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3 years ago

What to forces keep things In orbit

Physics

2 answers:

77julia77 [94]3 years ago

8 0

Gravity and the size of planets the more mass that planet has or density it hold more gravity

Aliun [14]3 years ago

6 0

Centripetal force.
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Help !! I hate science

Lorico [155]

Answer:

iron nail rusting

Explanation:

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3 years ago

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Mendeleevs periodic table was useful because it enabled scientists to predict properties of unknown (Blank)

mash [69]

I think it enabled scientists to predict properties of unknown elements.

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3 years ago

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A student is skateboarding down a ramp that is 6.0 m long and inclined at 180 with respect to the horizontal. The initial speed

dlinn [17]

Answer:

The speed at the bottom of the ramp is 2.6m/s

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

Notice that it is necessary to found the acceleration that can be done by means of Newton's second law:

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

<em>Force in the x axis:</em>

$F_{x} = W_{x}$

The component of the weight in the x axis can be gotten by means of trigonometric:

$\frac{OC}{H} = sen \theta$

$\frac{W_{x}}{W} = sen \theta$

$W_{x} = W sen \theta$

$W_{x} = mgsen \theta$

$F_{x} = mgsen \theta$ (3)

<em>Forces in the y axis: </em>

$F_{y} = N - W_{y}$ (4)

The component of the weight in the y axis can be gotten by means of trigonometric:

$\frac{AC}{H} = cos \theta$

$\frac{W_{y}}{W} = cos \theta$

$W_{y}= W cos \theta$

Remember that the weight is defined as:

$W = mg$

$W_{y}= mgcos \theta$

The normal force can be obtained from equation (4)

$N - W_{y} = 0$

$N = W_{y}$

$N = mgcos \theta$

Therefore, equation 4 can be rewritten as:

$F_{y} = mgcos \theta - mgcos \theta$

$F_{y} = 0$ (5)

Then, replacing equation 3 and equation 5 in equation 2 it is gotten:

$mgsen \theta + 0 = ma$

$mgsen \theta = ma$ (6)

However, a can be isolated from equation 6

$a = \frac{mgsen \theta}{m}$

$a = gsen \theta$ (7)

Finally, equation 7 can be replaced in equation 2:

$v_{f} = \sqrt{v_{i}^{2} + 2d(gsen \theta)}$

$v_{f} = \sqrt{(2.6m/s)^{2} + 2(6.0m)(9.8m/s^{2})sen 180^{\circ})}$

$v_{f} = 2.6m/s$

Hence, the speed at the bottom of the ramp is 2.6m/s

7 0

3 years ago

A child in a boat throws a 6-kg package horizontally to the right with a speed of 8 m/s.

victus00 [196]

Answer:

0.5 m/s

Explanation:

Using conservation of momentum, P1=P2, the system starts off with zero momentum as nothing is moving. But in the second part of the equation(P2) the child throws the package to the right. By Newton's third law, the child and the boat should move to the left. Plugging in what we know, 0= -90v + 6kg*8m/s. Solving for v you will get 0.5 meters per second. The mass is 90kg as that the is mass of the child and boat combined. I also made it negative as the boat and child move left (I designated this as the negative direction) and the package's momentum is positive as it is moving to the right.

4 0

3 years ago

Please help i’ll mark as brainliest

Anestetic [448]

Answer:

give me more information so I can help

6 0

4 years ago