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Katena32 [7]
3 years ago
13

What two things can happen in the life of a massive star after it passes from the supernova stage?

Physics
1 answer:
kupik [55]3 years ago
4 0
Medium stars become dwarf stars, large stars become either neutron (pulsar) stars or black holes
You might be interested in
A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on
Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

F_f = ma (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg

And we can fidn the acceleration by using the formula:

a=\frac{v-u}{t}

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N

Where the negative sign means the direction of the force is opposite to the motion of the box.

6 0
3 years ago
A car speedometer has a 4% uncertainty. What is the range of possible speeds (in km/h) when it reads 110 km/h?
Kruka [31]
4% of 110 is 4.4. So the possible range of speeds is the interval from 110-4.4 till 110+4.4.
105.6 till 114.4
4 0
3 years ago
How do we calculate the value of work?
Katarina [22]
The percent complete is calculated by dividing the quantity of material progressed at a point in time by the total quantity required for the project. The resulting percent is multiplied by the current agreed committed value of the material item to obtain the VOWD for that item.
7 0
3 years ago
Read 2 more answers
A cylindrical tungsten filament 14.0 cmcm long with a diameter of 1.00 mmmm is to be used in a machine for which the temperature
Advocard [28]

Answer:

Resistivity ρ=1.12 x 10^-4 Ωm

Explanation:

ρ= RA/l, where R is resistance, A is cross sectional area and l is length

A=πr^2

Note Current is given R is proportion to temperature and inversely proportional to Current R=(20+273)/14*10^-2 =2000Ω

⇒ρ=R*πr^2/l all length in metre.

8 0
3 years ago
A horizontal 953 N merry-go-round of radius 1.68 m is started from rest by a constant horizontal force of 73.9 N applied tangent
solniwko [45]

Answer:

K.E=365.2 J

Explanation:

Given data

Weight w =953 N

radius r=1.68 m

F=73.9 N

t=2.55 s

g=9.8 m/s²

To find

Kinetic Energy K.E

Solution

From the moment of inertia

I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}

The angular acceleration is given as

a=T/I\\a=\frac{FR}{I}\\ a=\frac{(73.9)(1.68)}{137.232}\\a=0.905rad/s^{2}

The angular velocity is given as

w=at\\w=(0.905)(2.55)\\w=2.31rad/s

So the Kinetic Energy is given as

K.E=(1/2)Iw^{2}\\ K.E=(1/2)(137.232)(2.31)^{2}\\ K.E=365.2J

3 0
3 years ago
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