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Vika [28.1K]
2 years ago
9

Find the change in velocity of a 369 g hockey puck subject to the force shown below.

Physics
1 answer:
Elena-2011 [213]2 years ago
3 0

Answer:

The change in velocity is 15.83 [m/s]

Explanation:

Using the Newton's second law we have:

ΣF = m*a

The force in the graph is 185 N, therefore:

185=0.369*a\\Where\\a=acceleration made it by the force [m/s^2]

a=501.35[m/s^2]

Now using the following kinematic equation:

V^{2}=Vi^{2} + 2*a*(x-xi) \\where\\V=final velocity [m/s]\\Vi= initial velocity [m/s] = 0 the hockey disk is in rest when receives the hit.\\ x = Final position [m] = 0.4 m\\xi = initial position [m] = 0.15m\\

Now replacing the values:

V^{2}=0 + 2*501.35*(0.4-0.15)\\ \\V= 15.83[m/s]

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3 years ago
The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr
WARRIOR [948]
In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
y_n= \frac{n \lambda D}{a} (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
\lambda is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem, 
D=37.0 cm=0.37 m
\lambda=530 nm=5.3 \cdot 10^{-7} m
while the distance between the first and the fifth minima is
y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m (2)

If we use the formula to rewrite y_5, y_1, eq.(2) becomes
\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m
Which we can solve to find a, the width of the slit:
a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}=  1.57 \cdot 10^{-3} m=1.57 mm

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3 years ago
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2. Different forms of energy are related because energy cannot be created or destroyed. they can all be transformed into from one form to another.

Explanation:

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