Answer:

Explanation:
information we know:
Total force: 
Weight: 
distance: 
vertical component of the force: 
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In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).
horizontal component: 
vertical component: 
but from the given information we know that 
so, equation these two
and 

and we know the force
, thus:

now we clear for 

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:


whith this horizontal component we calculate the work to move the crate a distance of 4 m:

the work done is W=173.48J
Answer:
The inner planets are closer to the Sun and are smaller and rockier. The outer planets are further away, larger and made up mostly of gas. The inner planets (in order of distance from the sun, closest to furthest) are Mercury, Venus , Earth and Mars.
Explanation:
Here we can use momentum conservation as in this type of collision there is no external force on it

now here we can say




now here we can say


now by coefficient of restitution
for elastic collision we know that e = 1


now by solving the two equation


also we know that

so final speed of the nail is 6.875 m/s
Answer:
The work flow required by the compressor = 100.67Kj/kg
Explanation:
The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .
The work flow can be determined using the equation:
M1h1 + W = Mh2
U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2
Workflow = P2alpha2 - P1alpha1
Workflow = (h2 -U2) - (h1 - U1)
Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)
Workflow = ( 193.191 - 92.519)Kj/kg
Workflow = 100.672Kj/kg
Answer:
Hence the pressure is 
Explanation:
Given data
Q=1500 J system gains heat
ΔV=- 0.010 m^3 there is a decrease in volume
ΔU= 4500 J internal energy decrease
We know work done is
W= Q- ΔU
=1500-4500= -3000 J
The change in the volume at constant pressure is
ΔV= W/P
there fore P = W/ΔV= -3000/-0.01= 3×10^5
Hence the pressure is 