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3 years ago

A. Use the graph and the element made in question 2 to determine the mass of the star.

Physics

1 answer:

Minchanka [31]3 years ago

4 0

Black balls with blue waffles

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A car traveling at 21 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?

diamong [38]

The formula for acceleration is the velocity times the inverse of time so it would be 21 times 1/13. So roughly 0.0769... is the acceleration(m/s^2).

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3 years ago

The pH of acid rain would be _____. greater than 7 less than 7 exactly 7 exactly 14

Nookie1986 [14]

Less than 7, Ph of 7 is neutral Acid Rain would be between 4 and 0.

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3 years ago

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What is the energy equivalent of an object with a mass of 1.83 kg?

xxMikexx [17]

To determine the energy equivalent of an object, we use the famous equation of Einstein which is E=mc^2 where m is the mass of the object and c is the speed of light (3x10^8 m/s). We calculate as follows:

E = mc^2
E = 1.83 kg (3x10^8 m/s)^2
E = 1.647x10^17 J

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3 years ago

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PLEASE HELP

Sergeu [11.5K]

The vertical component of the initial velocity is $v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is $v_0_x = \frac{x}{t}$

The horizontal displacement when the object reaches maximum height is $X = \frac{xy}{gt^2} + \frac{x}{2}$

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

$y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is calculated as;

$x = v_0_xt\\\\v_0_x = \frac{x}{t}$

The time to reach to the maximum height is calculated as;

$T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T = \frac{1}{g} (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t$

The horizontal displacement when the object reaches maximum height is calculated as;

$X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}$

Learn more here: brainly.com/question/20689870

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3 years ago

3. A cat walks 0.220km North, then 0. 120 km South in a time of 400 seconds. whats the displacement and average velocity?

Andru [333]

Answer:

The rate at which velocity changes with respect to a change in time is called. acceleration.

Explanation:

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3 years ago