Which of the answers shows three examples of a physical change

What is the magnitude of a vector with components (15 m, 8 m)?

il63 [147K]

The answer is 17 m because you have to add the 15 m and the 8 m together to get the answer so it will be like this 17x17 = 15x15 + 8x8 got it?

8 0

2 years ago

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Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia

Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

Wn = 9.14 x 10¹⁷ N

3 0

2 years ago

An ice skater glides for two meters across ice is work done or no work done

garik1379 [7]

Work is done. work=forcexdisplacement. the ice skater glides 2 meters (displacement), so yes.

5 0

2 years ago

You push on the top edge of a 1.8 m tall solid circular cylinder of iron which is 4.00 cm in diameter. You push with a horizonta

Citrus2011 [14]

Answer:

$\triangle x=3.2*10^-^5 m$

Explanation:

From the question we are told that

Height of circular cylinder is $h= 1.8m$

Diameter of cylinder $D=4cm=>0.04m$

Horizontal Force $F=900N$

$Y = 10.0 *10^1^0 N/m^2 \\B = 9.0 * 10^1^0 N/m^2\\S = 4.0 * 10^1^0 N/m^2\\$

Generally the formula for shear modulus is mathematically represented by

$G=\frac{\tau}{\gamma}$

Where

$G=Shear modulus\\\tau= shear\ stress\\\gamma=shear strain$

$G=\frac{F/A}{\triangle x/L}$

$G=\frac{900/\pi r^2}{\triangle x/1.8}$

$4.0*10^1^0=\frac{900/\pi r^2}{\triangle x/1.8}$

$4.0*10^1^0=\frac{900}{\pi r^2} *\frac{1.8}{\triangle x}$

${\triangle x} =\frac{1620}{\pi r^2* 4.0*10^1^0}$

$\triangle x=3.2*10^-^5 m$

5 0

2 years ago

Help needed Can a body be accelerating if it is moving in circle

belka [17]

Answer:

When a body is moving on a circle it is accelerating because centripetal acceleration is always acting on it towards the center.

Please see the attached picture...

From the above diagram,we can say the acceleration is always acting on the body when it moves in a circle.

Hope this helps.......

Good luck on your assignment.......

5 0

2 years ago

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