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igor_vitrenko [27]
3 years ago
8

If the graph shown is a position-time graph of an object moving at constant velocity, what is the velocity of the object? Assume

that the units of position are meters and the units of time are seconds.

Physics
1 answer:
belka [17]3 years ago
5 0

Answer:

C.) v = 50 m/s

Explanation:

The relationship between position vs. time graph and velocity is that the derivative (slope) of the position vs. time graph gives you velocity. In other words, find the slope to get the velocity.

Select two arbitrary points. I'll choose (3s, 225m) and (0s, 75m).

Now use the slope equation:

v = \frac{y_{2} - y_{1} }{x_{2} - x_{1}} =\frac{225 - 75}{3-0} = 50 m/s

v = 50 m/s

Hopes this helps!

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An open organ pipe is 1.6m long. If the speed of sound is 343m/s, what are the pipes: a) fundamental , b) 1st overtone , & c
Yakvenalex [24]

Answer:

a) 107.1875 Hz

b) 214.375 Hz

c) 321.5625 Hz

Explanation:

L = length of the open organ pipe = 1.6 m

v = speed of sound = 343 m/s

f = fundamental frequency

fundamental frequency is given as

f = \frac{v}{2L}

inserting the values

f = \frac{343}{2(1.6)}

f = \frac{343}{2(1.6)}

f = 107.1875 Hz

b)

first overtone is given as

f' = 2f

f' = 2 (107.1875)

f' = 214.375 Hz

c)

first overtone is given as

f'' = 3f

f'' = 3 (107.1875)

f'' = 321.5625 Hz

3 0
3 years ago
A block of ice of mass 4.30 kg is placed against a horizontal spring that has a force constant k = 250 N/m and is compressed a d
OleMash [197]

Answer:

W = 0.060 J

v_2 = 0.18 m/s

Explanation:

solution:

for the spring:

W = 1/2*k*x_1^2 - 1/2*k*x_2^2

x_1 = -0.025 m and x_2 = 0

W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2

W = 0.060 J

the work-energy theorem,

W_tot = K_2 - K_1 = ΔK

with K = 1/2*m*v^2

v_2 = √2*W/m

v_2 = 0.18 m/s

8 0
3 years ago
which of the following cannot be increased by using a machine of some kind? work, force, speed, torque
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Explanation:

Work cannot be increased by using a machine of some kind.

8 0
2 years ago
A bookcase has a base of 1 m long and 0.5 m wide. It has a mass of 300kg. Find the pressure it exerts on the floor in kPa.
Snezhnost [94]
1m*0.5m=0.5m^2\\
g=10\frac{m}{s^2}\\
300kg*10\frac{m}{s^2}=3000N\\
3000N--0.5m^2\\
\ \ x\ \ --\ 1\\
x=6000Pa=60kPa
4 0
2 years ago
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