Answer:
Spring constant in N / m = 6,000
Explanation:
Given:
Length of spring stretches = 5 cm = 0.05 m
Force = 300 N
Find:
Spring constant in N / m
Computation:
Spring constant in N / m = Force/Distance
Spring constant in N / m = 300 / 0.05
Spring constant in N / m = 6,000
The solution to the problem is as follows:
<span>First, I'd convert 188 mi/hr to ft/s. You should end up with about ~275.7 ft/s.
So now write down all the values you know:
Vfinal = 275.7 ft/s
Vinitial = 0 ft/s
distance = 299ft
</span>
<span>Now just plug in Vf, Vi and d to solve
</span>
<span>Vf^2 = Vi^2 + 2 a d
</span><span>BTW: That will give you the acceleration in ft/s^2. You can convert that to "g"s by dividing it by 32 since 1 g is 32 ft/s^2.</span>
We know average speed =total distance/time taken
So avg speed=(85+63)/(5+5)=14.8km/hr
Answer:

Explanation:
<u>Vertical Launch Upwards</u>
In a vertical launch upwards, an object is launched vertically up without taking into consideration friction with the air.
If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

The tennis ball was thrown straight up with a speed of v0=22.5 m/s. The acceleration of gravity is g=9.81\ m/s^2, thus:


Answer:
measuring the zero intensity point, we can deduce the movement of the screen.
The distance from the center of the pattern to the first zero is proportional to the distance to the screen,
Explanation:
The expression for the diffraction phenomenon is
a sin θ = m λ
for the case of destructive interference. In general the detection screen is quite far from the grid, let's use trigonometry to find the angles
tan θ = y / L
in these experiments the angles are small
tan θ = sin θ / cos θ = sin θ
sunt θ = y / L
we substitute
a
= m λ
y = m L λ / a
therefore, by carefully measuring the zero intensity point, we can deduce the movement of the screen.
The distance from the center of the pattern to the first zero is proportional to the distance to the screen, so you can know where the displacement occurs, it should be clarified that these displacements are very small so the measurement system must be capable To measure quantities on the order of hundredths of a millimeter, a micrometer screw could be used.