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V125BC [204]
3 years ago
5

An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the

level ground 150m downrange from the launch point throws an apple straight up with the minimum intial speed necessary to meet the path of the arrow.
(a) what is the initial speed of the apple?
(b) at what time after the arrow launch should the apple be thrown so that the arrow hits the apple?
Physics
1 answer:
garri49 [273]3 years ago
5 0

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = \dfrac{150}{45 cos 50^0}

t  =  5.19 s

a) height of arrow

s = u t +\dfrac{1}{2}gt^2

s = v sin \theta \times t+\dfrac{1}{2}gt^2

s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = \dfrac{u}{g}=\dfrac{30.29}{9.81}

                                             = 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s

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Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

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mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}

The electrostatic force between them is

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According to the question

1 % of Fg = Fe

0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}

2.927 \times 10^{35}=9 \times10^{9}q^{2}

3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0
3 years ago
You are in Paris, 60 m up in the Eiffel Tower. If you throw a euro downward at a velocity of 2.0 m/s, how long would it take the
kondor19780726 [428]

Answer:

t = 3.29 seconds

Explanation:

It is given that,

Height of the Eiffel tower is 60 m

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It will move under the action of gravity in the downward direction. Firstly, we can find the final velocity as follows :

v^2-u^2=2ad\\\\v=\sqrt{u^2+2ad} \\\\v=\sqrt{(2)^2+2\times 9.81\times 60} \\\\v=34.36\ m/s

Let t is the time taken by the euro to hit the ground. It can be calculated as :

v=u+at\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{34.36-2}{9.81}\\\\t=3.29\ s

Hence, it will take 3.29 seconds to hit the ground.

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At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (7.00 m/s2)i hat + (6.00 m/s2)j. It m
Elina [12.6K]

Answer:

r = 3519.55 m

Explanation:

We know that the acceleration of a particle in a circular motion is directed towards the center of the circumference and has magnitude:

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Where r is the radius of the circumference and ω is the angular velocity.

From the two acceleration vectors we find that their magnitude is

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Now we need to calculate the angular velocity to obtain the radius. Since t2-t1 = 3s is less than one period we can be sure that the angular velocity is equals to the angle traveled between this time divided by 3 s.

The angle with respect to the x-axis  for the particle at t1 and t2 is:

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olga_2 [115]

Answer:

There may be excess charges in the interior of the wire

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There may be excess charges on the surface of the wire.

There is no net flow of mobile electrons inside the wire.

Explanation:

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3 0
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