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Vilka [71]
3 years ago
9

Stored energy and the energy of positions are ________________ energy

Physics
1 answer:
yKpoI14uk [10]3 years ago
3 0

Answer:

Gravitational Energy

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What is the charge on a hypothetical ion with 35 protons and 33 electrons?
kenny6666 [7]

+2 electron charges = 2x1.6x10^-19Coulombs

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What is:<br> the fastest car <br> the fastest animal<br> the fastest human<br> the fastest aeroplane
seropon [69]
The Hennessey Venom GT<span> is the fastest road car in the world.
</span>The fastest land animal<span> is the </span>Cheetah
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4 0
3 years ago
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In a $100$ meter track event, Alice runs at a constant speed and crosses the finish line $5$ seconds before Beatrice does. If it
Liono4ka [1.6K]

Answer:

10s

Explanation:

If it took Beatrice 25 seconds to complete the race

Distance = 100 meter

Beatrice speed = 100/25

                          = 4m/s

If Alice runs at a constant speed and crosses the finish line $5$ seconds, she must have completed the race in 20s (25 -5).

Her speed where constant

= 100/20

= 5 m/s

It would take Alice

= 50/5

= 10s

It would take Alice 10s to run $50$ meters.

5 0
3 years ago
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 ray of light travels across a liquid-to-glass interface. if the indices of refraction for the liquid and glass are, respective
prisoha [69]
When it comes to optics, Snell's law is the basic formula to be used. If you notice, when light hits the water, the light does not travel in the same direction. After, it hits the water, it changes in angle. Light becomes refracted. This is observed when your hands tend to become bigger if you place it underwater. The formula for Snell's Law is

n₁ sin θ₁ = n₂sin θ₂, where n is the index of refraction. This depends on the type of medium. For example, for air, n=1. The parameters θ₁ is the angle of incidence, and θ₂ is the angle of refraction. Critical angle is the incident angle needed so that the refract angle is 90°. So, modifying the equation:

n₁ sin θcrit = n₂sin 90°, since sin 90°=1,
sin θcrit = n₂/n₁
θcrit = sin ⁻¹ (n₂/n₁)

Since liquid comes first before glass, n₁=1.75 and n₂=1.52. Substituting,
θcrit = sin ⁻¹ (1.52/1.75)
θcrit = 60.29°

7 0
3 years ago
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