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V125BC [204]
3 years ago
7

What type of objects exert gravitational pull? (5 points) Only living objects Only space objects All objects having mass All sub

stances that are visible
Physics
1 answer:
Yanka [14]3 years ago
6 0

Answer:

Every object in the universe that has mass exerts a gravitational pull, or force, on every other mass. The size of the pull depends on the masses of the objects. You exert a gravitational force on the people around you, but that force isn't very strong, since people aren't very massive.

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What energy powers the star throughout its life?
aleksley [76]
<span>It is called Nuclear fusion. Stars are controlled by atomic combination in their centers, generally changing over hydrogen into helium. The generation of new components by means of atomic responses is called nucleosynthesis. A star's mass figures out what another kind of nucleosynthesis happens in its center.</span>
3 0
3 years ago
What is the density of a rock with a mass of 10 grams and a volume of 5 ml
Talja [164]
Density= mass/volume
So the answer is 2g/ml
Hope this helps!
3 0
3 years ago
Beyond what point must an object be squeezed for it to become a black hole
Naddik [55]

Answer:

Condition

\frac{m}{R} \geq \frac{c^{2} }{2*g}

Explanation:

Basically black hole is an object from which light rays can not escape it means   to go out from gravitational field , that body should thrown with speed greater then light.

Let's do some calculation

Gravitational potential at surface =-\frac{G*M*m}{R}

If we give kinetic energy equal to magnitude of Potential energy as on surface it will escape.

\frac{m*v^{2} }{2} =-\frac{G*M*m}{R}

⇒\frac{m}{R} =\frac{c^{2} }{2*G}

It will be more better for black hole if above ratio (analogous to density ) is more then above calculated

4 0
3 years ago
550 g of water at 105°C is poured into an 855 g aluminum container with an initial temperature of 11°C. The specific heat of alu
nexus9112 [7]

Answer:

54.22 kJ

Explanation:

In this case, we need to calculate the heat. The expression to use is:

Q = m * C * ΔT

Now, with the specific heat of water (4186 J/ kg K), we can calculate the temperature in which this occurs.

So:

Q = 0.550 * 4186 * (105 - T)

Q = 2302.3 (105 - T)

Q = 241,741.5 - 2302.3T

Now with the Aluminium:

Q = 0.855 * 900 * (T - 11)

Q = 769.5T - 8464.5

Now, with both equations, we solve for the final temperature:

769.5T - 8464.5 = 241,741.5 - 2302.3T

(2302.3 + 769.5)T = 241,741.5 + 8464.5

3071.8T = 250,206

T = 81.45 K

This is the temperature which the change occurs. Now, let's determine the amount of heat from water to Al:

Q = 241,741.5 - 2302.3(81.45)

Q = 54,219.17 J or simply 54.22 kJ.

5 0
3 years ago
the fireman wishes to direct the flow of water from his hose to the fire at b. determine two possible angles u1 and u2 at which
Lelechka [254]

The two possible angles obtained by using the qudratic equation are;

θ_{1} = 15.10° and θ2 = 73.51°

Given, speed of water = v_{A} = 50ft/s

For the motion along x direction, time period can be calculated as follows:

s_{x} = (v_{A}) _x_{} } t

35 = (50 × cosθ) t

t = 0.64 / cosθ

For the motion in y direction, an equation can be obtained as follows:

s_{y} = (v_{A})_{y}  t +\frac{1}{2} (a_{y} )t^{2}

s_{y} = (-v_{A}sinθ) }  t +\frac{1}{2} (a_{y} )t^{2}

Plugging in the values we get:

-20 = (-50_sinθ) }  t +\frac{1}{2} (-32.2} )t^{2}

-20 = -32tanθ - 10.304sec^{2}θ

Upon solving the above quadratic equation, we get,

tanθ = 0.27 , -3.38

Therefore,

tanθ_{1} = 0.27

θ_{1} = 15.10°

and, tanθ_{2} = -3.38

θ_{2} = 73.51

Learn more about quadratic equation here:

brainly.com/question/17177510

#SPJ4

8 0
10 months ago
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