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Which force does not operate at a distance of 1 m?

sleet_krkn [62]

Answer:

A

Explanation:

8 0

3 years ago

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Felipe believes that whenever the Moon is in the position that is shown from above (top view) in Diagram A, the Moon always look

taurus [48]

Answer:he is incorrect

Explanation:he is wrong because if the sunlight is on the opposite side of the earth than the moon, then the sunlight would be blocked by the earth, making the moon dark, because the moon gets its light from the light that is reflected of the surface from the sun

8 0

3 years ago

A steel piano string of mass per unit lenght 4×10-⁴kgm-¹ was struck, if the speed of the sound on the string is 300ms-¹. Calcula

Snezhnost [94]

Answer:

36 N

Explanation:

Velocity of a standing wave in a stretched string is:

v = √(T/ρ),

where T is the tension and ρ is the mass per unit length.

300 m/s = √(T / 4×10⁻⁴ kg/m)

T = 36 N

7 0

3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago

13.An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane tra

Murrr4er [49]

Answer:

The answer is 3.33m

Explanation:

The acceleration "a" is constant.

Acceleration is the variation of velocity over time,

$\frac{dv}{dt} = a$ .

solving the last equation

$\int_{v_0}^v dv = a\int_0^t dt \rightarrow v-v_0 = at$ ,

where $v_0=0$ because the airplane starts from rest.

Once again, velocity is the variation of distance over time.

$\frac{dx}{dt} = at \rightarrow \int_{x_0}^x dx = a\int_0^t t\ dt$

then

$x- x_0 = \frac{1}{2}at^2$

where $x_0=0$ if we consider the end of the runway as the initial point (this step is for simplicity but you can let it expressed, it's going to cancel anyway).

If $x=1.11\ m$ at $t=1s$ , then

$a = \frac{2x}{t^2} = 2.22\ m/s^2$

and the final expression for the distance is

$x = 1.11 t^2$ .

If t = 2s, x = 4.44 m. Which means thad the additional distance is

$x(2s) - x(1s) = 4.44 - 1.11 = 3.33\ m$

8 0

4 years ago