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3 years ago

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

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: A honeybucket man is carrying his load. He has a pole 2 m long with a bucket hanging from each end. The buckets have a mass of

nasty-shy [4]

Answer:

Explanation:

Using the principle of moment, assuming the rod is uniform rod of mass 1 kg

the center of mass of the rod will be at 1 m

assuming the system is in equilibrium,

clockwise moment = anticlockwise moment

let the distance of the man shoulder be x from the center of gravity and also is the pivot point

total mass of bucket + mass of honey = 2kg + 3 kg = 5 kg for rear bucket and

2kg + 5 kg = 7 kg for front bucket

( 5kg × ( 1+x)) + ( 1 kg × x) = 7 kg × ( 1 - x)

5 + 5 x + x = 7 - 7x

5 + 6x = 7 - 7x

6x + 7x = 7 - 5

13x = 2

x = 2 / 13 = 0.154 m

the honeybucket man's shoulder is 0.154 m from the center of the pole ( forward ).

8 0

3 years ago

A GPS signal travels at the speed of light 300,000,000 m/s. If it takes .05s to reach a phone on the surface of the Earth, how f

lesantik [10]

Nncncnnfkxkxkdnddbdndndkd

6 0

2 years ago

A young hockey player stands at rest on the ice holding a 1.3-kg helmet. The player tosses the helmet directly in front of him w

navik [9.2K]

Answer

given,

initial speed of hockey player= 0 m/s

mass of the helmet, m = 1.3 Kg

initial speed of the helmet, u = 0 m/s

final speed of the helmet, v = 6 m/s

recoil speed of the hockey player, v' = 0.25 m/s

we need to calculate the mass of the hockey player, M = ?

using conservation of momentum

m u + M u' = M v' + m v

initial speed of ice skater is zero

1.3 x 0 + M x 0 = M x (-0.25) + 1.3 x 6

negative sign is taken because recoil velocity is in opposite direction

0 = -0.25 M + 7.8

0.25 M = 7.8

M = 31.2 Kg

Hence, the mass of the young hockey player is equal to 31.2 Kg

5 0

2 years ago

A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete

ale4655 [162]

Answer:

The last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

$cos X=\frac{x^2-y^2-z^2}{-2yz}$

$cos X=\frac{y^2 + z^2-x^2}{2yz}$

substituting the values in the equation we get,

$cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}$

$cos X=0.8951$

X = 26.47°

similarly,

$cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}$

$cos Y=0.649$

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

7 0

3 years ago

List the 4 things technology can be.

inessss [21]

Technology can be the four things listed below...

Phones
Computers
TVs
Playstation or Xbox

6 0

3 years ago

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