever due to friction

3. visual estimation of height of the beanbag

5. position of the fulcrum for the lever affecting transfer of energy

Explanation:

Edge 2021

8 0

3 years ago

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If the mass of each ball is 12 kg and they experience a gravitational force of magnitude 2.00

VLD [36.1K]

Answer:

${2 \times 10} - 7 = 12 {}^{2} \div x \\ x = 2 \times 10 - 7 \div 12 {}^{2}$

I'm just in jss2 but I read physics. this is what I think

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2 years ago

Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1

lesantik [10]

Answer:

a) $\lambda=6.63\times10^{-31}m$

b) $\lambda=6.63\times10^{-39}m$

c) $\lambda=9.97\times10^{-11}m$

d) $\lambda=4.03\times10^{-36}$ m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

$\lambda=\frac{h}{P}=\frac{h}{mv}$

with h the Plank's constant ( $6.63\times10^{-34}\frac{m^{2}kg}{s}$ ) and P the momentum of the object that is mass (m) times velocity (v).

a) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}$

$\lambda=6.63\times10^{-31}m$

b) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}$

$\lambda=6.63\times10^{-39}m$

c) $\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}$

$\lambda=9.97\times10^{-11}m$

d) $\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}$

$\lambda=4.03\times10^{-36}$ m

e) $\lambda=\frac{6.63\times10^{-34}}{(74*0)}$

λ=∞

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3 years ago

A 10 kg turkey, He kicks the 0.5 kg ball with a force of 50N for 0.2 seconds and the ball flies straight away horizontally from

Harman [31]

Answer:

a. 20m/s

b.50N

c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.

Explanation:

a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:

$F\bigtriangleup t=m\bigtriangleup v\\\\50N\times 0.2s=0.5kg\times \bigtriangleup v\\\\\bigtriangleup v=2(50N\times0.2)\\\\=20m/s$

Hence, the velocity of the ball after the kick is 20m/s

b.The force felt by the turkey:

#Applying Newton's 3rd Law of motion, opposite and equal reaction:

-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.

c. Using the law of momentum conservation:

-Due to ther external forces exerted on the turkey, it remains stationery.

-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.

-Momentum is not conserved due to these external forces.

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3 years ago