The average velocity of the object in moving from point-B to point-C is
(the straight-line distance and direction from point-B to point-C)
divided by
(the time the object takes to make the trip) .
Answer:
the size, length, or amount of something, as established by measuring.
Explanation:
Answer:
move upward with increasing speed
I draw fbd diagram , see in figure
when lift going upward -
F - mg = ma
f = m( a + g)
scale so highest reading
The final momentum of the body is equal to 120 Kg.m/s.
<h3>What is momentum?</h3>
Momentum can be described as the multiplication of the mass and velocity of an object. Momentum is a vector quantity as it carries magnitude and direction.
If m is an object's mass and v is its velocity then the object's momentum p is:
. The S.I. unit of measurement of momentum is kg⋅m/s, which is equivalent to the N.s.
Given the initial momentum of the body = Pi = 20 Kg.m/s
The force acting on the body, Pf = 25 N
The time, Δt = 4-0 = 4s
The Force is equal to the change in momentum: F ×Δt = ΔP
25 × 4 = P - 20
100 = P - 20
P = 100 + 20 = 120 Kg.m/s
Therefore, the final momentum of a body is 120 Kg.m/s.
Learn more about momentum, here:
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<h2>
Answer:</h2>
<em>1.33 x 10⁻ ⁴ T outwards.</em>
<em></em>
<h2>
Explanation:</h2>
The equation for the magnetic force (F) on a wire whose length is L and carrying a current I in a magnetic field (B) that is uniform is given by;
F = ILB sin θ ---------------------(i)
Where;
θ = angle between the direction of the current and that of the magnetic field.
From the question,
F = 4.0 × 10⁻² N
I = 12A
L = 25m
θ = 90°
<em>Substitute these values into equation(i) and solve as follows;</em>
4.0 × 10⁻² = 12 x 25 x B x sin 90°
4.0 × 10⁻² = 300 x B x 1
4.0 × 10⁻² = 300B
0.04 = 300B
B = 
B = 0.000133
B = 1.33 x 10⁻ ⁴ T
To get the direction of the magnetic field, the right-hand rule is used.
If the right hand fingers are positioned in the correct order specified by the right hand rule, then it would be seen that the magnetic field is directed outwards.
Therefore, the magnitude and direction of the magnetic field at this location is <em>1.33 x 10⁻ ⁴ T outwards.</em>