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nignag [31]
2 years ago
13

Two positive charges of 1 mC and 10 mC are separated by a distance of 10 m. Find the

Physics
1 answer:
horrorfan [7]2 years ago
3 0

Answer:

900N, charges repel

Explanation:

F = KQq/d²

K = 9 × 10^9Nm²/C²

Q= 10mC = 10 × 10^-3C

q = 1mc = 1 × 10^-3C

d = 10m

F = ?

Force = (9 × 10^9 × 10 × 10^-3 × 1 × 10^-3)/10²

= 9 × 10²

= 900N

It will be an electrostatic force of repulsion since like charges(two positive charges) repel

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How does a helicopter fly ? ( I need it explained in the view of physics)
Zielflug [23.3K]

Answer:

because blades of helicopter create a high pressure difference between lower and higher portion of blade so due to this pressure difference helicopter became able to fly.

Explanation:

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3 0
2 years ago
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Why is it that an object can accelerate while
posledela

For the same reason that you can skate around a curve at constant speed but not with constant velocity.

The DIRECTION you're going is part of your velocity, but it's not part of your speed.

If the DIRECTION changes, that's a change of velocity.

The object doesn't have to change speed to have a different velocity. A change of direction is enough to do it.

And any change of velocity is called acceleration.

3 0
2 years ago
traveling at about 30 mph, how many feet will the average driver cover from the time they see the danger until they hit the brak
barxatty [35]

Answer: 75 ft

Explanation:

Breaking distance = Speed²/ 20

= 30²/20

= 45 feet

Stopping distance = Speed + braking distance

= 30 + 45

= 75 ft

5 0
2 years ago
After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48
masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

15min*\frac{1hr}{60min}=0.25hr

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

V=\frac{distance}{time}

the plane traveled a distance to the north of 48km so the velocity is:

V=\frac{48km}{0.25hr}

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

V_{x}=42km/hr*cos(-19^{o} )=39.71 i

and

V_{y}=42km/hr*sin(-19^{o} )=-13.67 j

So the velocity of the wind can be expressed as a vector as:

V_{wind}=(39.71i - 13.67j)km/hr

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

V_{plane x}=V_{plane/wind x}+V_{wind x}

V_{plane/wind x}=V_{plane x}-V_{wind x}

V_{plane/wind x}=(0-39.71 i)km/hr

V_{plane/wind x}= -39.71 i km/hr

and

V_{plane y}=V_{plane/wind y}+V_{wind y}

V_{plane/wind y}=V_{plane y}-V_{wind y}

V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr

V_{plane/wind x}= 205.67 j km/hr

So the velocity of the plane with respect to the wind can be rewritten as:

V_{plane/wind x}= (-39.71i + 205.67 j) km/hr

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

speed=\sqrt{(-39.71)^{2}+(205.67)^{2}  }

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0
3 years ago
5.1C A fluid flows steadily through a pipe with a uniform crosssectional area. The density of the fluid decreases to half its in
prisoha [69]

The options are;

a. V2 equals 2V1.

b. V2 equals (V1)/2.

c. V2 equals V1.

d. V2 equals (V1)/4.

e. V2 equals 4V1.

Answer:

Option A: V2 equals 2V1

Explanation:

Since the flow is steady, then we can say;

mass flow rate at input = mass flow rate at output.

Formula for mass flow rate is;

m' = ρVA

Thus;

At input;

m'1 = ρ1•V1•A1

At output;

m'2 = ρ2•V2•A2

So, m'1 = m'2

Now, we are told that the density of the fluid decreases to half its initial value.

Thus; ρ2 = (ρ1)/2

Since m'1 = m'2, then;

ρ1•V1•A1 = (ρ1)/2•V2•A2

Now, the pipe is uniform and thus the cross section doesn't change. Thus;

A1 = A2

We now have;

ρ1•V1•A1 = (ρ1)/2•V2•A1

A1 and ρ1 will cancel out to give;

V1 = (V2)/2

Thus, V2 = 2V1

5 0
2 years ago
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