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EleoNora [17]
4 years ago
8

What is the net force on the loop from the wire if the loop carries a clockwise current i2 = 50 a?

Physics
1 answer:
natima [27]4 years ago
5 0
WE are given
I2 = 50 A
The formula for the net force is
F = I2 L B
where I2 is the current
L is the length
and
B is the magnetic field
If we know L and B, we can substitute the values into the formula and solve fore the force
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Answer:

h = 69.6 m

Explanation:

Data:

  • Vo = 25 m/s
  • t = 2.0 s
  • g = 9.8 m/s²
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Formula:

  • \boxed{\bold{h=V_{0}*t+\frac{g*(t)^{2}}{2}}}

Replace and solve:

  • \boxed{\bold{h=25\frac{m}{s}*2.0\ s+\frac{9.8\frac{m}{s^{2}}*(2.0\ s)^{2}}{2}}}
  • \boxed{\bold{h=50\frac{m}{s^{2}}+\frac{9.8\frac{m}{s^{2}}*4\ s^{2}}{2}}}
  • \boxed{\bold{h=50\ m+\frac{39.2\ m}{2}}}
  • \boxed{\bold{h=50\ m+19.6\ m}}
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The building has a height of <u>69.6 meters.</u>

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3 years ago
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In this electric motor, an electric current causes the coil to rotate, changing
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4 years ago
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4 0
3 years ago
A 900 N crate is being pulled across a level floor by a force F of 340 N at an angle of 23° above the horizontal. The coefficien
Olegator [25]

Answer:

0.958891203 m/s²

Explanation:

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\mu = Coefficient of friction = 0.25

Force of friction acting on the force applied

f=\mu N\\\Rightarrow f=0.25\times 900\\\Rightarrow f=225\ N

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F=340cos 23\\\Rightarrow F=312.97165\ N

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F_n=F-f\\\Rightarrow F_n=312.97167-225\\\Rightarrow F_n=87.97167\ N

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a=\frac{F_n}{m}\\\Rightarrow a=\dfrac{87.97167}{\dfrac{900}{9.81}}\\\Rightarrow a=0.958891203\ m/s^2

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