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3 years ago

Question 7 of 25

Physics

2 answers:

Crank3 years ago

5 0

Answer: 350 ms

Explanation:

Just took the quiz:)

tino4ka555 [31]3 years ago

5 0

Answer:

B. 350

Explanation:

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Can someone solve this problem and explain to me how you got it

likoan [24]

Answer:

Explanation:

Coulomb's law states that the force of attraction or repulsion between any two charges is proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between the charges

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question 7:

Given: $F=1.9*10^{-29}N$ , $q1=q2=1.6*10^{-19}C$

By coulomb's law,

$1.9*10^{-29}=9*10^{9}*\frac{1.6*10^{-19}*1.6*10^{-19}}{r^{2}}$

⇒ $r=3.5m$

Question 6:

Given: $q1=q2=-1.5*10^{-6}C$ , $r=0.28m$

By coulomb's law,

$F=9*10^{9}*\frac{1.5*10^{-6}*1.5*10^{-6}}{0.28^{2}}$

⇒ $F=0.26N$

3 0

3 years ago

What do you call the curved upper surface of the water that comely appears when using a graduated cylinder

algol [13]

The curved surface of water is called the meniscus

6 0

3 years ago

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What do borrowers use to secure a mortgage loan? Check all that apply.

Musya8 [376]

Answer:

a house or a land is used to secure a mortgage loan

5 0

2 years ago

An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed o

Vesnalui [34]

Answer:

a) 75.5 degree relative to the North in north-west direction

b) 309.84 km/h

Explanation:

a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it

So the pilot should head to the West-North direction at an angle of

$cos(\alpha) = 80/320 = 0.25$

$\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0$ relative to the North-bound.

b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is

$320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h$

7 0

3 years ago

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I just want to shift to another country to learn like London so who can tell me the progress

scoray [572]

Answer:

by shift do you mean travel

Explanation:

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3 years ago