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Ivan
2 years ago
5

What would make oppositely charged objects attract each other more? increasing the positive charge of the positively charged obj

ect and increasing the negative charge of the negatively charged object decreasing the positive charge of the positively charged object and decreasing the negative charge of the negatively charged object increasing the distance between the positively charged object and the negatively charged object maintaining the distance between the positively charged object and the negatively charged object
Physics
2 answers:
lapo4ka [179]2 years ago
8 0

Answer:

A :)

Explanation:

got a 100%

kykrilka [37]2 years ago
3 0

Answer:

A.

Explanation:

I did the test :D

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Assume that two of the electrons at the negative terminal have attached themselves to a nearby neutral atom. There is now a nega
devlian [24]

Answer: its negative

Explanation: becase it is

5 0
3 years ago
Cheetahs can accelerate to a speed of 19.6 m/s in 2.45 s and can continue to accelerate to reach a top speed of 27.6 m/s . Assum
marta [7]

Answer:

  • v_{top} = 61.96 \frac{mi}{h}

Explanation:

To express the cheetah's top speed in miles per hour, we just need to find the conversion factor.

We know that the top speed is

v_{top} = 27.7 \frac{m}{s}

So, we want to obtain miles from meters and hours from seconds.

<h3>miles from meters</h3>

First we write the equivalence:

1609.34 \ m = 1 \ mi

Now, we can divide by 1609.34 meters on both sides:

\frac{1609.34 \ m}{ 1609.34 \ m} = \frac{1 \ mi}{ 1609.34 \ m}

The left sides equals 1, so

1 = \frac{1 \ mi}{ 1609.34 \ m}

And this is our conversion factor from meters to miles. Now, we can multiply our top speed by this conversion factor, as the conversion factor equals one, and is dimensionless, the physical meaning will be the same.

v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}

v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}

v_{top} = 0.0172120 \frac{mi}{s}

This is the top speed in miles per second, now, for obtaining miles per hour:

<h3>hours from seconds</h3>

We can do pretty much the same, first, the equivalence:

1 \ h = 3600 \ s

as the seconds are dividing in the velocity, we know divide by 1 hour.

\frac{1 \ h}{ 1 \ h} = \frac{3600 \ s}{ 1 \ h}

1 = \frac{3600 \ s}{ 1 \ h}

and know we just multiply our top speed by this conversion factor

v_{top} = 0.0172120 \frac{mi}{s}  \frac{3600 \ s}{ 1 \ h}

v_{top} = 61.96 \frac{mi}{h}

8 0
3 years ago
A man of 60kg moves in a lift of constant velocity 5m/s .What is the reactive force acting on the man by the elevator?
Flauer [41]

Answer:

588 N

Explanation:

Since the 60 kg is moving at a constant velocity there is no acceleration. In order for the system to be balanced, both the normal force and the force of gravity must be equal. In this case the man has a mass of 60 kg. So to find the force you multiply mass by gravitys constant (9.81). And you end up with an answer of 588.6 but I rounded to 588.

4 0
2 years ago
Read 2 more answers
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia
Eduardwww [97]

Answer:

From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km

So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr

So runner towards the west will be

distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8

So equating east and west time we have

x/9= (11-x)/8

8x=99-9x

17x=99

x=5.92 km

That is the distance covered by runner towards the east and he will meet the runner toward the west at

6-5.92=0.08 km west of the flagpole.

7 0
3 years ago
A spring-loaded gun can fire a projectile to a height h if it is fired straight up. if the same gun is pointed at an angle of 45
Naily [24]
When firing straight up:
v^2 = u^2 - 2gh, where v = final velocity = 0, u = initial velocity, g = gravitational acceleration, h = maximum height attained.

Then,
0 = u^2 - 2gh
u = Sqrt (2gh) ---- (1)

When firing at 45°,
Initial velocity, U = u Sin 45 = Sqrt (2gh)·Sin 45

Maximum height, H = U^2*(Sin Ф)^2/2g

substituting;
H = [Sqrt (2gh)·Sin 45]^2*(Sin 45)^2]/2g
H = [2gh*(Sin 45)^2*(Sin 45)^2]/2g
H = [h*(Sin 45)^4] = h/4

Therefore, maximum height when the gun fires at 45° is a quarter of maximum height when the gun fires vertically up.
6 0
3 years ago
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