Answer:
Explanation:
Given
Initial speed 
distance traveled before coming to rest 
using equation of motion
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
for 
using same relation we get
divide 1 and 2 we get
So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed
Answer:
the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm
Explanation:
Given the data in the question;
wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m
Index of refraction; n = 1.35
Now, the thinnest thickness of the soap film can be determined from the following expression;
= ( λ / 4n )
so we simply substitute in our given values;
= ( 463 × 10⁻⁹ m ) / 4(1.35)
= ( 463 × 10⁻⁹ m ) / 5.4
= ( 463 × 10⁻⁹ m ) / 4(1.35)
= 8.574 × 10⁻⁸ m
= 85.74 × 10⁻⁹ m
= 85.74 nm
Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm
Answer:
Going from earth to the sun a probe would encounter the next layers in order:
- Corona
- Transition Region
- Chromosphere
- Photosphere
- Convection Zone
- Radiative Zone
- Core
A brief description of them:
Corona is the outermost layer and it cannot be seen with the naked eye, is starts at about 2100 km from the surface of the sun and it has no limit defined.
Transition Region is between the corona and the chromosphere, it has an extension of about 100km
The chromosphere is between 400 km from the surface of the sun to 2100 km. In this layer the further you get away from the sun it gets hotter.
The photosphere is the surface of the sun, the part that we can see, and extends from the surface to 400km.
The convection zone is where convection happens, hot gas rises, cools and rises again.
Radiative Zone is where the photons try to rise to move to higher layers.
The core of the Sun is where nuclear fusion occurs due to the very high temperatures.
Answer:
<em>Second option</em>
Explanation:
<u>Linear Momentum</u>
The linear momentum of an object of mass m and speed v is
P=mv
If two or more objects are interacting in the same axis, the total momentum is
Where the speeds must be signed according to a fixed reference
The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right
The second cart of mass m goes to the right at a speed v
The total momentum before the impact is
The total momentum after the collision is negative, both carts will join and go to the left side
The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven
The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either
The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".
The second option is the correct one because the mass 
has a negative momentum and then the sum of both masses keeps being negative
Answer: a) 2.5 * 10^14, b) t = 1.2*10^-8 s, c) F = 2.2775 * 10^-15 N
Explanation: Since it starts from rest, initial velocity = 0, final velocity (v) = 3*10^6 m/s, distance covered (s) = 1.80cm = 1.80/100 = 0.018m
Since the force on the electron is constant, it acceleration will be constant too thus making newton's laws of motion valid.
Question a)
To get the acceleration, we use the formulae that
v² = u² + 2as
But u = 0
v² = 2as
(3*10^6)² = 2*a*(0.018)
9* 10^12 = 0.036*a
a = 9 * 10^12 / 0.036
a = 250 * 10^12
a = 2.5 * 10^14 m/s².
Question b)
To get the time, we use
v = u + at
But u = 0
v = at
3*10^6 = 2.5 * 10^14 * t
t = 3*10^6 / 2.5*10^14
t = 1.2*10^-8 s
Question c)
To get the force, we use the formulae below
F = ma
F = 9.11*10^-31 * 2.5 * 10^14
F = 22.775 * 10^-17
F = 2.2775 * 10^-15 N
