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svet-max [94.6K]

2 years ago

9

A basketball player shoots toward a basket 5.6 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor a

t an angle of 60 o above the horizontal, what must the initial speed be if it were to go through the basket?

1 answer:

vaieri [72.5K]2 years ago

7 0

Answer:

9.58 m/s

Explanation:

x = 5.6 m

y = 3 m

θ = 60°

Let u be the initial velocity of projection.

The general equation of projectile path is given by

$y = x tan\theta -\frac{gx^{2}}{2u^{2}Cos^{2}\theta }$

$3 = 5.6\times tan60 -\frac{9.8\times 5.6 \times 5.6}{2\times u^{2}Cos^{2}60}$

$\frac{614.66}{u^{2}}=6.7$

u = 9.58 m/s

thus, the velocity of projection is given by 9.58 m/s.

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Compare and contrast electric potential energy and electric potential difference? Explain.

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Answer:

<u><em>Electric Potential Energy:</em></u>

The energy that is needed to move a charge against an electric firld is called Electric Potential Energy

<u><em>Electric Potential Difference:</em></u>

The amount of work done in carrying a unit charge from one point to an other in an electric field is called Electric Potential Difference.

<u><em>Relation:</em></u>

Relation between Electric potential and electrical potential energy is given by

$\delta V=\frac{PE}{q}$

Here PE represents Electric potential energy

and $\delta V$ is Electric potential difference

it means electric potential difference is the difference in electric potential energy divided by the charge.

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2 years ago

A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates are separated by a d

Snowcat [4.5K]

Answer:

The energy stored is 1.4 x 10^-9 J.

Explanation:

Side of square, L = 10 cm = 0.1 m

Distance, d = 2 mm = 0.002 m

Electric field, E = 4000 V/m

The energy stored in the capacitor is

$U = 0.5 C V^2$

The capacitance is given by

$C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J$

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2 years ago

4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two

Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=mV_{o}+MU_{o}$ (2)

$p_{f}=(m+M)V_{f}$ (3)

$m=1200 kg$ is the mass of the first car

$V_{o}=20 m/s$ is the velocity of the first car, to the North

$M=1400 kg$ is the mass of the second car

$U_{o}=-22 m/s$ is the mass of the second car, to the South

$V_{f}$ is the final velocity of both cars after the collision

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg}$ (6)

Finally:

$V_{f}=-2.61 m/s$ (7) This is the resulting velocity of the wreckage, to the south

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2 years ago

How would a chart of mid-17th century religious beliefs differ from this chart?

tino4ka555 [31]

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A chart of mid-17th-century religious beliefs differs from this chart in that Fewer people would identify as an atheist because people were not willing to share alternative religious beliefs publically.

In the 1600s, people did not have total freedom of speech if they had any. The church had a tremendous influence in the life of people and religious beliefs defined societies and families. The Church exerted its power and influence in many aspects of the people's lives and something out of the purview of the church or different to the religious beliefs of the Church was considered to be sacrilegious. The Church prosecuted people for being against the Church, so people of that time preferred to say that they were religious people supporting the church. Being an atheist was not really an option in the 17th century.

<u><em>Hope this helps!!</em></u>

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3 years ago

Read 2 more answers