When two forces act in the same direction on an object, the net force is found by ....................the forces.
2 answers:
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a) Vector product: |d1 × d2| = (34.2 m) k
b) Scalar product: d1 · d2 = -0.874 m
c) (d1 + d2) · d2 = 16.1 m
d) Component of d1 along direction of d2: -0.21 m
Explanation:
a)
In this part, we want to calculate
|d1 × d2|
Which is the vector product between the two displacements d1 and d2.
The two vectors are:
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
The vector product of two vectors and
is also a vector which has components:
We notice immediatly that in this problem ,the two vectors d1 and d2 lie in the x-y plane, so they do not have components in zero. Therefore, the vector product has only one component, which is the one in z, and it is:
Therefore, the vector product of d1 and d2 is:
|d1 × d2| = (34.2 m) k
b)
In this case, we want to calculate
d1 · d2
Which is the scalar product between the two displacements.
The scalar product of two vectors and
is a scalar given by:
In this problem,
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
Therefore, the scalar product between the two vectors is:
c)
In this case, we want to calculate
(d1 + d2) · d2
Which means that first we have to calculate the resultant displacement d1 + d2, and then calculate the scalar product of the resultant vector with d2.
Given two vectors and
, the resultant vector is also a vector given by
In this case,
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
So the resultant vector is
So
And calculating the scalar product with d2, we find:
d)
The component of a vector a along another vector b is given by
where is the scalar product between and b
is the magnitude of vector b
In this problem, we have the two vectors
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
We want to find the component of d1 along the direction of d2.
We already calculated the scalar product of the two vectors in part b):
d1 · d2 = -0.874 m
The magnitude of a vector b is given by
So, for vector d2,
Now we can calculate the component of d1 along d2:
Learn more about operations with vectors:
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Answer:
The temperature change is -633.15K
Explanation:
If we considered the expansion as a reversible one (to be adiabatic is one of the requirements), the work done by expansion can be written as:
Where 2 and 1 subscripts mean the final and the initial state respectively. The equation negative sign says that for an expansion of the gas, the system is making work, so the energy is going out of the system.
Using the ideal gas equation, it is possible to change volume and pressure by temperatures:
So,
This result makes sense considering that the volume increases, so it is expected that the temperature decreases.