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3 years ago

A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.

Physics

1 answer:

kow [346]3 years ago

3 0

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, $h_{max}$ , is given by the following equation;

$h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}$

Therefore, substituting the known values for the pebble, we have;

$h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439$

Therefore, the maximum height of the pebble projectile, $h_{max}$ ≈ 6.4 m.

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3 years ago

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garik1379 [7]

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3 years ago

Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a w

amm1812

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
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$v = \lambda * f (1)$

Replacing by the givens, and solving for f, we get:

$f =\frac{c}{\lambda} =\frac{3e8m/s}{0.035m} = 8.57e9 Hz (2)$

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$v =\frac{d}{t} (3)$

where v= c= speed of light in vacuum = 3*10⁸ m/s

d= distance between mountaintops = 52 km = 52*10³ m

Solving for t, we get:

$t = \frac{d}{c} = \frac{52e3m}{3e8m/s} = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)$

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2 years ago

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yuradex [85]

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3 years ago

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