Answer:
1. 0 J
2. 7500 J
3. 7500 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) of car = 600 Kg
Initial velocity (v₁) of car = 0 m/s
Final velocity (v₂) of car = 5 m/s
Original kinetic energy (KE₁) =?
Final kinetic energy (KE₂) =?
Work used =?
1. Determination of the original kinetic energy.
Mass (m) of car = 600 Kg
Initial velocity (v₁) of car = 0 m/s
Original kinetic energy (KE₁) =?
KE₁ = ½mv₁²
KE₁ = ½ × 600 × 0²
KE₁ = 0 J
Thus, the original kinetic energy of the car is 0 J.
2. Determination of the final kinetic energy.
Mass (m) of car = 600 Kg
Final velocity (v₂) of car = 5 m/s
Final kinetic energy (KE₂) =?
KE₂ = ½mv₂²
KE₂ = ½ × 600 × 5²
KE₂ = 300 × 25
KE₂ = 7500 J
Thus, the final kinetic energy of the car is 7500 J
3. Determination of the work used.
Original kinetic energy (KE₁) = 0
Final kinetic energy (KE₂) = 7500 J
Work used =?
Work used = KE₂ – KE₁
Work used = 7500 – 0
Work used = 7500 J
Answer:
Explanation:
Direction of velocity of electromagnetic wave is given by the formula
E = E j ( vector form )
B = - Bi ( vector form )
Direction of velocity = direction of vector E X B
= E X B
= E j x -Bi
= - EB -k
v = EBk
So the direction of velocity will be along z direction.
Answer:
Low-temperature blackbody
Explanation:
There are 3 types of blackbody temperatures.
Low-temperature blackbody
High temperature extended area blackbody
High-temperature cavity blackbody
A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.
Natural Frequency
Hope that helped!