The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin
θ.
The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.
If the <u>air resistance is ignored</u>, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.
Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
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The kinetic theory of gases is a simple, historically significant model of the thermodynamic behavior of gases, with which many principal concepts of thermodynamics were established. The model describes a gas as a large number of identical submicroscopic particles, all of which are in constant, rapid, random motion
To create the shapes, stars are arranged on a piece of cardboard in the desired configuration. If the stars are placed in a smiley face pattern on the cardboard, for example, they will explode into a smiley face in the sky. In fact, you may see several smiley faces in the sky at one time.
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Answer:
The lungs get rid of carbon dioxide and water vapor. The liver gets rid of bile, which, in addition to breaking down fats, is partially made up of the breakdown of red blood cells. The kidneys get rid of toxins from the blood. The large intestine gets rid of undigested food
Answer:
Explanation:
a. The equation of Lorentz transformations is given by:
x = γ(x' + ut')
x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.
x' = 0
t' = 5.00 s
u =0.800 c,
c is the speed of light = 3×10⁸ m/s
Then,
γ = 1 / √ (1 - (u/c)²)
γ = 1 / √ (1 - (0.8c/c)²)
γ = 1 / √ (1 - (0.8)²)
γ = 1 / √ (1 - 0.64)
γ = 1 / √0.36
γ = 1 / 0.6
γ = 1.67
Therefore, x = γ(x' + ut')
x = 1.67(0 + 0.8c×5)
x = 1.67 × (0+4c)
x = 1.67 × 4c
x = 1.67 × 4 × 3×10⁸
x = 2.004 × 10^9 m
x ≈ 2 × 10^9 m
Now, to find t we apply the same analysis:
but as x'=0 we just have:
t = γ(t' + ux'/c²)
t = γ•t'
t = 1.67 × 5
t = 8.35 seconds
b. Mavis reads 5 s on her watch which is the proper time.
Stanley measured the events at a time interval longer than ∆to by γ,
such that
∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)
c. According to Stanley,
dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m
which is the same as in part (a)
