Answer:
The speed of the car, v = 21.69 m/s
Explanation:
The diameter is = 48.01 m
Therefore, the radius of the loop R = 24.005 m
Weight at the top is n = mv^2/R - mg
Since the apparent weight is equal to the real weight.
So, mv^2/R - mg = mg
v = √(2Rg)
v = √[2(24.005 m)(9.8 m/s^2)]
The speed of the car, v = 21.69 m/s
Answer: In or discussion of earthquake we want to answer the following questions side has moved to the left, we say that the fault is a left-lateral strike-slip fault. Surface waves behave like S-waves in that they cause up and down and side. Aftershocks are particularly dangerous because that can cause further .
Explanation: Have a nice day!!!
Answer:
No he should not attempt the pass
Explanation:
Let t be the time it takes for the car to pass the truck. The driver should ONLY attempt to pass when the distance covered by himself plus the distance covered by the oncoming car is less than or equal 400 m (a near miss)
At acceleration of 1m/s2 and a clear distance of 10 + 20 + 10 = 40 m, we can use the following equation of motion to estimate the time t in seconds
Within this time frame, the first car would have traveled a total distance of the clear distance (40m) plus the distance run by the truck, which is
8.94 * 25 = 223.6m
So the total distance traveled by the first car is 223.6 + 40 = 263.6m
The distance traveled by the 2nd car within 8.94 s at rate of 25m/s is
8.94 * 25 = 223.6 m
So the total distance covered by both cars within this time frame
223.6 + 263.6 = 487.2m > 400 m
So no, he should not attempt the pass as we will not clear it in time.
Answer:
d’= (0.561 i ^ - 0.634 j ^) m
, d’= 0.847 m
, 48.5 south east
Explanation:
This is a displacement exercise, one of the easiest methods to solve it is to decompose the displacements in a coordinate system. Let's start with beetle 1
Let's use trigonometry to break down your second displacement
d₂ = 0.89 m θ = 32 north east
sin θ = / d₂
d_{2y} = d2 sin 32
d_{2y} = 0.89 sin 32
d_{2y} = 0.472 m
cos 32 = d₂ₓ / d₂
d₂ₓ = d₂ cos 32
d₂ₓ = 0.89 cos 32
d₂ₓ = 0.755 m
We found the total displacement of the beetle 1
X axis
d₁ = 0.58 i ^
Dₓ = d₁ + d₂ₓ
Dₓ = 0.58 + 0.755
Dₓ = 1,335 m
Axis y
D_{y} = d_{2y}
D_{y} = 0.472 m
Now let's analyze the second beetle
d₃ = 1.37 m θ = 35 north east
Sin (90-35) = d_{3y} / d₃
d_{3y} = d₃ sin 55
d_{3y} = 1.35 sin 55
d_{3y} = 1,106 m
cos 55 = d₃ₓ / d₃
d₃ₓ = d₃ cos 55
d₃ₓ = 1.35 cos 55
d₃ₓ = 0.774 m
They ask us what the second displacement should be to have the same location as the beetle 1
Dₓ = d₃ₓ + dx’
D_{y} = d_{3y} + dy’
dx’= Dₓ - d₃ₓ
dx’= 1.335 - 0.774
dx’= 0.561 m
dy’= D_{y} - d_{3y}
dy’= 0.472 - 1,106
dy’= -0.634 m
We can give the result in two ways
d’= (0.561 i ^ - 0.634 j ^) m
Or in the form of module and address
d’= √ (dx’² + dy’²)
d’= √ (0.561² + 0.634²)
d’= 0.847 m
tan θ = dy’/ dx’
θ = tan⁻¹ dy ’/ dx’
θ = tan⁻¹ (-0.634 / 0.561)
θ = -48.5
º
This is 48.5 south east
Answer:
N₂/N₁ = 45
Explanation:
The relation of the number of turns with the voltage is given by the EMF equation of transformer:
(1)
<em>where V₁: voltage of the primary winding, V₂: voltage of the secondary winding, N₁: number of turns in the primary winding, and N₂: number of turns in the secondary winding </em>
Assuming that the primary winding is connected to the input voltage supply, and using equation (1), we can calculate the ratio of N₂ to N₁:
So, the ratio N₂/N₁ is 45.
Have a nice day!