Answer:
Two types of energy include potential and kinetic energy, which causes motion, which in turn is caused by a force.
Question:
A high ____will have a short wavelength
Answer:
That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength. Light waves have very, very short wavelengths
Explanation:
Hope it help
Answer: High voltage transmission minimizes energy losses during electricity transmission.
Explanation:
When electricity is to be transmitted over a long distance, high voltage transmission is preferred to minimize energy losses due to heat.
The higher the transmission voltage, the lower the current and the lesser the resitance in the wire and the lesser the energy lost due to heat during transmission.
High voltage transmission in kilowatts enables light weight cables to be used for long distance electricity transmission.
This explains why, even though only 110 V may be required at home by some appliances, but electricity is transmitted in kilovolts and often require a stepdown transformer
D = 497.4x10⁻⁶m. The diameter of a mile of 24-gauge copper wire with resistance of 0.14 kΩ and resistivity of copper 1.7×10−8Ω⋅m is 497.4x10⁻⁶m.
In order to solve this problem we have to use the equation that relates resistance and resistivity:
R = ρL/A
Where ρ is the resistivity of the matter, the length of the wire, and A the area of the cross section of the wire.
If a mile of 24-gauge copper wire has a resistance of 0.14 kΩ and the resistivity of copper is 1.7×10⁻⁸ Ω⋅m. Determine the diameter of the wire.
First, we have to clear A from the equation R = ρL/A:
A = ρL/R
Substituting the values
A = [(1.7×10⁻⁸Ω⋅m)(1.6x10³m)]/(0.14x10³Ω)
A = 1.9x10⁻⁷m²
The area of a circle is given by A = πr² = π(D/2)² = πD²/4, to calculate the diameter D we have to clear D from the equation:
D = √4A/π
Substituting the value of A:
D = √4(1.9x10⁻⁷m²)/π
D = 497.4x10⁻⁶m
Answer:
Forces 1 and 2 cannot have equal magnitudes.
Force 1 exceeds force 2.
Explanation:
Numbering the given statements:
- The force of the horse pulling on the cart.
- The force of the cart pulling on the horse.
- The force of the horse pushing on the road
.
- The force of the road pushing on the horse.
- When a horse pulls a cart along a flat road starting from the state of rest, the horse applies a force on the ground in downward-back direction diagonally and as its reaction it moves forward. During this action there is also a reaction force by the road on the horse which exceeds the force of the horse on the ground and as a result the horse moves forward.
- For pulling the cart the horse applies the force on the cart in the forward direction and as a reaction the cart applies a force on the horse in the backward direction which is less than the force of the horse on the cart resulting in the forward motion of the cart along with the horse.