A test car is driving toward a solid crash-test barrier with a speed of 46 mi/h. Two seconds prior to impact, the car begins to
1 answer:
Answer:
we have to measure distances and time, possibly with an automated system since the values are very small
Explanation:
For this exercise we can use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m - mv₀
In the exercise they indicate that the final speed is zero
F t = - m v₀
F = -m v₀ / t
With this equation we can find what measurements should be carried out.
To find the speed with which the car collides with the wall, less measure the displacement and its time during the braking process before reaching the wall and from here find the speed with which it reaches the wall.
During the impact, we must find the time that the vehicle is in contact with the wall in the first approach is equal to the time that the car takes to reach the final speed of zero.
In summary we have to measure distances and time, possibly with an automated system since the values are very small
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Answer:
Minimum diameter of the camera lens is 22.4 cm
The focal length of the camera's lens is 300cm
Explanation:
y = Resolve distance = 0.3 m
h = Height of satellite = 100 km
λ = Wavelength = 550 nm
Angular resolution
From Rayleigh criteria
Minimum diameter of the camera lens is 22.4 cm
Relation between resolvable feature, focal length and angular resolution
The focal length of the camera's lens is 300cm
Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s
The speed of a electron that is accelerated from rest through an electric potential difference of 120 V is
We know, that the energy of the system is always conserved.
Using the Law of Conservation of energy,
U=0
Here, K is the kinetic energy and U is the potential energy.
Now, substituting the formula of U and K, we get:
=0------(1)
Here,
m is the mass of the electron
v is the speed of the electron
q is the charge on the electron
V is the potential difference
Let and
represent the final and initial speed.
Here, =0
Solving for , we get:
=6.49m/s
To learn more about the conservation of energy, refer to:
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