Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.
Ek = mv^2 / 2 — multiply both sides by 2
2Ek = mv^2 — divide both sides by m
2Ek / m = V^2 — switch sides
V^2 = 2Ek / m — plug in values
V^2 = 2*30J / 34kg
V^2 = 60J/34kg
V^2 = 1.76 m/s — sqrt of both sides
V = sqrt(1.76)
V = 1.32m/s (roughly)
Answer:
Anaerobic
Explanation:
aerobic: <em>relating to, involving, or requiring free oxygen.</em>
anaerobic: <em>relating to, involving, or requiring an absence of free oxygen.</em>
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Answer:
(i) C=3Co
(ii) V=Vo/3
(iii) Decrease
Explanation:
(i) capacitance increase K times.
(ii) As charge remain constant so by using Q=CV
you will get potential decrease by K time.
(iii) Decrease due to induced electric field inside
dielectric material..
E=Eo-Eind
Explanation:
heyaaa hope it helps ☺️✌️
When block is pushed upwards along the inclined plane
the net force applied on the block will be given as
here we know that
m = 75 kg
now plug in all values into this
now for finding the power is given as
Answer:
the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
Explanation:
a) Kinetic energy of block = potential energy in spring
½ mv² = ½ kx²
Here m stands for combined mass (block + bullet),
which is just 1 kg. Spring constant k is unknown, but you can find it from given data:
k = 0.75 N / 0.25 cm
= 3 N/cm, or 300 N/m.
From the energy equation above, solve for v,
v = v √(k/m)
= 0.15 √(300/1)
= 2.598 m/s.
b) Momentum before impact = momentum after impact.
Since m = 1 kg,
v = 2.598 m/s,
p = 2.598 kg m/s.
This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,
u = 2.598 kg m/s / 8 × 10⁻³ kg
= 324.76 m/s.
Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
