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3 years ago

you dribble a basketball while walking on a basketball court list the pairs of action reaction forces in this situation

1 answer:

4 0

There is the force of gravity on you and the ball by the Earth. The other half of the pair would be you and the ball applying the force of gravity on the Earth.

When you dribble the ball the floor applies a normal force on the ball to keep it from going through the floor. The ball applies a normal force back on the floor. There is also a normal force on you when you walk.

There is an applied force by your hand on the ball when you hit the ball down. The ball also applies the same force on your hand in the opposite direction.

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For Jane to see an image, light must enter her eyes. What specifically is entering her eyes when she sees an image?

Sergio [31]

Energy
because once the light hits her eyes energy flows through her body so the answer is A energy

3 0

3 years ago

Read 2 more answers

A resonance tube can be used to measured the speed of sound in air. A tuning fork is held above the opening of the tube and stru

LenaWriter [7]

Answer:

330.24 Hz

Explanation:

Given:

Frequency, f = 320 Hz

L1 = 25.8 cm

L2 = 78.4 cm

L3 = 131.1 cm

Let the wavelength be λ

Then, L1 which is the length of the column of air is λ/4.

λ/4 = 25.8 cm

λ = 25.8 × 4 = 103.2 cm = 1.032 m

Then, speed of sound in air is:

v = λ f

⇒ v = 1.032 × 320 Hz

⇒ v = 330.24 m/s

7 0

3 years ago

A 400 gram sample of alcohol at 10o C [Ca=2.64 J/g*oC] is mixed with 400 grams of warm water [Cw=4.186 J/g*Co] at 88o C. Assume

alukav5142 [94]

The heat lost by the water will be equivalent to the energy gained by the alcohol. Thus:
maCaΔT = -mwCwΔT
400 x 2.64 x (T - 10) = 400 x 4.186 x (88 - T)
T = 57.8 °C

8 0

4 years ago

A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho

Lady_Fox [76]

Answer:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>

The ball rotates 6.78 revolutions.

Explanation:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>

At the bottom the ball has the following angular speed:

$\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s$

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

$sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m$

To find the revolutions we need the time, which can be found using the following equation:

$v_{f} = v_{0} + at$

$t = \frac{v_{f} - v_{0}}{a}$ (1)

So first, we need to find the acceleration:

$v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}$ (2)

By entering equation (2) into (1) we have:

$t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}$

Since it starts from rest (v₀ = 0):

$t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s$

Finally, we can find the revolutions:

$\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev$

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!

3 0

3 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago