Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.
So now we show it > 
Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.
Which the formula for constant acceleration would be > 
The initial velocity is 50mi/h 
When it stops the final velocity is 
Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27
Then we substitute the values in....
So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 
So this means that the car traveled in feet 70.8 ft before it came to a stop.
<h2>
Answer: 745.59 nm</h2>
Explanation:
The diffraction angles 
when we have a slit divided into 
parts are obtained by the following equation:
(1)
Where:
is the width of the slit
is the wavelength of the light
is an integer different from zero
Now, the first-order diffraction angle is given when 
, hence equation (1) becomes:
(2)
We know:
In addition we are told the diffraction grating has 750 slits per mm, this means:
Solving (2) with the known values we will find :
(3)
(4)
Knowing 
:
>>>This is the wavelength of the light, wich corresponds to red.
Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
Answer:
The slope of a velocity graph represents the acceleration of the object. So, the value of the slope at a particular time represents the acceleration of the object at that instant.
