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3 years ago

After a massive-star supernova, what is left behind?.

1 answer:

8 0

Answer: A Nebula is left behind. A spectacular explosion in which a star ejects most of its mass in a violently expanding cloud of debris.

Hope this helps!

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Using the right amount of significant figures, calculate the answer to the following problem, 215.5+101.02555

cestrela7 [59]

Answer:

This is how I figured it out:

215.5 rounded to one significant figure is 200
101.02555 rounded to one significant figure is 100.
200 + 100 = 300.

Hope this helps!

Explanation:

7 0

3 years ago

An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. Howeve

Vlad1618 [11]

Answer:

$v=0.9833\ c$

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

$\text{Density} = \frac{m}{lwh}$

Given :

Side, b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 $kg/m^3$

So,

$8100=\frac{3.3}{l \times 0.13 \times 0.13}$

$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$

l = 0.024 m

Then for relativistic length contraction,

$l= l' \sqrt{1-\frac{v^2}{c^2}}$

$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$

$0.184= \sqrt{1-\frac{v^2}{c^2}}$

$0.033= 1-\frac{v^2}{c^2}}$

$\frac{v^2}{c^2}= 0.967$

$\frac{v}{c}=0.9833$

$v=0.9833\ c$

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

3 0

3 years ago

A block of mass 0.08 kg is pushed against a spring with spring constant k=31 N/m. The spring is compressed 0.15 meters from its

ELEN [110]

Answer:

1.11 meters

Explanation:

As the spring is compressed, elastic potential energy is built up in the spring. The total elastic potential energy can be found using the following formula

Ep = 1/2 x k x s²

where k = 31 N/m (spring constant)

s = 0.15 m (compression)

Ep = 3.4875 J

When the block of mass is released, the elastic potential energy (Ep) is converted to kinetic energy (Ek). From this we can find the initial velocity of the mass of block after release

Ek = 1/2 x m x u²

where Ek = Ep = 3.4875J

m = 0.08 kg (mass of block)

u = unknown (initial velocity)

u = 2.9526 m/s

Now that we know the initial velocity we need to find the deceleration of the mass of block due to friction. We will first find the force of friction from the following formula

F = ∪ x m x g

where F = unknown (frictional force)

∪ = 0.4 (coefficient of friction)

m = 0.08 kg (mass of block)

g = 9.81 m/s² (acceleration due to gravity)

F = 0.31392 N

From this force we calculate the deceleration based on the following formula

F = m x a

where F = 0.31392 (frictional force)

m = 0.08 kg (mass of block)

a = unknown (acceleration)

a = -3.924 m/s² -

*the negative sign is due to this value being deceleration

Now to find the total distance traveled we use the equation for motion

v² = u² + 2as

where v = 0 (final velocity)

u = 2.9526 m/s (initial velocity

a = -3.924 m/s² (deceleration due to friction)

s = unknown (distance traveled)

s = 1.11 meters

3 0

3 years ago

A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?

katrin [286]

Answer:

k = 25.18 N/m

Explanation:

Simple Harmonic Oscillator

It consists of a weight attached to one end of a spring being allowed to move forth and back.

If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:

$\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}$

If the period is known, we can find the value of the constant by solving for k:

$\displaystyle k=m\left(\frac{2\pi}{T}\right)^2$

Substituting the given values m=5 Kg and T=2.8 seconds:

$\displaystyle k=5 \left(\frac{2\pi}{2.8}\right)^2$

k = 25.18 N/m

5 0

3 years ago

An increase in resistance causes the flow of electric current to

Zolol [24]

An increase in voltage causes the flow of electric current to INCREASE. An increase in resistance causes the flow of electric current to DECREASE.

100% right

6 0

3 years ago

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