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maria [59]
3 years ago
8

Q: A spring of original length 10 cm, Stretches to 12 cm, when a Force of 40 N is applied to It. what is the extension of the sp

ring when a Force of 26 N, is applied?​​
Physics
1 answer:
Paul [167]3 years ago
4 0

Answer:

4

Explanation:

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The average speed between earth and the sun is 1.50 x10^8 km. Calculate the average speed of the Earth in its orbit in kilometer
cluponka [151]

Answer:

The average speed of the earth in its orbit is 29.86km/s

Explanation:

The average distance between the Earth and the Sun is 1.50x10^{8} km.

The average speed of the earth in its orbit can be found by the next equation :

v = \frac{2 \pi r}{T}  (1)

Where r is the radius and T is the period.

In this case, the orbit of the Earth can be considered as a circle

(r = 1.50x10^{8}km) instead of an ellipse.

It takes 1 year to the Earth to make one revolution around the Sun. Therefore, its period will be 365.25 days.

Notice that to express the period in terms of seconds, the following is needed:

365.25d . \frac{86400s}{1d} ⇒ 31557600s

Then, equation 1 can be used:

v = \frac{2 \pi (1.50x10^{8}km)}{31557600s}

v = 29.86km/s

7 0
3 years ago
Read 2 more answers
According to Kepler's Second Law the radius vector drawn from the Sun to a planet Multiple Choice is the same for all planets. s
Mademuasel [1]

Answer:

sweeps out equal areas in equal times.

Explanation:

As we know that there is no torque due to Sun on the planets revolving about the sun

so we will have

\tau_{net} = 0

now we have

\frac{dL}{dt}= 0

now we also know that

Area = \frac{1}{2}r^2d\theta

so rate of change in area is given as

\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt}

so we will have

\frac{dA}{dt} = \frac{1}{2}r^2\omega

\frac{dA}{dt} = \frac{L}{2m}

since angular momentum and mass is constant here so

all planets sweeps out equal areas in equal times.

4 0
3 years ago
Two ways that weak nuclear forces and electromagnetic forces are simaliar
STALIN [3.7K]
Have very short ranges
6 0
3 years ago
Blocks A (mass 2.00 kg) and B (mass 10.00 kg, to the right of A) move on a frictionless, horizontal surface. Initially, block B
olga2289 [7]

Answer:dont care

Explanation:

5 0
3 years ago
Help with the two questions above? Correct answers?
LenKa [72]

(6) first choice: the frequency appears higher and wavelength is shorter.

The car approaches a stationary observer and so the sound will appear to have shorter wavelength. This creates an effect of its siren to sound with higher frequency than it would do if both were stationary.

(7) The Doppler formula for frequency in the case of a stationary observer and source approaching it is as follows:

f_O = \frac{v}{v-v_s}\cdot f= \frac{343\frac{m}{s}}{(343-25)\frac{m}{s}}\cdot 400Hz \approx 431Hz

The wavelength is then

\lambda = \frac{343\frac{m}{s}}{431Hz}\approx 0.80 m

The third choice "0.80m; 431Hz" is correct

7 0
3 years ago
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