The force of friction is 19.1 N
Explanation:
According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

where
is the net force
m is the mass
a is the acceleration
The bag is moving at constant speed, so its acceleration is zero:

Therefore the net force is zero as well:

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

where
is the horizontal component of the applied force, with
F = 22.5 N

is the force of friction
And solving for
, we find

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Answer:
Jari
Explanation:
The question requires to know who is traveling faster. This is done by comparing the gradients. The steeper the slope (high gradient), the faster the speed and vice versa.
From Jari's line, the starting point is (0, 0) and another point is (6, 7)
The gradient being change in y to change in x
Change in y=7-0=7
Change in x=6-0=6
Slope is 7/6
For Jade, first point is (0, 10) then another point is (6, 16)
Change in y=16-10=6
Change in x=6-0=6
Slope is 6/6=1
Clearly, 7/6 is greater than 6/6 or 1 hence Jari is faster than Jade
Answer and Explanation:
curents i = 2.9 A
i ' = 4.4 A
the magnitude (in T.m) of the path integral of B.dl around the window frame = μo * current enclosed
= μo* ( i '- i )
Since from Ampere's law
where μ o = permeability of free space = 4π * 10 ^-7 H / m
plug the values we get the magnitude (in T.m) of the path integral of B.dl = ( 4π*10^-7 ) (2.9+4.4)
= 1.884 * 10^-6 Tm
Answer:
U = 1794.005 × 10⁶ J
Explanation:
Data provided;
Capacitance of the original capacitor, C = 1.27 F
Potential difference applied to the original capacitor, V = 59.9 kV
= 59.9 × 10³ V
Now,
The Potential energy (U) for the capacitor is calculated as:
Potential energy of the original capacitor, U =
× C × V²
on substituting the respective values, we get
U =
× 1.27 × ( 59.9 × 10³ )²
or
U = 1794.005 × 10⁶ J