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shusha [124]
3 years ago
9

A ball is traveling horizontally when it clears a volleyball net that is 1.7m high. What maximum {horizontal) speed (in m/s) can

it have if it is to fall within 5 meters of the net? Please input your answers as positive values with two decimal places.
Physics
1 answer:
Firlakuza [10]3 years ago
7 0

Answer:Answer

Answer:

(a) The speed must be greater than 20 m/s

(b) The maximun speed must be 21.6 m/s

Explanation:

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Calculate the net force experienced by Mr. Ferrari if he has a mass of 100 kg and an acceleration of 3.5 m/s/s.
saveliy_v [14]
The answer is 350. the formula for net force is f=ma which means you multiply the mass times he acceleration. 100 times 3.5 is 350
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3 years ago
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a minibus drives 238 miles with a constant speed of 84 mph and another 80 miles with a constant speed of 60 mph. how much time i
Aloiza [94]

Answer:

4 hours ten minutes

Explanation:

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3 years ago
HELP OR I WILL DIEJSNCSDKKS.
iogann1982 [59]

Answer:

Increase

Explanation:

because it makes it makes sense lol

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A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. A 5.0-µF capacitor is similarly ch
IceJOKER [234]

Answer:

Explanation:

Given that,

First Capacitor is 10 µF

C_1 = 10 µF

Potential difference is

V_1 = 10 V.

The charge on the plate is

q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC

q_1 = 100 µC

A second capacitor is 5 µF

C_2 = 5 µF

Potential difference is

V_2 = 5V.

Then, the charge on the capacitor 2 is.

q_2 = C_2 × V_2

q_2 = 5µF × 5 = 25 µC

Then, the average capacitance is

q = (q_1 + q_2) / 2

q = (25 + 100) / 2

q = 62.5µC

B. The two capacitor are connected together, then the equivalent capacitance is

Ceq = C_1 + C_2.

Ceq = 10 µF + 5 µF.

Ceq = 15 µF.

The average voltage is

V = (V_1 + V_2) / 2

V = (10 + 5)/2

V = 15 / 2 = 7.5V

Energy dissipated is

U = ½Ceq•V²

U = ½ × 15 × 10^-6 × 7.5²

U = 4.22 × 10^-4 J

U = 422 × 10^-6

U = 422 µJ

6 0
4 years ago
. Egbert is in a rowboat four miles from the nearest point on Egbert a straight shoreline. He wishes to reach a house 12 miles f
swat32

Answer:

t=\frac{13}{3} =4.33\ mi.hr^{-1}

Explanation:

Given:

  • Distance from the shore, s=4\ mi
  • distance of house from the shore, h=12\ mi
  • speed of rowing, v_r=3\ mi.hr^{-1}
  • speed of walking, v_w=4\ mi.hr^{-1}

<em>For the least amount of time to reach the house one must row at the nearest point on the shore and then walk from there.</em>

<u>Now the time taken to reach the shore:</u>

t_s=\frac{s}{v_s}

t_s=\frac{4}{3}\ mi.hr^{-1}

<u>Time taken in walking to house from the shore:</u>

t_h=\frac{h}{v_w}

t_h=\frac{12}{4}

t_h=\frac{12}{4}

t_h=3\ mi.hr^{-1}

<u>Therefore total time taken:</u>

t=t_h+t_s

t=\frac{13}{3} =4.33\ mi.hr^{-1}

6 0
3 years ago
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