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pentagon [3]
4 years ago
8

You are in a spaceship moving in a straight line at constant speed. You cannot see out of the ship. Assuming perfectly uniform m

otion, what can you do to prove you are moving?
Physics
1 answer:
Dimas [21]4 years ago
8 0

Answer:

There is no experiment to prove that you are in motion

Explanation:

A frame of reference which has constant velocity is known as an inertial frame of reference. Motion is relative. One can detect one's motion only when one observes change in position with respect to a fixed body.

Thus, if you are in a spaceship moving at a constant speed in a straight line and unable to look outside, you would not be able to prove that you are moving. Everything within the spaceship would have same speed. If you will throw any object within the spaceship, then the parameters measured by you would also not show that the spaceship is in motion.

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Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
kifflom [539]

Answer:

a)   E = 1.58 10²¹ J , b) Oil = 4,236 107 liter ,  e)   T = 54.3 C

Explanation:

a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface

     I = P / A

     A = 4π r²

in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m

     I = P / A

     I = P / 4π r²

let's calculate

     I = 3,828 10²⁵/4 pi (1.5 10¹¹)²

     I = 1.3539 10²W / m² = 135.4 W / m2

the energy that reaches the disk of the Earth is

    E = I A

the area of ​​a disc

    A = π r²

    E = I π r²

where r is the radius of the Earth 6.37 10⁶ m

     E = 135.4 π(6.37 10⁶)

     E = 1,726 10¹⁶ W

This is the energy per unit of time that reaches Earth

    t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s

     

    E = 1,826 10¹⁶ 86400

     E = 1.58 10²¹ J

b) for this part we can use a direct proportions rule

      Oil = 1.58 10²¹ (1 / 37.3 10⁶)

      Oil = 4,236 10⁷ liter

c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law

       P = σ A e T⁴

       T = \sqrt[4]{P/Ae}

nos indicate the refect, therefore the amount of absorbencies

       P_absorbed = 0.7 P

let's calculate

       T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)

       T = RER (8,676 106)

       T = 54.3 C

b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere

4 0
3 years ago
A paintball is fired horizontally from a tower 45 m above the ground. If the paintball gun fires at 90 m/s… How long does it tak
ankoles [38]

Answer:

The time taken for the paint ball to hit the ground is t =  2.143 \ s

The distance of the landing point from the tower is d = 192.86 \  m

Explanation:

From the question we are told that

   The height of the tower is h =  45 \ m

    The speed of the paintball in the horizontal direction is  v_x =  90 \  m/s

Generally from kinematic equation we have that

      h =  ut + \frac{1}{2} at^2

Here u is the initial  velocity of the paintball in the vertical direction and the value is 0 m/s , this because the ball was fired horizontally

         a is equivalent to g  = 9.8\ m/s^2

        t is the time taken for the paintball to hit the ground

   So    

       45  =  0* t + \frac{1}{2} 9.8 * t^2

=>    t =  2.143 \ s

Generally the distance of its landing position from the tower is

         d = v  *  t

=>      d = 90 *  2.143

=>    d = 192.86 \  m

     

3 0
3 years ago
Please help...<br> Describe energy transformations in your own words.
Fynjy0 [20]
Energy transformations is a process of an energy form transforming into a new form of energy. Like thermal energy, mechanical, etc. An example; when we eat; it turns from chemical energy to mechanical or electrical to heat. Like a plug and to activate a toaster.
7 0
3 years ago
A sound wave of the form s = sm cos(kx - ?t + f) travels at 343 m/s through air in a long horizontal tube. At one instant, air m
Naddik [55]

Answer:

960.24 Hz

Explanation:

Here is the complete question

A sound wave of the form

S=Smcos(kx−ωt+Φ)

travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?

Solution

Given x₁ = 2.0 m, x₂ = 2.070 m, maximum positive displacement s = 6.00 nm at x₁, positive displacement s = 2.00 nm at x₂, velocity of wave v = 343 m/s, maximum positive displacement s₀ = 6.00 nm

Let t₀ = 0 at x₁ = 2.0 m for maximum displacement.

So, s = s₀cos(kx−ωt+Φ)

     6 = 6cos(2k - 0 + Φ) = 6cos(2k + Φ)⇒ cos(2k + Φ) = 6/6 = 1

cos(2k + Φ) = 1 ⇒ (2k + Φ) = cos⁻¹ (1) = 0 ⇒ 2k + Φ = 0

Let t₀ = 0 at x₂ = 2.070 m for displacement s = 2.00 nm.

So, s = s₀cos(kx−ωt+Φ)

     2 = 6cos(2.070k - 0 + Φ) = 6cos(2.070k + Φ)⇒ cos(2.070k + Φ) = 2/6 = 1/3

cos(2.070k + Φ) = 1/3 ⇒ (2.070k + Φ) = cos⁻¹ (1/3) = 70.53 ⇒ 2.070k + Φ = 70.53.

We now have two simultaneous equations.

2k + Φ = 0   (1)

2.070k + Φ = 70.53.    (2)

Subtracting (2) -(1)

2.070k - 2k = 70.53

0.070k = 70.53

k = 70.53/0.070 = 1007.554π/180 rad/m = 17.59 rad/m

k = 2π/λ ⇒ λ = 2π/k

and frequency, f = v/λ = v/2π/k = kv/2π = 17.59 × 343/2π = 960.24 Hz

4 0
3 years ago
What happens to the kinetic energy of a snowball as it rolls across the lawn and gains mass?
Delvig [45]
The kinetic energy will increase
6 0
3 years ago
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