Answer: 13.94 tons/s
Explanation:
On adding heat energy to a substance, the temperature would be changed by a particular amount. This relationship between heat energy and temperature is often different for each material. The specific heat, is a value that describes how they relate.
Heat energy = mass flow rate * specific heat * Δ T
Q = MC (ΔΦ)
Heat energy, Q= 3.5*10^8J
Mass flow rate, M= ?
Specific heat, C= 4184j/KgC
Change in temperature, ΔΦ= 6°C
M = Q/CΔΦ
M = (3.5*10^8)/4184*6
M = 13942kg/s
M = 13.94 tons/s
Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ =
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s
Answer with Explanation:
We are given that
Diameter of pipe,


Volume flow rate of the petroleum along the pipe=

By equation of continuity







1 m=100 cm
Wee can use here kinematics
as we know that

for shorter tree we know that


now since we know that other tree is twice high
So height of other tree is y = 39.2 m
now again by above equation



so the time taken is 2.83 s
Answer:
Explanation:
Rx = -28.2 units
Ry = 19.6 units
magnitude of R = √ [( - 28.2 )² + ( 19.6 ) ]
= √ ( 795.24 + 384.16 )
= 34.34 units
If θ be the angle measured counterclockwise from the +x-direction
Tanθ = 19.6 / - 28.2 = -0.695
θ = 180 - 34.8
= 145.2° .