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musickatia [10]
3 years ago
10

Go to your backyard or to your high school’s basketball court or tennis court and try out this experiment to see if you can pred

ict motion and inertia accurately. Bring a friend or classmate. Take along a ball.
Select two spots roughly 10 meters apart. Mark each of them clearly. The first spot will be the starting point and the second will be the target.
Ask a friend to run as fast as possible (sprint) carrying the ball from the starting point to the target. Have your friend release the ball when he or she is immediately over the target trying NOT to swing arms when dropping the ball.
Record the results in the table below.
Repeat the sprint at least five times. Record where the ball lands each time.
Next, ask your friend to jog to the target with the ball and drop it. In the table, again record where the ball lands. Do two trials.
For the final test, ask your friend to walk to the target and drop the ball. Also record these results in the table. Do two trials
Physics
1 answer:
Talja [164]3 years ago
5 0

Answer:1 Sprint 54 cm Beyond

2 Sprint 41 cm Beyond

3 Sprint 68 cm Beyond

4 Sprint 32 cm Beyond

5 Sprint 44 cm Beyond

6 Jog   31 cm Beyond

7 Jog         22 cm Beyond

8 Walk 17 cm Beyond and left

9 Walk 6 cm Beyond

Explanation:

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PLEASE ANSWER QUICK!!!
rusak2 [61]

I would say its D.

-hope this helps

3 0
3 years ago
Read 2 more answers
Is there a way to see the color of cosmic rays the human eye can't see
KATRIN_1 [288]
You would have to use a machine to convert the colors to one visible by humans or become some other species of animal with a larger light spectrum.
4 0
3 years ago
The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC
enyata [817]

Answer:

The y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

Explanation:

<u>Given:</u>

  • Electric field in the region, \vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.
  • Charge placed into the region, q = 10.4\ nC = 10.4\times 10^{-9}\ C.

where, \hat i,\ \hat j are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.

Thus, the y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

3 0
3 years ago
A wind turbine is rotating at 15 rpm under steady winds flowing through the turbine at a rate of 42,000 kg/s. The tip velocity o
zzz [600]

Answer:

a) 5.22 m/s

b) 31.4 %

Explanation:

f = rotating speed = 15 rpm = 15/60 =0.25 rps

m = Mass flow rate of air = 42000 kg/s

v = Tip velocity = 250 km/h = 250/3.6 = 69.44 m/s

W = Work output = 180 kW

A = Swept area of wind turbine

r = Radius of wind turbine

η = Efficiency

r=\frac{v}{2\pi f}\\\Rightarrow r=\frac{250\div 3.6}{2\pi 0.25}=\frac{250}{1.8\pi}

A=\pi r^2\\\Rightarrow A=\pi\left(\frac{250}{1.8\pi}\right)^2\\\Rightarrow A=\left(\frac{250}{1.8}\right)^2\frac{1}{\pi}

m=\rho V\\\Rightarrow m=\rho vA\\\Rightarrow v=\frac{m}{\rho A}\\\Rightarrow v=\frac{42000}{1.31 \left(\frac{250}{1.8}\right)^2\frac{1}{\pi}}\\\Rightarrow v=5.22\ m/s

∴ The average velocity of the air is 5.22 m/s

E=m\frac{v^2}{2}=42000\frac{5.22^2}{2}\\\Rightarrow E=572538.92

\eta=\frac{W}{E}=\frac{180000}{572538.92}\\\Rightarrow \eta =0.314

∴ Conversion efficiency of the turbine is 0.314 or 31.4 %

7 0
3 years ago
Read 2 more answers
A parallel-plate vacuum capacitor has 8.38 j of energy stored in it. the separation between the plates is 2.30 mm. if the separa
avanturin [10]
<h3><u>Answer;</u></h3>

= 4.19 Joules

<h3><u>Solution;</u></h3>

Energy stored in capacitor = U = 8.38  J

U =(1/2)CV^2

C =(eo)A/d

C*d=(eo)A=constant

C2d2=C1d1

C2=C1d1/d2

Initial separation between the plates =d1= 2.30mm .  

Final separation = d2 = 1.15 mm

But; Energy=U =(1/2)q^2/C  

U2C2 = U1C1

U2 =U1C1 /C2

U2 =U1d2/d1

Final energy = Uf = initial energy × d2/d1

                                = 8.38 ×1.15/2.30

                                = 4.19 Joules

Thus; The final energy = 4.19 Joules

6 0
4 years ago
Read 2 more answers
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