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Rudiy27
3 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]3 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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Answer:

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Explanation:

The magnitude of the electric field is defined as the force per charge on the test charge.

Since we define electric field as the force per charge, it will have the units of  force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).

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3 years ago
A projectile is fired with a velocity of 22 m/s at an angle of 25°. What is the vertical component of the velocity?
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Answer:

Vertical component of velocity is 9.29 m/s

Explanation:

Given that,

Velocity of projection of a projectile, v = 22 m/s

It is fired at an angle of 22°

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The vertical component of velocity is v sinθ

So, vertical component is given by :

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3 years ago
What is the 5 quantitative research problems?<br>​
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Answer:

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Explanation:

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Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the
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Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency f= 6.37\times10^{14}\ Hz

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Using formula of wavelength

\lambda=\dfrac{c}{f}

where, c = speed of light

f = frequency

Put the value into the formula

\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}

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d =\dfrac{m\times\lambda}{\sin\theta}

Where, m = number of fringe

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Here, \sin\theta =\dfrac{y}{l}

Put the value into the formula

d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}

d=40.11\times10^{-6}\ m

d = 40.11\ \mu m

Hence, The distance between the two slits is 40.11 μm.

7 0
3 years ago
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