Answer:
Explanation:
A plane flies due north (90° from east) with a velocity of 100 km/h for 2 hours.
With no wind, it will be 100*2 = 200 km north of its starting point.
But a steady wind blows southeast at 30 km/h at an angle of 315° from due east.
So the wind itself will blow the plane 30*2 = 60km at an angle of 315° from due east.
That is the same as 60*cos315° = 42.43km due east and 60*sin315° = -42.43km north.
Combining, the plane is at 42.43km due east and 200-42.43 = 157.57km due north from its starting point.
Answer: 750Kg
Explanation:
Recall that force is the product of the mass M, of an object moving at a uniform acceleration.
i.e Force = Mass x Acceleration
In this case, Mass = ?
Force = 3.00 x 10^3 N = (3.00 x 1000N)
= 3000N
Uniform acceleration = 4.00m/s^2
Force = Mass x Acceleration
3000N = Mass x 4.00m/s^2
Mass = (3000N/4.00m/s^2)
Mass = 750Kg (The SI unit of mass is kilograms)
Thus, the mass of the car is 750Kg
Physical Change
characteristic is the chemical bonds in the substance are unchanged. Because
a physical change is any change happens in an object but without involving
a change in its chemical substance. Example, Solid to liquid change or
also known as melting, liquid to gas change also known as evaporation, gas
to solid change also known as deposition, liquid to solid or
solidification, solid to gas or sublimation, and gas to liquid or
condensation. The physical form of a substance is change into a new form
but the chemical is unchanged.
Answer:
1-d
2- a
3-e
4-b
5-c
This is the correct sequence
Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa
