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2 years ago

If the distance between two objects is increased,the gravitational attraction between them will.....?

Physics

2 answers:

Archy [21]2 years ago

5 0

Attraction will decrease

tia_tia [17]2 years ago

4 0

The attraction will decrease. Hope I helped :)

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A plate carries a charge of 3.8 UC, while a rod carries a charge of 1.9 C. How many electrons must be transferred from the plate

frosja888 [35]

Answer:

$N_{electrons}=Q_{transfered}/q_{electron}=5.94*10^{18}electrons$

Explanation:

The total charge is distributed over the two objects:

$Q_{total}/2=(3.8*10^{-6}C+1.9C)/2=0.9500019C\\$

The plate and the rod must have $Q_{total}/2\\$ . So the charge transferred from the plate to the rod is:

$Q_{transfered}=3.8*10^{-6}C-Q_{total}/2=3.8*10^{-6}C-0.9500019C=-0.9499981C\\$

Number of electrons:

$N_{electrons}=Q_{transfered}/q_{electron}=-0.9499981C/(-1.6*10^{-19}C)=5.94*10^{18}electrons$

4 0

2 years ago

How many elections are shared between one nitrogen atom and one carbon atom?

Charra [1.4K]

Answer:

3 electrons from nitrogen and 3 from carbon while carbon already has a lone pair along with a negative charge (called cyanide)

8 0

3 years ago

Which global concern is indirectly responsible for all of the others? A) pollution B) deforestation C) resource depletion D) hum

vredina [299]

D) human overpopulation

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2 years ago

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Kiley went 5.7 km/h north and then went 5.8 km/h west. From start to finish, she went 8.1 km/h northwest.

TiliK225 [7]

<span>5.7 km/h north and 5.8 km/h west are instantaneous velocities, while 8.1 km/h is the average velocity.

This is because each value has a magnitude and direction so it is a velocity. Moreover, the 8.1 km/h is the resultant of the two velocities so it is the average while the other two are instantaneous.</span>

6 0

2 years ago

Read 2 more answers

A driver has a reaction time of 0.50 s , and the maximum deceleration of her car is 6.0 m/s^2 . She is driving at 20 m/s when su

QveST [7]

Answer:

The car stops after 32.58 m.

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 20 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -6 m/s²

Time taken by the car to stop

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20}{-6}\\\Rightarrow t=3.33\ s$

Total Time taken by the car to stop is 0.5+3.33 = 3.83 s

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=20\times 3.83+\frac{1}{2}\times -6\times 3.83^2\\\Rightarrow s=32.58\ m$

The car stops after 32.58 m.

Distance between car and obstacle is 50-32.58 = 17.42 m

4 0

2 years ago