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Jet001 [13]
3 years ago
12

How long will it take to go 150 km [E] traveling at 50 km/hr?

Physics
2 answers:
pantera1 [17]3 years ago
7 0
It will take 3 hours
Anit [1.1K]3 years ago
7 0

Answer:

SHTUP PErIODT POOH

Explanation:

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An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of
Gnoma [55]

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength = \frac{2L}{3}

The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength

Distance from center  = 0.25 × wavelength

Distance = 0.25 x \frac{2L}{3}

Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm

Distance = 0.25 x \frac{2 \times 39}{3}

Distance from the center to the nearest anti - node = 6.5 cm

Hence, the nearest distance to the anti - node from the center = 6.5 cm

So, option C is correct.

7 0
3 years ago
A test car travels in a straight line along the x-axis. The graph in the figure shows the car’s position x as a function of time
svetlana [45]
If it is s-t graph , point is c
if it is v-t graph , point is e
8 0
3 years ago
Read 2 more answers
Why are denser materials found closer to earths center
ICE Princess25 [194]
Denser materials tend to be closer to earths center due to their mass gravity is shown by the equation mg
Which stands for mass x gravity.
5 0
3 years ago
The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the
Eddi Din [679]

Answer:

x = D (M/M-m) 2.41

Explanation:

a) Let's apply Newton's second law to find the summation of force, where each force is given by the law of universal gravitation

        F = g m₁m₂ / r²

        Σ F = 0

       F1- F2 = 0

       F1 = F2

We set the reference system in the body of greatest mass (M) the planet

       F1 = g m₁ M / x²

       F2 = G m1 m / (D-x)²

      G m₁ M / x² = G m₁ m / (D-x)²

      M (D-x)² = m x²

      MD² -2MD x + M x² = m x²

     x² (M-m) -2MD x + MD² = 0

We solve the second degree equation

     x = [2MD  ±√ (4M²D² - 4 (M-m) MD²)] / 2 (M-m)

     x = {2MD ± 2D √ (M² + (M-m) M)} / 2 (M-m)

     x = D {M  ±  Ra (2M²-mM)} / (M-m)

    x = D (M ± M √ (2-m/M)) / (M-m)

    x = D (M / (M-m)) (1 ±√ (2-m/M)

Let's analyze this result, the value of M-m >> 1, so if we take the negative root, the value of x would be negative, it is out of the point between the two bodies, so the correct result must be taken with the positive root

 

    x = D (M / (M-m)) (1 + √2)

     x = D (M/M-m) 2.41

b) X = 2/3 D

     x = D (M/M-m) 2.41

     2/3 D = D (M/(M-m)) 2.41

     2/3 (M-m) = M 2.41

     2/3 M - 2/3 m = 2.41 M

     1.743 M = 0.667 m

     M/m = 0.667/1.743

     M/m =  0.38

3 0
3 years ago
What are the directions of movement for the Sagittarius, frontal, transverse planes ?
Dima020 [189]

Answer:

Squats involve flexion (forward motion) and extension (backward on the way up), so would fit into the sagittal plane. Frontal plane motion would include leaning from left to right as in sidebends and lateral raises, or perhaps you might picture jumping jacks for a good image of movement along the frontal plane.

5 0
3 years ago
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