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3 years ago

5

Two asteroids collide while traveling at constant velocities. Ignoring their gravitational attraction to each other, the forces

between them act:
1.) Only before the collision.

2.) Only during the collision.

3.) Only after the collision.

4.) Before, during, and after the collision.

1 answer:

Natasha2012 [34]3 years ago

6 0

Answer:

Only after the collision

Explanation:

Brainlists please

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A particle of mass 10 g and charge 72 μC moves through a uniform magnetic field, in a region where the free-fall acceleration is

sineoko [7]

Answer:

$-0.07163\hat k\ T$ or 0.07163 T into the page

Explanation:

m = Mass of particle = 10 g

a = Acceleration due to gravity = -9.8j m/s²

v = Velocity of particle = 19i km/s

q = Charge of particle = 72 μC

B = Magnetic field

Here the magnetic and gravitational forces on the particle are applied in the opposite direction so,

$F_b=F_g$

$F_b=qvBsin\theta\\\Rightarrow F_b=qvBsin90\\\Rightarrow F_b=72\times 10^{-6}\times 19000B$

$F_g=ma\\\Rightarrow F_g=0.01\times -9.8$

$72\times 10^{-6}\times 19000B=0.01\times -9.8\\\Rightarrow B=\frac{0.01\times -9.8}{72\times 10^{-6}\times 19000}\\\Rightarrow B=-0.07163\hat k\ T$

The magnetic field is 0.07163 T into the page

5 0

3 years ago

A plane travels 1743 KM in 2 hours 30 minutes. How fast was the plane traveling?

Advocard [28]

Answer:

$v=697.2km/h$

Explanation:

Hello.

In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:

$V=\frac{d}{t}$

The distance is clearly 1743 km and the time is:

$t=2h+30min*\frac{1h}{60min} =2.5h$

Thus, the velocity turns out:

$v=\frac{1743km}{2.5h}\\ \\v=697.2km/h$

Which is a typical velocity for a plane to allow it be stable when flying.

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3 years ago

There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

$F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})$

$F_{net}=0.208 N$

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3 years ago

What do electric forces between charges depend on

denis23 [38]

Answer:

On the magnitude of the charges, on their separation and on the sign of the charges

Explanation:

The magnitude of the electric force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

From the formula, we see that the magnitude of the force depends on the following factors:

- magnitude of the two charges

- separation between the charges

Moreover, the direction of the force depends on the sign of the two charges. In fact:

- if the two charges have same sign, the force is repulsive

- if the two charges have opposite signs, the force is attractive

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Which of the following was not a famous basketball player for the NBA? Select one: a. Wilt Chamberlain b. Lady Bird Johnson c. M

DiKsa [7]

Answer:

B Lady Bird Johnson

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