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3 years ago

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Which are examples of projectile motion?

1 answer:

erastovalidia [21]3 years ago

4 0

I think C and D are the best way to describe projectile motion since both left the ground and stayed aloft for a small or large amount of time. A and B are example of motion but not projectile

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An incident on route 12 summary?I need to describe what happens in the book in a short summary can anyone help?

Serga [27]

Answer: An Incident on Route 12 is presented here in a high quality paperback edition. This popular classic work by James H. Schmitz is in the English language, and may not include graphics or images from the original edition.

Explanation: I HOPE THAT HELPED

5 0

3 years ago

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is $v = 0.0002808 \ m/s$

Explanation:

From the question we are told that

The current on the copper is $I = 20 \ A$

The cross-sectional area is $A = 5.261 \ mm^2 = 5.261 *10^{-6} \ m^2$

The number of copper atom in the wire is mathematically evaluated

$n = \frac{\rho * N_a}Z}$

Where $\rho$ is the density of copper with a value $\rho = 8.93 \ g/m^3$

$N_a$ is the Avogadro's number with a value $N_a = 6.02 *10^{23}\ atom/mol$

Z is the molar mass of copper with a value $Z = 63.55 \ g/mol$

So

$n = \frac{8.93 * 6.02 *10^{23}}{63.55}$

$n = 8.46 * 10^{28} \ atoms /m^3$

Given the 1 atom is equivalent to 1 free electron then the number of free electron is

$N = 8.46 * 10^{28} \ electrons$

The current through the wire is mathematically represented as

$I = N * e * v * A$

substituting values

$20 = 8.46 *10^{28} * (1.60*10^{-19}) * v * 5.261 *10^{-6}$

=> $v = 0.0002808 \ m/s$

8 0

2 years ago

Without friction, what is the mass of an ball accelerating at 1.8 m/sec2 to which an

Juli2301 [7.4K]

Answer:

<h2>23.33 kg </h2>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{42}{1.8} = 23.3333... \\$

We have the final answer as

<h3>23.33 kg</h3>

Hope this helps you

4 0

2 years ago

Read 2 more answers

How was helium first discovered?

lora16 [44]

I think it’s the third one idk tho

3 0

3 years ago

Read 2 more answers

A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux

Mila [183]

Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

$\Phi = EA cos \theta$

where

E is the electric field

A is the cross-sectional area

$\theta$ is the angle between the direction of the electric field and the normal to A

The flux is maximum when $\theta=0^{\circ}$ , so we are in this situation and therefore $cos \theta =1$ , so we can write

$\Phi = EA$

Here we have:

$\Phi = 7.50\cdot 10^5 N/m^2 C$ is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

$A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2$

And so, we can find the magnitude of the electric field:

$E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C$

3 0

3 years ago