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labwork [276]
3 years ago
9

Answer plese lang po​

Physics
1 answer:
erastovalidia [21]3 years ago
4 0

Answer:

hi

Explanation:

how to convert gumbauan baboy to kung burger p

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A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
3 years ago
A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in se
Rudik [331]

Answer:

The acceleration of the object is -69.78 m/s²

Explanation:

Given;

postion of the particle:

x = 32t - 38\frac{t^3}{3} \\\\

The velocity of the particle is calculated as the change in the position of the  particle with time;

v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s

Acceleration is the change in velocity with time;

a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2

4 0
3 years ago
Which of the following statements best describes the balls motion ?
Schach [20]

Answer:

C

Explanation:

8 0
3 years ago
Read 2 more answers
Question 8: Cosmology (8 points)
STatiana [176]

Big bang happened about 13.7 billion years ago in our universe.

<h3>Describe the beginning of the universe according to the big bang theory?</h3>

According to the big bang theory, about 13.7 billion years ago, an explosive expansion began, expanding our universe outwards faster than the speed of light.

<h3>Describe the future of the universe according to the flat model?</h3>

According to the flat model, the universe is infinite and will continue to expand forever because the universe is expanding.

<h3>What is cosmic background radiation? </h3>

Cosmic background radiation is a weak radio-frequency radiation that is traveling through outer space in every direction. It is the residual radiation of the big bang, when the universe was very hot.

<h3>How do observations of the cosmic background radiation provide evidence to support the big bang theory? </h3>

The Big Bang theory predicts that the early universe was a very hot place and that as it expands, the gas within it cools. Thus the universe has all over radiation which is called the “cosmic microwave background".

Learn more about big bang here: brainly.com/question/10865002

#SPJ1

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2 years ago
How does a plasma differ from a gas
faltersainse [42]
<span>They are different and unique from the other states of matter. Plasma is different from a gas, because it is made up of groups of positively and negatively charged particles. In neon gas, the electrons are all bound to the nucleus. In neon plasma, the electrons are free to move around the system.

Hope this helps.
</span>
3 0
3 years ago
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