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3 years ago

14

A boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water. How long is he airborn

e before he hits the water?

1 answer:

svp [43]3 years ago

7 0

Answer:

45

Explanation:

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Which factors affect heat transfer between a warm and a cool substance?

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Answer: B

Explanation:

It's not the time it took to heat the substance, so that rules out A and C.

This means that we only have to choose between

B. the area of contact

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A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f

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To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

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$A =0.1256 m^2$

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$\mu=IA$

$\mu=2(0.1256)$

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Which formula can be used to find the X component of the resultant vector?

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What is the torque τb about axis b due to the force f⃗ ? (b is the point at cartesian coordinates (0,b), located a distance b fr

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Check the attached file for the solution.

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3 years ago

Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge

Serhud [2]

Answer:

FC vector representation

$F_{C} =(19.03i+13.78j)N$

Magnitude of FC

$F_{C}=23.495N$

Vector direction FC

$\beta=35.91$ degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

$\alpha = tan^{-1}(\frac{3}{4}) = 36.86$ degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

$F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2}$ Equation (1): Magnitude of the electric force of the charge qA over the charge qC

$F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2}$ Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

$k=8.99*10^9 \frac{N*m^2}{C^2}$

$q_{A}=2.70*10^{-4} C$

$q_{B}=-6.36*10^{-4} C$

$q_{C}=1.04*10^{-4} C$

$r_{AC} =3$

$r_{BC}=\sqrt{4^2+3^2} = 5$

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

$F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N$

$F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N$

Components of the FBC force at x and y:

$F_{BCx}=23.785 *Cos(36.86)=19.03N$

$F_{BCy}=23.785 *Sin(36.86)=14.27N$

Components of the resulting force acting on qC:

$F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N$

$F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N$

FC vector representation

$F_{C} =(19.03i+13.78j)N$

Magnitude of FC

$F_{C}= \sqrt{19.03^2+13.78^2} =23.495N$

Vector direction FC

$\beta = tan^{-1} (\frac{13.78}{19.03})=35.91$ degrees: angle that forms FC with the horizontal

7 0

3 years ago