Answer:
Efficiency of the machine = 75%
Explanation:
Given:
Input work = 8,000 J
Output work = 6,000 J
Find:
Efficiency of the machine
Computation:
Efficiency of the machine = [Output work / Input work]100
Efficiency of the machine = [6,000 / 8,000]100
Efficiency of the machine = 75%
Potential energy is simply energy that can be but isn't yet
it can also be changed by putting it into motion or setting it off by using force or enacting gravity by dropping it.
Answer:
6.4 J
Explanation:
m = mass of the bullet = 10 g = 0.010 kg
v = initial velocity of bullet before collision = 1.8 km/s = 1800 m/s
v' = final velocity of the bullet after collision = 1 km/s = 1000 m/s
M = mass of the block = 5 kg
V = initial velocity of block before collision = 0 m/s
V' = final velocity of the block after collision = ?
Using conservation of momentum
mv + MV = mv' + MV'
(0.010) (1800) + (5) (0) = (0.010) (1000) + (5) V'
V' = 1.6 m/s
Kinetic energy of the block after the collision is given as
KE = (0.5) M V'²
KE = (0.5) (5) (1.6)²
KE = 6.4 J
The electric force on the proton is:
F = Eq
F = electric force, E = electric field strength, q = proton charge
The gravitational force on the proton is:
F = mg
F = gravitational force, m = proton mass, g = gravitational acceleration
Since the electric force and gravitational force balance each other out, set their magnitudes equal to each other:
Eq = mg
Given values:
q = 1.60×10⁻¹⁹C, m = 1.67×10⁻²⁷kg, g = 9.81m/s²
Plug in and solve for E:
E(1.60×10⁻¹⁹) = 1.67×10⁻²⁷(9.81)
E = 1.02×10⁻⁷N/C
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Hope this helps!
