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3 years ago

Which will most likely bring clouds and precipitation that last for several days

Physics

1 answer:

Nostrana [21]3 years ago

8 0

C the answer is C fronts would most likely be responsible for several days of rain and clouds

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An object of mass 20kg is released from a height of 10 meters above the ground level. The kinetic energy of the the object just

Yuri [45]

Answer:

The answer is the object weighs 5 so 3-5 is 2

Explanation:

I took this test

7 0

3 years ago

A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha

lord [1]

Answer:

The value is $F_f = 46.935 \ N$

Explanation:

From the question we are told that

The magnitude of the horizontal force is $F = 92.7 \ N$

The mass of the crate is $m = 40.5 \ kg$

The acceleration of the crate is $a = 1.13 \ m/s$

Generally the net force acting on the crate is mathematically represented as

$F_{net} = F - F_f = ma$

Here $F_f$ is force of kinetic friction (in N) acting on the crate

$92.7 - F_f = 40.5 * 1.13$

=> $F_f = 46.935 \ N$

5 0

3 years ago

At room temperature, glass used in windows actually has some properties of a liquid. it has a very slow, viscous flow. (visco

tamaranim1 [39]

The fahrenheit temperature is 927965. It is calculated using the formula 515515 Degree Cx1.8+32=927965. The degree celcius and fahrenheit are two units two measure temperature. If the value is given in celcius it can be converted into fahrenheit using the above formula.

8 0

3 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for M, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

5 0

3 years ago

. A newly discovered planet has three times the mass and five times the radius of Earth. What is the ratio of the acceleration d

NikAS [45]

Answer:

0.12

Explanation:

The acceleration due to gravity of a planet with mass M and radius R is given as:

g = (G*M) / R²

Where G is gravitational constant.

The mass of the planet M = 3 times the mass of earth = 3 * 5.972 * 10^24 kg

The radius of the planet R = 5 times the radius of earth = 5 * 6.371 * 10^6 m

Therefore:

g(planet) = (6.67 * 10^(-11) * 3 * 5.972 * 10^24) / (5 * 6.371 * 10^6)²

g(planet) = 1.18 m/s²

Therefore ratio of acceleration due to gravity on the surface of the planet, g(planet) to acceleration due to gravity on the surface of the planet, g(earth) is:

g(planet)/g(earth) = 1.18/9.8 = 0.12

3 0

3 years ago