Answer:
Explanation:
= Velocity of one lump = 
= Velocity of the other lump = 
m = Mass of each lump = 
The collision is perfectly inelastic as the lumps stick to each other so we have the relation
The velocity of the stuck-together lump just after the collision is .
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
Answer:
θ = 22.2
Explanation:
This is a diffraction exercise
a sin θ = m λ
The extension of the third zero is requested (m = 3)
They indicate the wavelength λ = 630 nm = 630 10⁻⁹ m and the width of the slit a = 5 10⁻⁶ m
sin θ = m λ / a
sin θ = 3 630 10⁻⁹ / 5 10⁻⁶
sin θ = 3.78 10⁻¹ = 0.378
θ = sin⁻¹ 0.378
to better see the result let's find the angle in radians
θ = 0.3876 rad
let's reduce to degrees
θ = 0.3876 rad (180º /π rad)
θ = 22.2º
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<span>The two factors that act on parachutes are gravity and air resistance, which is also called drag. Gravity acts as a force to pull parachutes down to the surface of the Earth, while air resistance generates movement in the opposite direction of the falling parachute, and essentially pushes the parachute upward. hope this helps!:)</span>
