Help me please, ;) I could use it
2 answers:
Answer:
The solution(s) are in order with respect to the attachments
Joules ; 5. Adding the same amount of heat to two different objects will produce the same increase in temperature ; 2. Same speed in both ; 2. A
Explanation:
Diagram 1 ( Liquid Nitrogen ) : So as you can see, we want our units in Joules here, and can therefore multiply the mass of gaseous nitrogen and the latent heat of liquid nitrogen, to cancel the units kg, and receive our solution - in terms of Joules. Let's do it.
q ( energy removed ) = mass of nitrogen latent heat of liquid nitrogen,
q = 1.3 kg 2.01
10⁵ J / kg =
=
=
=
Joules =
kiloJoules = 2.613
10⁵Joules is the energy that must be removed
Diagram 2 : The same amount of heat does not necessarily mean the same increase in temperature for two different objects. The increase in temperature depends on the specific heat capacity of the substance. Therefore your solution is 5 ) Adding the same amount of heat to two different objects will produce the same increase in temperature.
Diagram 3 : The temperatures in both glasses are the same, and hence the molecules have the same average speed. Therefore your solution is 2 ) Same speed in both.
Diagram 4 : Glass A has more water molecules, and hence has more thermal energy. Your solution is 2 ) A.
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This question is incomplete, the complete question is;
The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.
(a) Determine the forces and and the couple
(b) Determine the sum of the moments about the right end of the beam.
(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?
Answer:
a)
the x-component of the force at A is = 0
the y-component of the force at A is = 400 N
the couple acting at A is; = 146 N-m
b)
the sum of the momentum about the right end of the beam is; ∑ = 0
c)
the equivalent force acting at the left end is; F = -400J ( N)
the couple acting at the left end is; M = - 146 N-m
Explanation:
Given that;
The sum of the forces acting on the beam is zero ∑f = 0
Sum of the moments about the left end of the beam is also zero ∑ = 0
Vector force acting at A, =
+
Now, From the image, we have;
a)
∑f = 0
- 600j + 200j = 0i + 0j
+
- 600j + 200j = 0i + 0j
+ (
- 400)j = 0i + 0j
now by equating i- coefficients'
= 0
so, the x-component of the force at A is = 0
also by equating j-coefficient
- 400 = 0
= 400 N
hence, the y-component of the force at A is = 400 N
we also have;
∑ = 0
- ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0
- 30 N-m - 228 N-m + 112 Nm = 0
- 146 N-m = 0
= 146 N-m
Therefore, the couple acting at A is; = 146 N-m
b)
The sum of the moments about right end of the beam is;
∑ = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)(
) +
∑ = (108 N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m
∑ = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m
∑ = 0
Therefore, the sum of the momentum about the right end of the beam is; ∑ = 0
c)
The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.
The equivalent force at the left end will be;
F = -600j + 200j (N)
F = -400J ( N)
Therefore, the equivalent force acting at the left end is; F = -400J ( N)
Also couple acting at the left end
M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)
M = -(30 N-m) + (112 N-m) - ( 228 N-m))
M = 112 N-m - 258 N-m
M = - 146 N-m
Therefore, the couple acting at the left end is; M = - 146 N-m
