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krok68 [10]
3 years ago
9

13. What is the net force in the free-body diagram below?

Physics
1 answer:
kobusy [5.1K]3 years ago
5 0
The answer might be either C or D. But I’m going toward D.
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A trip is taken that passes through the following points in order
IgorLugansk [536]

Answer:

The displacement from point B to point E is 25.0 m left

8 0
3 years ago
If the man has initial velocity of 0.4 m/s, acceleration is 0.343 m/s^2, a time of 5.8 seconds, and experimental final velocity
agasfer [191]

Answer:

The answer is going to be C.

Explanation:

Trust me. Im an expert in physics

6 0
3 years ago
A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune wi
sertanlavr [38]

Answer:

5.15348 Beats/s

4.55 mm

Explanation:

v_1 = Velocity of sound = 342 m/s

v_2 = Velocity of sound = 346 m/s

f_1 = First frequency = 440 Hz

Frequency is given by

f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz

Beat frequency is given by

|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s

Beat frequency is 5.15348 Hz

Wavelength is given by

\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m

Relation between length of the flute and wavelength is

\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m

At v = 346 m/s

\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m

L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m

Difference in length is

\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm

It extends to 4.55 mm

7 0
3 years ago
If you were to push a pile of books across a table, in which location would this be the easiest?
kati45 [8]

same amount is the answer

6 0
4 years ago
The solar mass of the Sun is 1. The orbital period of Jupiter is 11. 9 Earth years. What is the distance between Jupiter and the
cestrela7 [59]

The distance between Jupiter and the sun is 5.2 AU.

According to Kepler's third law, the square of the period of revolution of planets is proportional to the cube of their mean distances from the sun. From this; T^2 = r^3.

Now, we are told that the orbital period (T) is 11. 9 Earth years. We have to make the distance the subject of the formula.

r =T^2/3

r = (11.9)^2/3

r = 5.2 AU

Learn more: brainly.com/question/15207516

6 0
2 years ago
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