Find the work needed to lift a 20-N book 2 m.
1 answer:
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Answer:
284.4233 N/m
Explanation:
k = Spring constant
x = Compression of spring = 14.5 cm
U = Potential energy = 2.99 J
The potential energy of a spring is given by
Rearranging to get the value of k
The spring constant is 284.4233 N/m
The normal force acting on the object is 500 N in the upward direction
<u>Explanation:</u>
As George is applying a downward force, the normal force will be in the upward direction. The normal force will be exerted due to the acceleration due to gravity exerted on the object.
So, as per Newton's second law, the normal force acting on the object can be measured by the product of mass of the object and the acceleration due to gravity acting on the object.
But as the acceleration due to gravity is a downward acting acceleration and the normal force is a upward acting force, so the acceleration will be having a negative sign in the formula.
Here, acceleration due to gravity g = -10 m/s² and mass is given as 50 kg, then
Normal force = 50 × (-10) = -500 N
So, the normal force acting on the object is 500 N in the upward direction.
Answer:
Below
Explanation:
First draw the vectors that represent both electric fields.
E1 is the elictric field created by q1, E2 is the one created by q2.
● q1 is negative so E1 will point from P.
● q2 is positive so E2 will point out of P
(Picture below)
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The resulting electric field is equal to the sum of the two fields since both vectors are colinear.
Let E be the total field.
● E = E1 + E2
The formula of the electric field intensity is:
● E = K ×(q/d^2)
-K is Coulomb's constant
-d is the distance between the charge and the object ( here P)
-q is the charge
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● E1 = K × (q1/d1^2)
The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)
Coulombs constant is 9×10^9 m^2/C^2
● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]
● E1 = -359.43 N/C
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● E2 = K ×(q2/d^2)
The distance between q2 and P is 0.25 m.
● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]
● E2 = 463.68 N/C
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● E = E1 + E2
● E = -359.43+463.68
● E = 105.25 N/C
