Answer:
28.7%
Explanation:
efficiency = work output /work input × 100
Answer:
W_apparent = 93.1 kg
Explanation:
The apparent weight of a body is the weight due to the gravitational attraction minus the thrust due to the fluid where it will be found.
W_apparent = W - B
The push is given by the expression of Archimeas
B = ρ_fluide g V
ρ_al = m / V
m = ρ_al V
we substitute
W_apparent = ρ_al V g - ρ_fluide g V
W_apparent = g V (ρ_al - ρ_fluide)
we calculate
W_apparent = 980 50 (2.7 - 0.8)
W_apparent = 93100 g
W_apparent = 93.1 kg
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
Answer:
1.6675×10^-16N
Explanation:
The force of gravity that the space shuttle experiences is expressed as;
g = GM/r²
G is the gravitational constant
M is the mass = 1.0 x 10^5 kg
r is the altitude = 200km = 200,000m
Substitute into the formula
g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²
g = 6.67×10^-6/4×10^10
g = 1.6675×10^{-6-10}
g = 1.6675×10^-16N
Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N
Answer:
1.2 amps :)
Explanation:
A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?
Known:
Unknown:
I = V/R
= 12.0 V / 10.0 Ω
= 1.2 amps
