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2 years ago

Planet X has no atmosphere. This planet will most likely _____.

Physics

3 answers:

aniked [119]2 years ago

8 0

C) have many craters from meteoroids

TiliK225 [7]2 years ago

6 0

A) be too hot to support life
Hope dis helps! :3

eduard2 years ago

5 0

A) be too hot to support life

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1.Mention two uses of the concave mirror.

Anuta_ua [19.1K]

Answer:

1. telescope

$f = \frac{r}{2}$

$f = \frac{r}{2}$ f- focal length

$f = \frac{r}{2}$ f- focal length r- the radius of curvature of the mirror

$\frac{1}{f} = \frac{1}{p} + \frac{1}{l}$

p-the distance of the object from the vertex of the mirror

l-the distance of the figure from the vertex of the mirror

8 0

2 years ago

What is the magnitude of velocity for a 2000 kg car possessing 3000 kgm/s of momentum

Alex

For this case we have that by definition, the momentum is given by:

$p = mv$

Where,

m: mass
v: speed

Therefore, replacing values we have:

$3000 = (2000) v$

From here, we clear the value of the speed:

$v = \frac {3000} {2000}\\v = 1.5 \frac {m} {s}$

Answer:

The magnitude of velocity is:

$v = 1.5 \frac {m} {s}$

4 0

2 years ago

Read 2 more answers

Apply the concept of density to explain why oil floats on water.

Sauron [17]

Anything less dense than water will float, like oil. Anything more dense than water will sink, like rock.

6 0

2 years ago

Which of Newton's laws of motion best illustrates the principle of inertia?

andrew-mc [135]

Newton's first law of motion best illustrates the principle of inertia

5 0

2 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c

seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

$A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2$

So the volume flow rate along the pipe is

$\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s$

We can use the similar logic to find the cross-section area at the refinery

$A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2$

The radius of the pipe at the refinery is:

$A_r = \pi r^2$

$r^2 =A_r/\pi = 0.446/\pi = 0.141$

$r = \sqrt{0.141} = 0.377m$

So the diameter is twice the radius = 0.38*2 = 0.754m

6 0

2 years ago