Answer:
potential energy = mgh
= 400÷1000 × 10× 4÷100
= 0.4 × 10 × 0.04
=4/10 ×10×4/100
= 4/10 × 4/10
=16/100
= 0.16 joules
m1 (400) stretches 4cm
m1 (100g) stretches 1cm
so, m2(800g) stretches 8 cm
potential energy of m2 = mgh
= 800/1000 ×10×8/100
= 0.8 × 0.8
=8/10 ×8/10
= 64/100
=0.64 joules
Ratio of s1 to s2
16/100 ÷ 64/100
= 1:4 ( answer)
Answer:
3,237.78N
Explanation:
According to Newton's second law of motion
F = mass * acceleration
Since a = v-u/t
F = m(v-u)/t
Given
Mass m = 0.155kg
v = 50.0m/s
u = -44.0m/s
Time t = 0.00450secs
Substitute
F = 0.155(50-(-44))/0.00450
F = 0.155(50+44)/0.00450
F = 0.155(94)/0.00450
F = 14.57/0.00450
F = 3,237.78N
hence he hit the baseball with a force of 3,237.78N
Answer:
aaksj
Explanation:
a) the capacitance is given of a plate capacitor is given by:
C = \epsilon_0*(A/d)
Where \epsilon_0 is a constant that represents the insulator between the plates (in this case air, \epsilon_0 = 8.84*10^(-12) F/m), A is the plate's area and d is the distance between the plates. So we have:
The plates are squares so their area is given by:
A = L^2 = 0.19^2 = 0.0361 m^2
C = 8.84*10^(-12)*(0.0361/0.0077) = 8.84*10^(-12) * 4.6883 = 41.444*10^(-12) F
b) The charge on the plates is given by the product of the capacitance by the voltage applied to it:
Q = C*V = 41.444*10^(-12)*120 = 4973.361 * 10^(-12) C = 4.973 * 10^(-9) C
c) The electric field on a capacitor is given by:
E = Q/(A*\epsilon_0) = [4.973*10^(-9)]/[0.0361*8.84*10^(-12)]
E = [4.973*10^(-9)]/[0.3191*10^(-12)] = 15.58*10^(3) V/m
d) The energy stored on the capacitor is given by:
W = 0.5*(C*V^2) = 0.5*[41.444*10^(-12) * (120)^2] = 298396.8*10^(-12) = 0.298 * 10 ^6 J
After years of observation, Hubble made an extraordinary discovery. In 1923 he spotted a Cepheid variable star in what was known as the Andromeda Nebula. Using Leavitt's techniques, he was able to show that Andromeda was nearly 1 million light years away and clearly a galaxy in its own right, not a gas cloud.
