Answer:
We know that force applied per unit area is called pressure.
Pressure = Force/ Area
When force is constant than pressure is inversely proportional to area.
1- Calculating the area of three face:
A1 = 20m x 10 m =200 Square meter
A2 = 10 mx 5 m = 50 Square meter
A3 = 20m x 5 m = 100 Square meter
Therefore A1 is maximum and A2 is minimum.
2- Calculate pressure:
P = F/ A1 = 30 / 200 = 0.15 Nm⁻² ( minimum pressure)
P = F / A2 = 30 / 50 = 0.6 Nm⁻² ( maximum pressure)
Hence greater the area less will be the pressure and vice versa.
B.
The child is too old to be gaining something from the screen time.
Answer:
The work done is 360 J.
Explanation:
Given that,
Mass = 50 kg
Distance =3 m
We need to calculate the work done
The work done is equal to the product of force and displacement.
Using formula of work done
Where, F = force
D = distance
θ = Angle between force and displacement
Put the value into the formula
Hence, The work done is 360 J.
This problem has three questions I believe:
>
How hard does the floor push on the crate?
<span>We have to find the net
vertical (normal) Fn force which results from Fp and Fg.
We know that the normal component of Fg is just Fg, which is equal to as 1110N.
From the geometry, the normal component of Fp can be calculated:
Fpn = Fp * cos(θp)
= 1016.31 N * cos(53)
= 611.63 N
The total normal force Fn then is:
Fn = Fg + Fpn
= 1110 + 611.63
=
1721.63 N</span>
> Find the friction
force on the crate
<span>We
have to look for the net horizontal force Fh which results from Fp and Fg.
Since Fg is a normal force entirely, so we can say that the
horizontal component is zero:
Fh = Fph + Fgh
= (Fp * sin(θp)) + 0
= 1016.31 N * sin(53)
=
811.66 N</span>
> What is the minimum
coefficient of static friction needed to prevent the crate from slipping on the
floor?
We just need to compute the
ratio Fh to Fn to get the minimum μs.
μs = Fh / Fn
= 811.66 N / 1721.63 N
<span>=
0.47</span>
