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2 years ago

Considering how the predictions compare to the observations, what does this imply about the mass distributed in the galaxy?

Physics

1 answer:

Dafna1 [17]2 years ago

8 0

Answer:

The observations show a higher velocity than is predicted, mainly in the galaxy's outer regions, indicating that there is more mass in the exterior areas than we can see.

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a goalkeeper catches a 491 g soccer ball traveling horizontally at 29.4 m/s. if it took 2,218 n of force to stop the ball, how m

yarga [219]

The ball will take 2.551 seconds to reach its peak position.

We must know how long the balls are in the air before we can predict where they will fall. It will take 2 seconds for both balls to touch the ground.

<h3>How quickly does a ball drop?</h3>

The falling ball travels a distance of d = 12 9.8 (m/s2) t2, with a speed of v = 9.8 (m/s2) t as a function of time. The ball travels 4.9 m in a second. The falling ball's velocity is v = -9.8 (m/s2) t j, and its position is r = (4.9 m - 12 9.8 (m/s2) t2) j as a function of time.

To know more about balls visit:-

brainly.com/question/19930452

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6 0

1 year ago

The force on the spring is F0 and it stores elastic potential energy PEs0. If the spring displacement is tripled to 3x0, determi

Dimas [21]

Answer:

Explanation:

Let initial extension in the spring= x₀

Force on the spring = F₀

Let spring constant = k

Fo = k x₀

Fn = 3k x₀

Fn /Fo = 3

PEs0 ( ORIGINAL) =1/2 k x₀²

PEsn ( NEW) =1/2 k (3x₀)²

PEsn / PEs0 = 9

7 0

3 years ago

In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de

snow_tiger [21]

As we know that range of the projectile motion is given by

$R = \frac{v^2 sin(2\theta)}{g}$

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

$\theta, 90 - \theta$

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

5 0

3 years ago

bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving +1.72 m/s what is the veloci

spin [16.1K]

Answer: V1 = 3.559 - 0.744V2

Explanation: Given that the

bumper car A has

Mass M1 = (281 kg) moving with

Velocity U1 = 2.82 m/s

bumper car B has

Mass M2 = (209 kg) moving with

Velocity U2 = 1.72 m/s

Where U1, U2 are the initial velocity of the two cars

Since the collision is elastic, we will use the formula below,

M1U1 + M2U2 = M1V1 + M2V2

Substitute the values into the formula

281×2.28 + 209×1.72 = 281V1 + 209V2

640.68 + 359.48 = 281V1 + 209V2

1000.16 = 281V1 + 209V2

Make V1 the subject of formula

281V1 = 1000.16 - 209V2

V1 = 1000.16/281 - 209V2/281

V1 = 3.559 - 0.744V2

Therefore, the velocity of car A which

is V1 after the collision will be expressed as V1 = 3.559 - 0.744V2

6 0

3 years ago

A 12.0 kg rock slides off the edge of a bridge and falls into the water 25.0 meters below. what is the kinetic energy of the roc

Pavel [41]

During the fall, all the initial potential energy of the rock
$U=mgh$
has converted into kinetic energy of motion
$K= \frac{1}{2}mv^2$
where h is the initial height of the rock, m its mass, and v its velocity just before hitting the water. So, for energy conservation, we have
$U=K$
and so we can find the value of K, the kinetic energy of the rock just before hitting the ground:
$K=U=mgh = (12.0 kg)(9.81 m/s^2)(25.0 m)=2943 J$

7 0

3 years ago