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Rudiy27
2 years ago
15

Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?

Physics
1 answer:
frosja888 [35]2 years ago
3 0

Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

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H = 11 m, the vertical distance that the bullet falls.
Initial vertical velocity  = 0
Horizontal velocity  = 144.7 m/s

The time, t, taken to fall 11 m is given by
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4.9t² = 11
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If aerodynamic resistance is ignored, the horizontal distance traveled before the bullet hits the ground is
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Answer: 216.8 m
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2 years ago
Which missing item would complete this alpha decay reaction ? ________—> 228/88Ra + 4/2He A.224/96 Rn B. 232/90 Th C. 230/92
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A.224/96 Rn I think i'm right if i'm wrong i'm srry


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3 years ago
A satellite moves in a circular orbit a distance of 1.6×10^5 m above Earth's surface. (radius of Earth is 6.38 x 10^6m and its m
Rashid [163]

Answer:

The speed of the satellite is 7809.52 m/s        

Explanation:

It is given that,

Radius of Earth, r=6.38\times 10^6\ m

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We need to find the speed of the satellite. It is given by :

v=\sqrt{\dfrac{GM}{R}}

R = r + d

R=(6.38\times 10^6\ m+1.6\times 10^5\ m)=6540000\ m

So, v=\sqrt{\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 5.98\times 10^{24}\ kg}{6540000\ m}}

v = 7809.52 m/s

So, the speed of the satellite is 7809.52 m/s. Hence, this is the required solution.

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3 years ago
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faust18 [17]

Answer:

a, 1.775s

b, 17.04μC

c, 1.28s

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R = 1.25MΩ

C = 1.42µF

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q = 8.78 µC

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τ = (1.25*10^6) * ( 1.42*10^-6)

τ = 1.775s

q• = εC

q• = 12 * 1.42*10^-6

q• = 17.04*10^-6C

q• = 17.04μC

Time t =

q = q• [1 - e^(t/τ)]

t = τIn[q•/(q•-q)]

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8 0
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Answer:

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OR

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I=12/4=3amp

total current

I=6+3

9amp

7 0
2 years ago
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