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mr Goodwill [35]
2 years ago
6

Magnets are not hammered or rubbed why?​

Physics
1 answer:
Nutka1998 [239]2 years ago
8 0
Lose its magnetic properties
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Particles of atoms
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Answer:

Protons , neutrons and elections
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3 years ago
You drop a stone down off a bridge. You are able to count to 4.0 seconds when it finally hits the water. How high is the bridge?
mart [117]

Answer:

The height of the bridge is 78.4 m.

Explanation:

Given;

time of the stone motion off the bridge, t = 4.0 s

acceleration due to gravity, g = 9.8 m/s²

The height of the bridge is given by;

h = ut + ¹/₂gt²

where;

u is the initial velocity of the stone, u = 0

h = ¹/₂gt²

h = ¹/₂(9.8)(4)²

h = 78.4 m

Therefore, the height of the bridge is 78.4 m.

7 0
3 years ago
Calculate the gravitational attraction between two objects which masses are: mass A: 2.5kg, mass B: 5kg. The distance between th
jonny [76]

From the calculation, the gravitational force of attraction is  1.33 * 10^-14 N.

<h3>What is the gravitational force?</h3>

The gravitational force is an attractive force that acts between any two masses.

It is given by;

F = Gm1m2/r^2

F = 6.67 * × 10−11 * 2.5 * 5/(250)^2

F  = 83.4  × 10−11 /62500

F= 1.33 * 10^-14 N

Learn more about gravitational force:brainly.com/question/12528243

#SPJ1

4 0
2 years ago
Which statement correctly explains scientific theories?
Elan Coil [88]
<span>Scientific theories are tested and proven over time; they are then considered scientific laws.

Sometimes however, they are proven wrong, and so they do not become laws

hope this helps</span>
4 0
3 years ago
Read 2 more answers
A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of
Vladimir79 [104]

Answer:

m = 2.23 \times 10^{-32} kg

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force isa =   \frac{F}{m} = \frac{V^2}{r}

Gravitational forceF= \frac{Gm m}{d^2}

we know that

v = \frac{2\pi R}{T}

solving for

R = \frac{vT}{2\pi}

F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}

F = G\times \frac{\pi m}{(vT)^2}

a = \frac{v^2}{\frac{vT}{2\pi}}

a = \frac{2\pi v}{T}

we know that

f =ma

G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}

solving for m

m = \frac{2Tv^3}{\pi G}

m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}

m = 2.23 \times 10^{-32} kg

5 0
3 years ago
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