Answer: The work done is W = 3528J.
Explanation: Work (W) is an amount of energy used to do determined movement. To calculate it, we use the formula, as the movement is horizontal:
W = F.d, where
F is force on the object
d is distance it went through.
The force acting on the crate is due to friction.
= μ .
μ is the kinetic friction
is the normal force
There is no up and down movement, so , with g = 9.8m/s²
= μ . m . g
= 0.9 . 80 . 9.8
= 705.6 N
Substituing into the work formula, we have:
W = . d
W = 705.6 . 5
W = 3528 J
The work done to move the crate across a distance of 5m is 3528 J.
We could determine the acceleration using this formula
Joe accelerated from 35 m/s to 50 m/s in 50 seconds, plug in the numbers
a=
a =
a =
a = 0.3
The acceleration is 0.3 m/s²
Velocity = distance/time
v = (35)/(1/2)
v = 70 km/h
60 km/h for 25 minutes
25 minutes = 25/60 hour
distance= velocity * time
d =(60) * (25/60)
d = 25 km
ΔV = { V(initial) - V(final) } / time
v= (70-60) / (45/60)
average velocity = 13.33 km/h
avg veloticy = 3.7 m/s
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Answer:
Explanation:
Let the magnetic field be B = B₁i + B₂j + B₃k
Force = I ( L x B ) , I is current , L is length and B is magnetic field .
In the first case
force = - 2.3 j N
L = 2.5 i
puting the values in the equation above
- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]
= - 20 B₃ j + 20 B₂ k
comparing LHS and RHS ,
20B₃ = 2.3
B₃ = .115
B₂ = 0
In the second case
L = 2.5 j
Force = I ( L x B )
2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )
= - 20 B₁ k + 20B₃ i
2.3i−5.6k = - 20 B₁ k + 20B₃ i
B₃ = .115
B₁ = .28
So magnetic field B = .28 i + .115 B₃
Part A
x component of B = .28 T
Part B
y component of B = 0
Part C
z component of B = .115 T .
B. Conductors allow electricity to go through them only. Such as copper etc.