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3 years ago

The weight of a basketball with a mass of 0.5 kg is _N.

Physics

1 answer:

Mariana [72]3 years ago

3 0

Answer:

4.9N

Explanation:

The formula for weight is F = ma.

You plug in the mass of 0.5kg and an acceleration of 9.8 to get

F = (0.5)(9.8)

F = 4.9N

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ListenA person on a ledge throws a ball vertically downward, striking the ground below the ledge with 200 joules of kinetic ener

polet [3.4K]

Answer:

A. 200 J

Explanation:

The initial kinetic energy depends on the initial speed, while the gravitational potential energy depends on the height, both balls are thrown with the same initial speed and from the same height. Therefore, due to the law of conservation of energy, the balls must have the same mechanical energy (the sum of both energies) when both impact the ground. Since the potential energy is zero at this point, its final kinetic energy must also be the same.

8 0

3 years ago

Radio waves travel 300,000,000 m/s. The frequency is 101,700,000. ehats the wavelength

Nesterboy [21]

Answer:

we have formula of frequency :

frequency(f)= speed of sound(c)/wavelength(λ)

for wavelength we swipe it with frequency as follows

λ=c/f

λ=300,000,000/101,700,000

λ=2.949

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3 years ago

a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f

harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for $\Delta$ DAC:

$tan\theta = \frac{d}{x}$

⇒ $\theta = tan^{-1} \frac{d}{x}$

Now for the $\Delta$ BAC:

$tan\theta = \frac{d + h}{x}$

⇒ $\theta = tan^{-1} \frac{d + h}{x}$

Now, differentiating w.r.t x:

$\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]$

For maximum angle, $\frac{d\theta }{dx}$ = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]

0 = $\frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}$

$\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}$

After solving the above eqn, we get

x = $\sqrt{\frac{d}{d + h}}$

The observer should stand at a distance equal to x = $\sqrt{\frac{d}{d + h}}$

4 0

3 years ago

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

7 0

3 years ago

Which would have the longer orbital period: a moon 1 million km from the center of Jupiter, or a moon 1 million km from the cent

Harman [31]

Answer:

earth

Explanation:

The formula for the orbital period of the moon is given by

$T = 2\pi \sqrt{\frac{r}{g}}$

As the time period is inversely proportional to the square root of the acceleration due to gravity of the planet.

As the value of acceleration due to gravity on Jupiter is more than the earth, so the period of moon around the earth is large as compared to the period of the moon around the Jupiter when the distance is same.

5 0

3 years ago