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3 years ago

Which describes Newton's law of universal gravitation?

Physics

2 answers:

kaheart [24]3 years ago

8 0

Answer:

Gravitational force is directly proportional to the distance between objects.

Anna007 [38]3 years ago

3 0

Answer:all object attracts all other objects

Explanation:

All objects attracts all other objects

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The strength of a magnetic field is an indication of how many flux lines per square inch there are. This is called

Nat2105 [25]

Answer:

gauss

Explanation:

8 0

3 years ago

Please help with a simple Physics question!

zheka24 [161]

Compression- a region in a longitudinal (sound) wave where the particles are closest together. Rarefaction- a region in a longitudinal (sound) wave where the particles are furthest apart. Wave motion and particles.

Answer is B.

6 0

3 years ago

A ship maneuvers to within 2.50 x 10^3 m of an islands 1.80 x 10^3 m high mountain peak and fires a projectile at an enemy ship

Neporo4naja [7]

Answer:

Explanation:

Distance between ship and enemy ship

= 500 + 610

= 3110 m

Range of projectile

R = u² sin2θ / g

= (250x 250 sin 150) / 9.8

= 3188m

The projectile falls within a distance of 3188 - 3110 = 78 m from enemy ship

Height of mountain = 1800 m

We shall find the height of projectile when its horizontal displacement is 2500m

x = 2500 , y = ?

u = 2500 ,

y = x / cos θ - .5 g x² /u²cos² θ

$\frac{2500}{.2588} - \frac{.5 \times9.8\times2500\times2500}{250\times250\times.2588}$

9660 - 7315 m

= 2345 m

It is within 545 m from mountain peak .

8 0

3 years ago

Q6. When a girl steps on to a weighing machine, it shows a reading of 42kg.

tigry1 [53]

Answer:

92.5 pounds

Explanation:

5 0

2 years ago

Two very long uniform lines of charge are parallel and are separated by 0.300 m. each line of charge has charge per unit length

Alenkasestr [34]

linear charge density of system of two line charges is given as

$\lambda = 5.20 \muC/m$

now as we know that electric field due to a line charge at some distance from it is given by

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

so here we will first find the electric field of first line charge at the position of other line charge

$E = \frac{5.20 * 10^{-6}}{2 \pi * 8.85 * 10^{-12}* 0.300}$

$E = 312000 N/C$

now as we know that

$F = qE$

here q = charge on the line charge system at which force is required

E = electric field on that system of charge where force is required

now we can find the charge by

$q = \lambda * L$

$q = 5.20 * 10^{-6}* 0.05 = 0.26 * 10^{-6} C$

Now using the above formula

$F = qE$

$F = 0.26 * 10^{-6} * 312000$

$F = 0.0811 N$

so force on the part of wire is F = 0.0811 N

8 0

3 years ago