The force applied by the man is 60 N
Explanation:
We can solve this problem by applying Newton's second law, which states that:
(1)
where
is the net force acting on the child+cart
m is the mass of the child+cart system
a is their acceleration
In this problem, we have:
m = 30.0 kg is the mass

And there are two forces acting on the child+cart system:
- The forward force of pushing, F
- The force resisting the cart motion, R = 15.0 N
Therefore we can write the net force as

where R is negative since its direction is opposite to the motion
So eq.(1) can be rewritten as

And solving for F,

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Answer:
<em><u>The</u></em><em><u> </u></em><em>answer</em><em> </em><em>Is</em><em> </em><em>B.</em><em> </em>
Explanation:
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<em><u>follow</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>for</u></em><em><u> </u></em><em><u>more</u></em>
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Answer:
B. τ = 16 Nm
Explanation:
In order to find the torque exerted by the weight attached to the heel of man's foot, when his leg is stretched out. We use following formula:
τ = Fd
here,
τ = Torque = ?
F = Force exerted by the weight = Weight = mg
F = mg = (4 kg)(10 m/s²) = 40 N
d = distance from knee to weight = 40 cm = 0.4 m
Therefore,
τ = (40 N)(0.4 m)
<u>B. τ = 16 Nm</u>
Answer:
Explanation:
The specific heat of gold is 129 J/kgC
It's melting point is 1336 K
It's Heat of fusion is 63000 J/kg
Assuming that the mixture will be solid, the thermal energy to solidify the gold has to be less than that needed to raise the solid gold to the melting point. So,
The first is E1 = 63000 J/kg x 1.5 = 94500 J
the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J
Therefore, all solid is not correct. You will have a mixture of solid and liquid.
For more detail, the difference between E1 and E2 is 7812 J, and that will melt
7812/63000 = 0.124 kg of the solid gold
Answer:
Inner radius = 2 mm
Explanation:
In a coaxial cable, series inductance per unit length is given by the formula;
L' = (µ/(2π))•ln(R/r)
Where R is outer radius and r is inner radius.
We are given;
L' = 50 nH/m = 50 × 10^(-9) H/m
R = 2.6mm = 2.6 × 10^(-3) m
Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m
Plugging in the relevant values, we have;
50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)
Rearranging, we have;
(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)
0.25 = ln((2.6 × 10^(-3))/r)
So,
e^(0.25) = (2.6 × 10^(-3))/r)
1.284 = (2.6 × 10^(-3))/r)
Cross multiply to give;
r = (2.6 × 10^(-3))/1.284)
r = 0.002 m or 2 mm