Given conditions:
height of object = 7.5cmdistance of object from mirror = 14 cmfocus length = -7 cmimage distance = ?
Using mirror formula:
1/(focus length) = 1/(object distance) + 1/(image distance)
or, -1/7 = 1/14 + 1/(image distance)
or, image distance = -4.66cm (the image formed is a virtual image)
Also, magnification of image is:
image height /height of object = - image distance /object distance
or, image height = - image distance / object distance * height of object
or, image height = -(-4.66) / 14 * 7.5 = 2.49 = 3(nearest whole number)
The acceleration on surface of moon =1.67m/s^2
Weight =mass ×acceleration
=7×1.67
=11.69N
Answer:
Explanation:
change in the volume of the gas = 5.55 - 1.22
= 4.33 X 10⁻³ m³
external pressure ( constant ) P = 1 x 10⁵ Pa
work done on the gas
=external pressure x change in volume
= 10⁵ x 4.33 X 10⁻³
=4.33 x 10²
433 J
Using the formula
Q = ΔE + W , Q is heat added , ΔE is change in internal energy , W is work done by the gas
Given
Q = - 124 J ( heat is released so negative )
W = - 433 J . ( work done by gas is negative, because it is done on gas )
- 124 = ΔE - 433
ΔE = 433 - 124
= 309 J
There is increase of 309 J in the internal energy of the gas.
Answer:
vT = v0/3
Explanation:
The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R
Since F = F0 = F'
GMm/R² = m(v0)²/R
GM = (v0)²R (1)
Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R
Since F1 = F'
GMm/27R² = m(vT)²/3R
GM = 27(vT)²R/3
GM = 9(vT)²R (2)
Equating (1) and (2),
(v0)²R = 9(vT)²R
dividing through by R, we have
9(vT)² = (v0)²
dividing through by 9, we have
(vT)² = (v0)²/9
taking square-root of both sides,
vT = v0/3
Do the same thing you were doing