What is the acceleration of a 31 kg object pushed with a force of 42N
1 answer:
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Answer:
The magnitude of momentum of the airplane is .
Explanation:
Given that,
Mass of the airplane, m = 3400 kg
Speed of the airplane, v = 450 miles per hour
Since, 1 mile per hour = 0.44704 m/s
v = 201.16 m/s
We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,
or
So, the magnitude of momentum of the airplane is . Hence, this is the required solution.
Answer:
For situation (a)
net charge E = E₊₂ + E₋₅ + E₋₃
E = K(q/d²)
where K = 8.99e9
d = 5.7cm = 5.7e-2m
Therefore,
E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)
E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)
E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)
thus
E = E₊₂ + E₋₅ + E₋₃
= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)
= 7.88e6(x) + 7.88e6(y)
use Pythagorean theorem
I <em>E </em>I = = 1.242e6
∅ = =
= 45°
Thus for (a) net magnitude = 1.115e6 @ 45° above +x axis
for situation (b)
net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆
E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)
E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)
E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)
E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)
thus,
E = E₊₄ + E₊₁ + E₋₁ + E₊₆
= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)
= 7.88e5(x) + 7.88e5(y)
use Pythagorean theorem
I <em>E </em>I = = 1.242e6
∅ = =
= 45°
Thus for (a) and (b) the net magnitude = 1.242e6 @ 45° above +x axis
Explanation:
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