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Nikolay [14]
3 years ago
5

there were 5 seconds left in a basketball game. In order to run out the clock a player threw the ball directly up as hard as he

could. How hard would the player have to throw the ball to make sure that it would not land before time expired? Hint: d= -16t + vt+ h
Physics
1 answer:
spin [16.1K]3 years ago
7 0

Answer:

80

Explanation:

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A mineral deposit along a strip of length 8 cm has density s(x) = 0.01x(8 − x) g/cm for 0 ≤ x ≤ 8. Calculate the total mass of t
vazorg [7]

Answer:

8z

Explanation:

It is 8z

3 0
3 years ago
What is the resistance of a circuit with a voltage of 10 V and current of 5A is?
wolverine [178]

Resistance = (voltage) / (Current)

Resistance = (10 V) / (5 A)

Resistance = 2 ohms.

5 0
3 years ago
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What are the top two gasses in the earth’s atmosphere ?
ch4aika [34]

Answer:

Nitrogen and oxygen are by far the most common; dry air is composed of about 78% nitrogen (N2) and about 21% oxygen (O2). Argon, carbon dioxide (CO2), and many other gases are also present in much lower amounts; each makes up less than 1% of the atmosphere's mixture of gases.

7 0
3 years ago
An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
2 years ago
What happens when a light ray is parallel to the principal axis ?
Phantasy [73]

Answer:

Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens. ... These rays of light will refract when they enter the lens and refract when they leave the lens.

Hope this helps...

7 0
3 years ago
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