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2 years ago

A projectile is launched

1 answer:

4 0

The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

The given parameters;

initial horizontal velocity, vₓ = 16 m/s
initial vertical velocity, $v_y =0$
time interval 1 seconds

The components of the velocity can be horizontal or vertical velocity.

The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.

The vertical component of the velocity is calculated as;

$v_y = v_0_y -gt\\\\v_y = 0 - (1\times 9.8)\\\\v_y = -9.8 \ m/s$

The horizontal component of the velocity is constant since it is not affected by gravity.

The horizontal component of the velocity = 16 m/s

Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

Learn more here:brainly.com/question/20349275

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The orbital radius of the Earth (from Earth to Sun) is 1.496 x 10^11 m.

mrs_skeptik [129]

Explanation:

The orbital radius of the Earth is $r_1=1.496\times 10^{11}\ m$

The orbital radius of the Mercury is $r_2=5.79 \times 10^{10}\ m$

The orbital radius of the Pluto is $r_3=5.91 \times 10^{12}\ m$

We need to find the time required for light to travel from the Sun to each of the three planets.

(a) For Sun -Earth,

Kepler's third law :

$T_1^2=\dfrac{4\pi ^2}{GM}r_1^3$

M is mass of sun, $M=1.989\times 10^{30}\ kg$

So,

$T_1^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 1.496\times 10^{11}\\\\T_1=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times1.496\times10^{11}}\\\\T_1=2\times 10^{-4}\ s$

(b) For Sun -Mercury,

$T_2^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.79 \times 10^{10}\ m\\\\T_2=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.79 \times 10^{10}}\ m\\\\T_2=1.31\times 10^{-4}\ s$

$T_3^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.91 \times 10^{12}\\\\T_3=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.91 \times 10^{12}}\\\\T_3=1.32\times 10^{-3}\ s$

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2 years ago

PLEASE HELP! Explain how energy travels outward from the core and is emitted from the Sun. Include what happens in each layer of

xxMikexx [17]

Fusion occurs in the Sun's core, releasing energy that is transferred outward. Once in the radiative zone, gamma rays are transferred by radiation. They are converted to other types of photons, which move into the convective zone, where they are transferred by convection. Finally, energy is emitted from the photosphere.

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2 years ago

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An object has a kinetic energy of 275 j and a momentum of magnitude 25.0 kg · m/s. find the (a) speed and (b) mass of the object

Sergeu [11.5K]

Kinetic energy = momentum^2 / 2 x mass
Mass = (momentum^2/ Kinetic energy) / 2

Mass = (25^2 / 275) / 2
Mass = 1.136 kg.

momentum = mass x velocity

velocity = mass / momentum
velocity = 1.136 / 25
velocity = 0.04544 m/s

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2 years ago

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Which of the following is the best example of kinetic energy? A. A bowling ball rolling toward the pins B. A horse standing in a

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The answer is A. Only an object in motion has kinetic energy.

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3 years ago

What does saturns core look like?

HACTEHA [7]

Answer:

At Saturn's center is a dense core of metals like iron and nickel surrounded by rocky material and other compounds solidified by the intense pressure and heat. It is enveloped by liquid metallic hydrogen inside a layer of liquid hydrogen—similar to Jupiter's core but considerably smaller

Explanation:

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2 years ago