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heat = mass x spec heat x temp rise
40.5=15.4x10^-3xspec heatx11.2
Answer:
799.54 ft
Explanation:
Linear thermal expansion is:
ΔL = α L₀ ΔT
where ΔL is the change in length,
α is the linear thermal expansion coefficient,
L₀ is the original length,
and ΔT is the change in temperature.
Given:
α = 1.2×10⁻⁵ / °C
L₀ = 800 ft
ΔT = -17°C − 31°C = -48°C
Find: ΔL
ΔL = (1.2×10⁻⁵ / °C) (800 ft) (-48°C)
ΔL = -0.4608
Rounded to two significant figures, the change in length is -0.46 ft.
Therefore, the final length is approximately 800 ft − 0.46 ft = 799.54 ft.
Answer:
179.47m/s
Explanation:
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity
Substitute
7750(179)+72(230) = (7750+72)v
1,387,250+16560 = 7822v
1,403,810 = 7822v
v = 1,403,810/7822
v= 179.47m/s
Hence the final velocity of the probe is 179.47m/s
