Answer:
1.024 × 10⁸ m
Explanation:
The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.
So, ω = v₀/8RE
For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have
2v₀ = Rω
substituting ω = v₀/8RE into the equation, we have
2v₀ = v₀R/8RE
dividing both sides by v₀, we have
2v₀/v₀ = R/8RE
2 = R/8RE
So, R = 2 × 8RE
R = 16RE
substituting RE = 6.4 × 10⁶ m
R = 16RE
= 16 × 6.4 × 10⁶ m
= 102.4 × 10⁶ m
= 1.024 × 10⁸ m
Answer:
335°C
Explanation:
Heat gained or lost is:
q = m C ΔT
where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
Heat gained by the water = heat lost by the copper
mw Cw ΔTw = mc Cc ΔTc
The water and copper reach the same final temperature, so:
mw Cw (T - Tw) = mc Cc (Tc - T)
Given:
mw = 390 g
Cw = 4.186 J/g/°C
Tw = 22.6°C
mc = 248 g
Cc = 0.386 J/g/°C
T = 39.9°C
Find: Tc
(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)
Tc = 335
Answer:
0.1 m
Explanation:
It is given that,
Mass of the object, m = 350 g = 0.35 kg
Spring constant of the spring, k = 5.2 N/m
Amplitude of the oscillation, A = 10 cm = 0.1 m
Frequency of a spring mass system is given by :
Time period:
Answer:
Explanation:
Volume of block A = 10 x 6 x 1 = 60 cm³
Mass of block A = 630 g
density of mass A = mass / density
= 630 / 60 = 10.5g / cm³
Volume of block B = 5 x 5 x 3 = 75 cm³
Mass of block A = 604 g
density of mass A = mass / density
= 604 / 75 = 8.05 g / cm³
Since density of both A and B are less than that of mercury , both will float in mercury.
Physics
What is the momentum of a 1.5 × 103 kilogram van that is moving at a velocity of 32 meters/second? A. 46.9 kilogram meters/second B. 4.7 × 103 kilogram meters/second C. 4.85 × 102 kilogram meters/second D. 4.85 × 104 kilogram meters/second