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Olegator [25]
3 years ago
8

A water balloon slingshot launches its projectiles essentially from ground level at a speed of 25.5 m/s . (You can ignore air re

sistance.)
A: At what angle should the slingshot be aimed to achieve its maximum range?

B: If shot at the angle you calculated in Part A, how far will a water balloon travel horizontally?

C: For how long will the balloon be in the air?
Physics
1 answer:
Alex787 [66]3 years ago
8 0
The formulae for range and flight time for a given launch from ground level at angle θ are <span>
R=<span><span><span>v2</span>sin2θ</span>g</span></span>
and <span>
T=<span><span>2vsinθ</span>g</span></span>

these follow from the equations of motion and the parabolic/symmetric shape of flight without air resiatance you can see that you will max out on range at
<span><span>
θ=<span>45o</span></span></span>
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A plane mirror of circular shape with radius r=20cm is fixed to the ceiling. A bulb is to be placed on the axis of the mirror. A
KIM [24]

Answer:

0.75 m

Explanation:

Let's call the distance between the bulb and the mirror x.

The bulb and the length of the mirror form a triangle.  The mirror and the illuminated area on the floor form a trapezoid.  If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle.  This triangle and the small triangle are similar.  So we can say:

x / 0.4 = (3 + x) / 2

Solving for x:

2x = 0.4 (3 + x)

2x = 1.2 + 0.4 x

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So the bulb should located no more than 0.75 m from the mirror.

5 0
3 years ago
A stone is dropped into a well. The sound of the splash is heard 6.25 s later. What is the depth of the well? (Take the speed of
Naya [18.7K]

Answer:

depth of well is 163.30 m

Explanation:

Given data

speed of sound = 343 m/s

timer = 6.25 s

to find out

depth of well

solution

let us consider depth d

so equation will be

depth = 1/2 ×g ×t²    ..............1

and

depth = velocity of sound × time    .................2

here we have given time 6.25 that is sum of 2 time

when stone reach at bottom that time

another is sound reach us after stone strike on bottom

so time 1 + time 2 = 6.25 s

so from equation 1  and 2 we get

1/2 ×g ×t² = velocity of sound × time

1/2 ×9.8 × t1² = 343 × (6.25 - t1 )

t1 = 5.77376 sec

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height = 1/2 ×9.8 × (5.773)²

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3 0
3 years ago
4.
Akimi4 [234]

Answer:

c

Explanation:

plz make me brainliest i have answered

5 0
3 years ago
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

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3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0
3 years ago
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