Briefly explain the difference between being self-centered and
2 answers:
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Answer:
The correct option is;
C. The temperature of the air affects the speed of sound
Explanation:
The information given are;
The speed with which sound travels in the desert air = 358 m/s
The speed with which sound travels in the polar air = 330 m/s
The major difference between the desert air and the polar air is that the temperature of the desert air is hotter than the temperature of the air in the polar region. Therefore the speed of sound is affected by the air temperature.
The equation that gives the relationship of the speed of sound in air, , to temperature is presented as follows;
Which shows that the speed of sound in air, , rises as the temperature of the air rises.
Where;
= The temperature of the air in degrees Celsius
= The temperature of the air in degrees Kelvin.
Answer:
Speed of the ball relative to the boys: 25 km/h
Speed of the ball relative to a stationary observer: 35 km/h
Explanation:
The RV is travelling at a velocity of
Here we have taken the direction of motion of the RV as positive direction.
The boy sitting near the driver throws the ball back with speed of 25 km/h, so the velocity of the ball in the reference frame of the RV is
with negative sign since it is travelling in the opposite direction relative to the RV. Therefore, this is the velocity measured by every observer in the reference frame of the RV: so the speed measured by the boys is
v = 25 km/h
Instead, a stationary observer outside the RV measures a velocity of the ball given by the algebraic sum of the two velocities:
v = +60 km/h + (-25 km/h) = +35 km/h
So, he/she measures a speed of 35 km/h.
Answer:
Explanation:
Given that:
A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA
Here:
the initial power factor i.e cos θ₁ = 0.7 lag
θ₁ = cos⁻¹ (0.7)
θ₁ = 45.573°
Active power P = 1500 watts
Apparent power S = 2100 VA
What amount of vars must be added to bring the pf to 0.85
i.e the required power factor here is cos θ₂ = 0.85 lag
θ₂ = cos⁻¹ (0.85)
θ₂ = 31.788°
However; the initial reactive power = P×tanθ₁
the initial reactive power = 1500 × tan(45.573)
the initial reactive power = 1500 × 1.0202
the initial reactive power = 1530.3 vars
The amount of vars that must therefore be added to bring the pf to 0.85
can be calculated as: