The horizontal speed is going to be the cosine of the given speed, therefore, the horizontal speed is 19.15 m/s. To find the time, divide the 22 m distance by the velocity. This results in 1.131 seconds, which is in between C and D.
The lithosphere includes the brittle upper portion of the mantle and the crust, the outermost layers of Earth's structure. It is bounded by the atmosphere above and the asthenosphere (another part of the upper mantle) below. Although the rocks of the lithosphere are still considered elastic, they are not viscous
The time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.
<h3>Time of travel of the P-wave</h3>
In rock, S waves generally travel about 60% the speed of P waves, and the S wave always arrives after the P wave.
<h3>Relationship between speed and time</h3>
v ∝ 1/t
v₁t₁ = v₂t₂
t₁/t₂ = v₂/v₁
t₁/t₂ = 0.6v₁/v₁
t₁/t₂ = 0.6
t₁ = 0.6t₂
t₁ = 0.6 x 22 mins
t₁ = 13.2 mins
Thus, the time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.
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The centripetal force experienced by the towel is 55 N.
The given parameters;
- angular speed of the washing machine, ω = 200 rpm
- radius of the machine' drum, r = 0.5 m
- mass of the towel, m = 0.25 kg
The centripetal force experienced by the towel spinning along the walls of the drum is calculated as follows;
Fc = mrω²
where;
<em>Fc is the centripetal force</em>
<em>ω is angular speed in rad/s</em>
The angular speed in rad/s is calculate as;

The centripetal force experienced by the towel is calculated as;

Thus, the centripetal force experienced by the towel is 55 N.
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A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.
<h3>What is specific heat?</h3>
It is the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).
A metallic rod of mass 150 g (m) absorbs 82.5 cal of heat (Q) and its temperature raises from 20 °C to 25 °C. We can calculate the specific heat (c) of the metal using the following expression.
Q = c × m × ΔT
c = Q / m × ΔT
c = 82.5 cal / 150 g × (25 °C - 20 °C) = 0.11 cal/g.°C
A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.
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