Answer:
27.95[kW*min]
Explanation:
We must remember that the power can be determined by the product of the current by the voltage.
where:
P = power [W]
V = voltage [volt]
I = amperage [Amp]
Now replacing:
Now the energy consumed can be obtained mediate the multiplication of the power by the amount of time in operation, we must obtain an amount in Kw per hour [kW-min]
Answer:
<em>Choice: c. 6sec</em>
Explanation:
<u>Horizontal Launch
</u>
When an object is thrown horizontally with a speed (v) from a height (h), it describes a curved path ruled by gravity until it finally hits the ground.
The horizontal component of the velocity is always constant because no acceleration exists in that direction, thus:
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:
Where
To calculate the time the object takes to hit the ground, we use the same formula as for free-fall, since the time does not depend on the initial speed:
The marble rolls the edge of the table at a height of h=180 m, thus:
t = 6 sec
Choice: c. 6sec
Answer:
No, either driver can not hear a different frequency from the other car's horn than they would if the cars were stationary.
Explanation:
Either driver hear a different frequency from the other car's horn than they would if the cars were stationary if two cars are traveling in the same direction and with the same speed along a straight highway because neither driver experiences a Doppler shift
Answer:
Vd = 1.597 ×10⁻⁴ m/s
Explanation:
Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³
To find:
Drift Velocity Vd=?
Solution:
the formula is Vd = I/nqA (n is the number of charge per unit volume)
n = No. of electron in a mole ( Avogadro's No.) / Volume
Volume = Molar mass / density ( molar mass of Al =27 g)
V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³
n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)
n= 6.02 × 10 ²⁸
Now
Vd = (6A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.9×10⁻⁶ m²)
Vd = 1.597 ×10⁻⁴ m/s
Answer:
there is the increase the temperature of cold body and decrease the temperature of hot body