Answer:
Your answer would be C, Radio waves.
Explanation:
Explanation:
We Know That
POTENTIAL ENERGY= MASS*g*HEIGHT
When the objects are lifted to same height then the object with heavier mass would have the highest potential energy
.
Answer:
a) load in Newton is 96,138 b) 129.314mm
Explanation:
Stress = force/ area (cross sectional area of the bronze)
Force(load) = 294*10^6*327*10^-6 = 96138N
b) modulus e = stress/ strain
Strain = stress/ e = (294*10^6)/ (121*10^ 9) = 2.34* 10^ -3
Strain = change in length/ original length = DL/ 129
Change in length DL = 129 * 2.34*10^ -3 = 0.31347
Maximum length = change in length + original length = 129.314mm
The resistance expected of the heater is 50.1 ohms.
<h3>What is resistance?</h3>
Resistance can be defined as the opposition to the flow of electric current in an electric circuit. The S.I unit of resistance is Ohms (Ω).
To calculate the resistance of the heater, we use the formula below.
<h3>Formula:</h3>
- R = V²/P............. Equation 1
Where:
- R = Resistance of the heater
- P = Power of the heater
- V = Voltage supplied to the heater
From the question,
Given:
- V = 480 V
- P = 4.6 kW = 4600 W
Substitute these values into equation 1
- R = (480²)/4600
- R = 50.1 ohms.
Hence, the resistance expected of the heater is 50.1 ohms.
Learn more about resistance here: brainly.com/question/17563681
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F × t = m( v - v₀ )
we substitute
F × 0.00265 = 1.75( 0 - 5.25 )
F × 0.00265 = 1.75( - 5.25 )
F × 0.00265 = -9.1875
F = -9.1875 / 0.00265
F = -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F = - F
we substitute
F = - ( -3466.98 N )
F = 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N
