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3 years ago

How are mass and brightness related? Daddies!

Physics

2 answers:

Delvig [45]3 years ago

5 0

Answer:

ftw

Explanation:

daddies?

maria [59]3 years ago

3 0

They aren’t. Brightness comes from the frequency of waves to your eye

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If you throw a baseball while riding a bike, the ball's speed relative to the ground is the sum of the bicycle's speed and the s

Ostrovityanka [42]

Take for example driving by with a cake in your hand, then dropping it while going 30 mph. It will not drop directly down, it will gradually go in the direction you were driving while falling.
This is true I believe, if I'm interpreting correctly.

5 0

2 years ago

Suppose an astronaut has landed on Planet * Fully equipped, the astronaut has a

gayaneshka [121]

From the calculations, the value of the acceleration due to gravity is 0.38 m/s^2.

<h3>What is weight?</h3>

The weight of an object is obtained as the product of the mass of the body and the acceleration due to gravity.

Thus;

When;

mass = 120 kg

weight = 46 N

acceleration due to gravity = 46 N/120 kg

=0.38 m/s^2

Learn more about acceleration due to gravity :brainly.com/question/13860566

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7 0

1 year ago

6.) What is

Feliz [49]

Answer:

100 V

Explanation:

Hi there!

Ohm's law states that $V=IR$ where V is the voltage, I is the current and R is the resistance.

Plug the given information into Ohm's law (R=50, I=A)

$V=IR\\V=(2)(50)\\V=100$

Therefore, the voltage across this current is 100 V.

I hope this helps!

6 0

2 years ago

Suppose a 1300 kg car is traveling around a circular curve in a road at a constant

Scrat [10]

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :

$F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N$

So, the force has a magnitude of 4212 N

4 0

2 years ago

An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5

Inessa05 [86]

Answer:

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

distance traveled by the airplane, s = 46,100 m

final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

v² = u² + 2as

$2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2$

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²

5 0

2 years ago