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Vlad [161]
2 years ago
10

All of the following are possible transition points that occur during adulthood except __________.

Physics
2 answers:
leonid [27]2 years ago
7 0

Answer:

B

Explanation:

.....................

IceJOKER [234]2 years ago
6 0

Answer:

B. Rapid Physical Growths

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Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
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Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

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We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

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2 years ago
A building contractor says that a certain building foundation should support a pressure greater than 13,000 pa if the area of th
Bess [88]

Answer:

d. 1.69 * 10^6 N

Explanation:

Pressure is defined as force divided by area.

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2 years ago
El foco de un automóvil, previsto para operar con una batería de 12 V, es de 40 W. a) ¿Qué resistencia eléctrica tiene su filame
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Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao

Explanation:

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2 years ago
A 50.0-kg crate is being pulled along a horizontal smooth surface. The pulling force is 10.0 n and is directed 20.0° above the h
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Answer:

0.188 m/s^2

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Assuming the crate does not lift above the ground and remains along the floor, then its acceleration will be in the horizontal direction. Therefore, we can use Newton's second law to find its acceleration:

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m is the mass of the crate

a_x is the acceleration

Here we have:

m = 50.0 kg

F_x = F cos \theta = (10.0 N)(cos 20.0^{\circ})=9.4 N is the component of the pulling force along the horizontal direction

Solving for the acceleration,

a_x = \frac{F_x}{m}=\frac{9.4}{50.0}=0.188 m/s^2

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3 years ago
Which of the following is an example of velocity?
pav-90 [236]
C :) cause a direction has nothing do do with velocity !
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