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3 years ago

Physical properties of minerals graphic organizer

Physics

1 answer:

Nadusha1986 [10]3 years ago

4 0

The answer is in the attachment
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5. Tại sao khi lặn ta luôn cảm thấy tức ngực và càng lặn sâu thì cảm giác tức ngực càng tăng? A. Ap suất của nước giảm B. Ap s

natali 33 [55]

Answer:

c and d

Explanation:

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7 0

2 years ago

An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion,

Ne4ueva [31]

A) f = 1.8 rev/s = 2 Hz
T = 1 / f = 0.55s

B) not really sure..srry

C) T = 2 pi √ ( L / g ) 
0.57 = 2 x 3.14 x √ ( 0.2 / g )

g = 25.5 m/s²

Hope this helps a little at least.. :)

5 0

3 years ago

Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb

Maslowich

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

Period = T = 88 days

Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

Where v is the velocity. r is the radius and w is the angular velocity.

And we know that:

w = 2*pi*f

where f is the frequency, and:

T = 1/f.

Then we can write:

w = 2*pi/T

and our equation becomes:

ar = r*(2*pi/T)^2

Now we solve this for r.

First we need to use the same units in both equations, so we want to write T in seconds.

T = 88 days,

A day has 24 hours, and one hour has 3600 seconds:

T = 88*24*3600 s =7,603,200s

Then:

3.96x10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

r = (3.96x10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

5 0

3 years ago

The block slides on a horizontal frictionless surface. The block has a mass of 1.0kg and is pushed 5.0N at 45°. What is the magn

Firlakuza [10]

Answer:

$3.5\:\mathrm{m/s^2}$

Explanation:

Newton's 2nd law is given as $\Sigma F = ma$ .

To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.

Using trigonometry to find the horizontal component of the force:

$\cos 45^{\circ}=\frac{x}{5},\\\frac{\sqrt{2}}{{2}}=\frac{x}{5},\\x=\frac{5\sqrt{2}}{2}$

Use this horizontal component of the force to solve for for the acceleration of the object:

$\frac{5\sqrt{2}}{2}=1.0\cdot a,\\a=\frac{5\sqrt{2}}{2}\approx \boxed{3.5\:\mathrm{m/s^2}}$

8 0

2 years ago

This will give brainliest

Troyanec [42]

Answer:

It is likely to be option B: 50 degrees and 50 degrees

Hope it helps.......... pls mark as brainliest.

6 0

2 years ago