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Nesterboy [21]
3 years ago
15

Fast-moving glaciers that can move up to 6 kilometers per year

Physics
2 answers:
Mademuasel [1]3 years ago
8 0

its surging due to other brain ly questions

AlladinOne [14]3 years ago
7 0
The answer for this question would be the term SURGING. Surging is the fast-moving glacier that can move up to 6 kilometers per year. I<span>t flows more quickly, sometimes moving 10 to 100 times faster than it normally does. This is one of the classification of a glacier aside from the normal type. Hope this answers your question.</span>
You might be interested in
A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m3 , undergoes a constant-pressure expansion at
Gnom [1K]

Answer:

Work: 4.0 kJ, heat: 4.25 kJ

Explanation:

For a gas transformation at constant pressure, the work done by the gas is given by

W=p(V_f -V_i)

where in this case we have:

p = 2 bar = 2\cdot 10^5 Pa is the pressure

V_i = 0.1 m^3 is the initial volume

V_f = 0.12 m^3 is the final volume

Substituting,

W=(2\cdot 10^5)(0.12-0.10)=4000 J = 4.0 kJ

The 1st law of thermodynamics also states that

\Delta U = Q-W

where

\Delta U is the change in internal energy of the gas

Q is the heat absorbed by the gas

Here we know that

\Delta U = +0.25 kJ

Therefore we can re-arrange the equation to find the heat absorbed by the gas:

Q=\Delta U + W = 0.25 kJ + 4.0 kJ = 4.25 kJ

7 0
3 years ago
A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and
cricket20 [7]

Answer:

The frequency f = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = \frac{d}{2} =\frac{44}{2} = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = \frac{d}{2} =\frac{3.0}{2} = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = \frac{d}{2} = \frac{2.0}{2} = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that  the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2

pgy_1=\frac{1}{2}pv^2_2 +pgy_2

v_2=\sqrt{2g(y_1-y_2)}

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = \sqrt{2*9.8*(0.35-0)}

v₂ = \sqrt {6.86}

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

v_3 = \frac{v_2r_2^2}{v_3^2}

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = (2.62  m/s)* (\frac{1.5mm^2}{1.0mm^2} )

v₃ = 0.0589 m/s

∴ velocity  of the fluid in the cylinder =  0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

f=\frac{v_s}{4(h-v_3t)}

where;

v_s = velocity of sound

h = height of the fluid

v₃ = velocity  of the fluid in the cylinder

f=\frac{343}{4(0.40-(0.0589)(0.4)}

f= \frac{343}{0.6576}

f = 521.59 Hz

∴ The frequency f = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})

\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)

\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}

where;

v_s (velocity of sound) = 343 m/s

v₃ (velocity  of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s  when the cylinder has been filling for 4.0 s.

8 0
3 years ago
Djejfje kdjejfjeofjod fjdidjskd.jdjdjdjdp
Mariana [72]

YUHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH gbghdwqygfywefyugsadfyuyuqwbhfbds       hehrwuhuaijkfa

5 0
3 years ago
Since friction is a force,<br> what unit is friction<br> measured In???
dimaraw [331]
It would be measured in newtons :)
4 0
3 years ago
Read 2 more answers
What is the need of force in our life​
melamori03 [73]

Answer:

A force is a push or a pull and it affects our daily lives because without force,people would not be able to open and close stuff or lift up our arms or legs .....or anything, for that matter.

Explanation:

brainly

3 0
3 years ago
Read 2 more answers
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