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3 years ago

Which force does not operate at a distance of 1 m?

Physics

2 answers:

sleet_krkn [62]3 years ago

8 0

Answer:

Explanation:

liberstina [14]3 years ago

4 0

The Force that doesn’t operate at the distance of 1m is Weak Nuclear

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WILL BE BRAINLIEST HELP!!!

ivolga24 [154]

Answer:

(8.0÷6.0)×10[(-2)-(-3)]

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3 years ago

How to find a planet’s gravitational field strength using its radius?

grin007 [14]

The gravitational field strength is approximately equal to 10 N.

Explanation:

Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.

So the gravitational field strength will be equal to the gravitational force acting on the object.

The formula for gravitational field strength is

$g = \frac{F}{m}$

Here g is the gravitational field strength, m is the mass of the object placed on the surface and F is the gravitational force acting on the object.

Since, the mass of any object placed on the surface of earth will be negligible compared to the mass of Earth, so the mass of the object is considered as 1 kg.

Then the g = F

And $F =\frac{GMm}{r^{2} }$

Here G is the gravitational constant, M is the mass of Earth and m is the mass of the object placed on the surface, while r is the radius of the Earth.

$g = F = \frac{6 \times 10^{24} \times 6.67 \times 10^{-11} \times 1}{(6.6 \times 10^{6}) ^{2} }$

$g = 0.977 \times 10^1= 9.77\ N$

So, the gravitational field strength is approximately equal to 10 N.

5 0

3 years ago

The table below shows the average wind speeds of four hurricanes in Florida. Florida Hurricanes Hurricane | Average Wind Speed (

Anna [14]

Answer:d

Explanation:

3 0

3 years ago

Read 2 more answers

Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l

Masja [62]

Answer:

a) I2 = 3 (o-10) / (o- 30) , b) h ’/h= 3 (o-10) / o (o-30)

Explanation:

The builder's equation is

1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

1 / i1 = 1 / f - 1 / o

1 / i1 = 1/15 - 1 / o

i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

o2 = d-i1

We write the builder equation

1 / f2 = 1 / o2 + 1 / i2

1 / i2 = 1 / f2 -1 / o2

1 / i2 = 1 / f2 - 1 / (d-i1)

1 / i2 = 1/20 - 1 / (d-i1) (1)

Let's evaluate the last term

d-i1 = d - 15 o / (o-15)

d-i1 = (d (o-15) - 15 o) / (o-15)

d- i1 = (30 or -30 15 -15 o) / (o-15)

d-i1 = (15 or - 450) / (o- 15)

d-i1 = = (15 or -450) / (o-15)

replace in 1

1 / i2 = 1/20 - (or - 15) / (15 or -450)

1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

1 / i2 = (-5 or -150) / (15 or -150)

1 / i2 = (or -30) / (3 or - 30)

I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

m = h ’/ h = - i / o

h ’= - h i / o

In our case

h ’= h i2 / o

h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0

3 years ago

A rifle is fired and recoils when the bullet leaves the gun. This is an example of Newton's third law. The acceleration of the b

sweet [91]

The acceleration of the bullet is called ACTION.

Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.


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5 0

3 years ago

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