No, you are at a constant rate which means that you are always at 40mph
Answer:
Explanation:
The mass of the block is 0.5kg
m = 0.5kg.
The spring constant is 50N/m
k =50N/m.
When the spring is stretch to 0.3m
e=0.3m
The spring oscillates from -0.3 to 0.3m
Therefore, amplitude is A=0.3m
Magnitude of acceleration and the direction of the force
The angular frequency (ω) is given as
ω = √(k/m)
ω = √(50/0.5)
ω = √100
ω = 10rad/s
The acceleration of a SHM is given as
a = -ω²A
a = -10²×0.3
a = -30m/s²
Since we need the magnitude of the acceleration,
Then, a = 30m/s²
To know the direction of net force let apply newtons second law
ΣFnet = ma
Fnet = 0.5 × -30
Fnet = -15N
Fnet = -15•i N
The net force is directed to the negative direction of the x -axis
Answer:
A.B = -38
Explanation:
A = 2i + 9j and B = -i - 4j.
So, A.B = (2i + 9j).(-i - 4j)
= 2i.(-i) + 2i.(-4j) + 9j.(-i) + 9j.(-4j)
= -2i.i - 8i.j - 9j.i - 36j.j
since i.i = 1, j.j = 1, i.j = 0 and j.i = 0, we have
A.B = -2(1) - 8(0) - 9(0) - 36(1)
A.B = -2 - 0 - 0 - 36
A.B = -38
Answer: 0.0163
Explanation: fn=n x v/2L
Fundamental frequency =f1
1 x (343/2L) = 10500
Rearrange the equation
L= v/(2xFn)
L= 343/(2x10500)
L=0.0163
Answer:
Magnetic field, B = 0.275 T
Explanation:
Given that,
Length of the wire, L = 35 cm = 0.35 m
Current carried in the wire, I = 2.6 A
The segment makes an angle of 53∘ with the direction of the magnetic field, 
Magnetic force, F = 0.2 N
To find,
The magnitude of the magnetic field.
Solution,
The magnetic force acting on the wire is given by :
is the angle between the length of wire and the magnetic field.
B = 0.275 T
Therefore, the magnitude of the magnetic field is 0.275 T. Hence, this is the required solution.
