What creates the magnetic field in an electromagnet?
2 answers:
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Answer:
b) d = 997.9 m , c) D = (952.9 i ^ +45 k ^) m , D = 953.6 m and θ = 2.7º
Explanation:
In this exercise we must add the distance traveled, remembering that the displacement is a vector and the distance a scalar.
a) The displacement scheme is the woman walks in her building A to descend to the lower floor 4.0 m, we assume that this includes the vertical displacement, until reaching the street level, the displacement is vertical in this part.
Being on the street, she travels the 6.5 blocks to reach the building where she works, they indicate that each block is 146.6 m, this movement is horizontal.
Upon reaching building B, she goes up to his office on the 14th floor where she travels 5.5m on each floor, it is assumed that the distance to go up to the upper floor is included, this displacement is vertical
b and c) Let's find the distance traveled and the displacement
in building A
Zₐ = 8 * 4.0
Zₐ = -32.0 m k ^
the vector k ^ indicates that the displacement is vertical and the negative sign that it is descending
on the street
= 6.5 146.6
X_{ab} = 952.9 m i ^
the vector i ^ indicates that the displacement is the x-axis, we assume that the axis is in the direction of the displacement
in building B
= 14 * 5.5
Z_{b} = 77 m k ^
displacement in the vertical axis and in the positive direction
now we calculate the distance traveled,
d = Zₐ + X_{ab} + Z_{b}
d = -32 + 952.9 + 77
d = 997.9 m
note that this value is a scalar
Let's calculate the displacement,
Z axis
= Zₐ + Z_{b}
Z_{total} = -32 + 77
Z_{total} = 45 m k ^
X axis
X_{total} = X_{ab}
X_{total} = 952.9 m i ^
we can give the result in two ways
a) D = X_total i ^ + Z_total k ^
D = (952.9 i ^ +45 k ^) m
b) in module form and angles
Let's use the Pythagorean theorem
D² = +
D = √(952.9² + 45²)
D = 953.6 m
We use trigonometry
tan θ = Z / X
θ = tan⁻¹ (Z / X)
θ = tan⁻¹ (45 / 952.9)
θ = 2.7º
this angle is measured from the positive side of the x axis towards the z axis
Answer:
A) 8,368 J
B) ) 0.893 J/gºC
Explanation:
A)
- The heat gained by the water can be obtained solving the following equation:
- where cw = specific heat of water = 4.184 J/gºC
- m= mass of water = 1,000 g
- ΔT = 2ºC
- Replacing these values in (1) we get:
B)
- Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative: -8,368 J.
- Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:
⇒
- which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.