Answer:
Area=1.5(1.5)=2.25m^2
Force of gravity=10N
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}
⟼Pressure=
Area
Force
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}
⟼Pressure=
2.25
10
\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}
⟼Pressure=4.4Pa
Answer:
Final temperature of the aluminum = 41.8 °C
Explanation:
We have the equation for energy
E = mcΔT
Here m = 55 g = 0.055 kg
ΔT = T - 27.5
Specific heat capacity of aluminum = 921.096 J/kg.K
E = 725 J
Substituting
E = mcΔT
725 = 0.055 x 921.096 x (T - 27.5)
T - 27.5 = 14.31
T = 41.81 ° C = 41.8 °C
Final temperature of the aluminum = 41.8 °C
v = initial velocity of launch of the stone = 12 m/s
θ = angle of the velocity from the horizontal = 30
Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.
v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s
a = acceleration of the stone = - 9.8 m/s²
t = time of travel = 4.8 s
Y = vertical displacement of stone = vertical height of the cliff = ?
using the kinematics equation
Y = v₀ t + (0.5) a t²
inserting the values
Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²
Y = - 84.1 m
hence the height of the cliff comes out to be 84.1 m
Answer:
Separation of the Earth into layers (crust, mantle, inner core, and outer core) was largely caused by gravitational differentiation (separating different constituents at temperature where materials are liquid or plastic, owing to differences in density) early in Earth's history.
Explanation:
hoped it helped!!
Answer:
Explanation:
Given data
length=100mm
Diameter=5mm
Thermal conductivity=5 W/m.K
Power=50 W
Temperature=25°C
The temperature of heater surface follows from the rate equation written as:
Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium
Substitute the given values
![S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B2%5Cpi%20%280.1m%29%7D%7Bln%5B%5Cfrac%7B4%2A0.1m%7D%7B0.005m%7D%20%5D%7D%5C%5C%20S%3D0.143m)
The temperature of heater is then:
The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.
