Complete question:
A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa
Answer:
The average pressure exerted on the surface by the block is 9655.17 Pa
Explanation:
Given;
density of the lead, ρ = 1.13 x 10⁴ kg/m³
mass of the lead block, m = 20 kg
surface area of the area of the block, A = 2.03 x 10⁻² m²
Determine the force exerted on the surface by the block due to its weight;
F = mg
F = 20 x 9.8
F = 196 N
Determine the pressure exerted on the surface by the block
P = F / A
where;
P is the pressure
P = 196 / (2.03 x 10⁻²)
P = 9655.17 N/m²
P = 9655.17 Pa
Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa
Answer:
It would be 2600
Explanation:
M/S stands for meters per second. If it moved 1 meter for 2600 seconds, than it would be 2600. You just multiply 2600 by 1! I hope this helps :D
Answer:
d. All of these
Explanation:
work is said to be done when a force is applied to an object through a certain distance. the SI unit of workdone is joules or newton per meter
mathematically
workdone = force x distance.
from the answers, work is being done because there is force applied in a certain distance.
- from wagon is used to carry vegetables from a garden.
- pulley is used to get water from a well.
- hammer is used to remove a nail from a wall.
Earth's gravity pulls air as close to the surface as possible. As altitude increases, the amount of gas molecules in the air decreases—the air becomes less dense than air nearer to sea level.
Answer:Orbital period =21.22hrs
Explanation:
given that
mass of earth M = 5.97 x 10^24 kg
radius of a satellite's orbit, R= earth's radius + height of the satellite
6.38X 10^6 + 3.25 X10^7 m =3.89 X 10^7m
Speed of satellite, v= 
where G = 6.673 x 10-11 N m2/kg2
V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)
V =10,241082.2
v= 3,200.2m/s
a) Orbital period
= 
V= 
T= 2
r/ V
= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s
=76,385.1 s
60 sec= 1min
60mins = 1hr
76,385.1s =hr
76,385.1/3600=21.22hrs