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>
Answer:
W
Explanation:
= Temperature of the room = 22.0 °C = 22 + 273 = 295 K
= Temperature of the skin = 33.0 °C = 33 + 273 = 306 K
= Surface area = 1.50 m²
= emissivity = 0.97
= Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴
Rate of heat transfer is given as
W
Answer:
0.5 m
Explanation:
Givens:
ym1 = 2.5 mm
ym2 = 4.5 mm
Ф_1=π / 4
Ф_2=π / 2
We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is
Ym = (ym1 + ym2)cos(Ф_2/2)
By substitution we have
Ym= (0.025 + 0.045)cos(π/4) = 0.496 m
The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore
Ym^2=(ym1^2+ym2^2)
So we have Ym=√0.025^2+0.045^2
= 0.5 m
Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² = 
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:
Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,
Now, plug in for 'σ' and 
for 
and solve for the electric field. This gives,
Therefore, the electric field between the plates has a magnitude of 5308.34 N/C
Answer:
B ) Ascend using my buddy alternative air source / make an emergency Ascent
Explanation:
From the description it can be seen his buddy is close by of which he can easily use the alternative air source. Also we can see that he is closer to the water surface than his buddy, of which controlled emergency swimming ascent is highly favourable in this condition.
