Answer:
6.86 × 10²⁴ kg
Explanation:
The mass of the earth m = density of earth, ρ × volume of earth, V
m = ρV
The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m
Since m = ρV
m = ρ4πr³/3
So, substituting the values of the variables into the equation for the mass of the earth, m, we have
m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3
m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3
m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3
m = 6546105.64378π × 10¹⁸ kg/3
m = 20565197.400122 × 10¹⁸ kg/3
m = 6855065.8 × 10¹⁸ kg
m = 6.8550658 × 10²⁴ kg
m ≅ 6.86 × 10²⁴ kg
Answer:Kinetic energy is the energy of motion. All moving objects have kinetic energy. When an object is in motion, it changes its position by moving in a direction: up, down, forward, or backward. 3. A force is a push or pull that causes an object to move, change direction, change speed, or stop.
Explanation: Not sure if that's what you meant but that's the answer I can give you.
Answer:
Both of them reach the lake at the same time.
Explanation:
We have equation of motion s = ut + 0.5at²
Vertical motion of James : -
Initial velocity, u = 0 m/s
Acceleration, a = g
Displacement, s = h
Substituting,
s = ut + 0.5 at²
h = 0 x t + 0.5 x g x t²
Vertical motion of John : -
Initial velocity, u = 0 m/s
Acceleration, a = g
Displacement, s = h
Substituting,
s = ut + 0.5 at²
h = 0 x t + 0.5 x g x t²
So both times are same.
Both of them reach the lake at the same time.
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .
Answer:
0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.
Explanation:
First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:
P₁ = F/A
where,
P₁ = Pressure exerted by empty rack = ?
F = Force exerted by empty rack = Weight of Empty Rack = 40 lb
A = Base Area = 452.4 in²
Therefore,
P₁ = 40 lb/452.4 in²
P₁ = 0.088 psi
Now, we calculate the pressure exerted by the rack along with the coat.
P₂ = F/A
where,
P₂ = Pressure exerted by rack filled with coats= ?
F = Force exerted by filled rack = Weight of Filled Rack = 65 lb
A = Base Area = 452.4 in²
Therefore,
P₂ = 65 lb/452.4 in²
P₂ = 0.144 psi
Now, the difference between both pressures is:
ΔP = P₂ - P₁
ΔP = 0.144 psi - 0.088 psi
<u>ΔP = 0.056 psi</u>
