First, we will get the distance traveled before the driver applied the brakes.
distance = velocity * time
distance = 25*0.34 = 8.5 m
Now, we will calculated the distance that the car traveled after the driver applied the brakes. To do this, we will use the equation of motion:
<span>vf^2 = vi^2 + 2*a*d where:
</span>vf = zero, vi = 25 m/s and a = -7 m/s^2
Note: The negative sign is only to show deceleration
d = <span> 1/2*(625) /(7) = 44.6428 m
The total stopping distance =</span> 8.5 + 44.6428 = 53.1428 m
Answer:
Explanation:
Power = Energy/time
-Don't have energy so I'm gonna solve for it
Gravitational Potential Energy = mass x gravity x height
= 60 kg x 9.8 m/s2 x 5m
= 2940 J
Power = Energy/time
=2940 J/10 s
= 294 W
Answer:

Explanation:
The acceleration of an object is given by:

where
v is the final velocity
u is the initial velocity
t is the time interval it takes for the velocity to change from u to v
For the rocket in this problem,
u = 20,000 m/s
v = 24,000 m/s
t = 55.0 - 5.0 = 50.0 s
Substituting,

Answer with Explanation:
We are given that
Angle of incidence,
Angle of refraction,
a.Refractive index of air,
We know that


b.Wavelength of red light in vacuum,

Wavelength in the solution,

c.Frequency does not change .It remains same in vacuum and solution.
Frequency,
Where 
Frequency,
d.Speed in the solution,

Answer:
26.9 Pa
Explanation:
We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:
(1)
where
is the cross-sectional area of the 1st section of the pipe
is the cross-sectional area of the 2nd section of the pipe
is the velocity of the 1st section of the pipe
is the velocity of the 2nd section of the pipe
In this problem we have:
is the velocity of blood in the 1st section
The diameter of the 2nd section is 74% of that of the 1st section, so

The cross-sectional area is proportional to the square of the diameter, so:

And solving eq.(1) for v2, we find the final velocity:

Now we can use Bernoulli's equation to find the pressure drop:

where
is the blood density
are the initial and final pressure
So the pressure drop is:
