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USPshnik [31]
3 years ago
8

How do weathering process affect deserts?

Physics
1 answer:
Damm [24]3 years ago
4 0
Water and wind cause mechanical weathering and produce angular rocks, sheer canyon walls, and pebble-covered surfaces.
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What is the universe made of? Astronomers face an embarrassing conundrum: they don't know what 95% of the universe is made of​
rewona [7]

Answer:

This what they all been waiting for

I guess so

They been waiting for this sh,it for a long time didn't they

I'ma give it everything I got

Ayo Dougie park that X6 around the corner

Aye I'm just feeling my vibe right now

I'm feeling myself

Explanation:

3 0
4 years ago
The first law of thermodynamics can be given as ________.
MrRa [10]
The first law of thermodynamics can be written as
\Delta U = Q-W
where
\Delta U is the variation of internal energy of the system
Q is the amount of heat absorbed by the system
W is the work done by the system on the surrounding.

Using this form, the sign convention for Q and W becomes:
Q > 0 --> heat absorbed by the system (because it increases the internal energy)
Q < 0 --> heat released by the system (because it decreases the internal energy)
W > 0 --> work done by the system (for instance, an expansion: when the system expands, it does work on the surrounding, and so the internal energy decreases, this is why there is a negative sign in the formula Q-W)
W < 0 --> work done by the surrounding on the system (for instance, a compression: when the system is compressed, the surrounding is doing work on the system, and so the internal energy of the system increases)
7 0
3 years ago
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
Lubov Fominskaja [6]

Answer:

The moment of inertia is I=\frac{M}{12} a^{2}

Explanation:

The moment of inertia is equal:

I=\int\limits^a_b {r^{2} } \, dm

If r is -\frac{a}{2}

and dm=\frac{M}{a} dr

I=\int\limits^a_b {r^{2}\frac{M}{a}  } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}

I=\frac{M}{a} \int\limits^a_b {r^{2}  } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\  I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}

7 0
3 years ago
A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0
3 years ago
PLEASE HELP MEE PLEASE I BEG
TEA [102]

Explanation: (I think)

Plug your values into the momentum equation.

So m1= 63kg

m2 = 10 kg

V1 = 12 m/s

And then plug in your values and solve for your unknown (v2)

8 0
3 years ago
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