Can some one please answers number 4
1 answer:
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Answer:
v = 15 m / s
Explanation:
In this exercise we are given the position function
x = 5 t²
and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval
let's look for the displacements
t = 0 x₀ = 0 m
t = 3 = 5 3 2
x_{f} = 45 m
we substitute
v = 15 m / s
Answer:
The magnitude of the induced Emf is
Explanation:
The width of the truck is given as 79inch but we need to convert to meter for consistency, then
The width= 79inch × 0.0254=2.0066 metres.
Now we can calculate the induced Emf using expresion below;
Then the
Where B= magnetic field component
L= width
V= velocity
=(40*10^-6) × (42) × (2.0066)
=0.003371V
Therefore, the magnitude emf that is induced between the driver and passenger sides of the truck is 0.003371V
Answer:
(a). The resultant of these forces is 1216.55 N.
(b). The direction of the resultant forces is 80.53°.
Explanation:
Given that,
First force = 1200 N
Second force = 200 N
(a). We need to calculate the resultant of these forces
Using cosine law
Put the value into the formula
The resultant of these forces is 1216.55 N.
(b). We need to calculate the direction of the resultant forces
Using formula of direction
Put the value into the formula
Hence, (a). The resultant of these forces is 1216.55 N.
(b). The direction of the resultant forces is 80.53°.