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2 years ago

14

Select all that apply. _____ is a place where the crust of the lithosphere has fractured. A fault A subduction zone An oceanic t

rench A river

2 answers:

Kazeer [188]2 years ago

6 0

The only right answer from these available options is a fault.

liberstina [14]2 years ago

3 0

A fault? yes !A fault is a fracture with a significant rock mass movement as a result.

A subdution zone? No, although it is a place where the tectonic place meet (so are likely to fracture, but do not necessarily need to fracture).

An oceanic trench? No, although a trench does mark where the tectonic plates meet

A river? no!

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

2 years ago

The intensity at a distance of 6.0 m from a source that is radiating equally in all directions is 6.0 × 10-10 w/m2 . what is the

satela [25.4K]

The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
$I= \frac{P}{A}$ (1)
where
P is the power
A is the area

In our problem, the intensity is $I=6.0 \cdot 10^{-10} W/m^2$ . At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
$A=4 \pi r^2 = 4 \pi (6.0 m)^2 = 452.2 m^2$

And so if we re-arrange (1) we find the power emitted by the source:
$P=IA = (6.0 \cdot 10^{-10}W/m^2)(452.2 m^2)=2.7 \cdot 10^{-7} W$

3 0

2 years ago

How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

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2 years ago

A 35-g block of ice at -14°C is dropped into a calorimeter (of negligible heat capacity) containing 400 g of water at 0°C. When

kompoz [17]

Answer:

Total mass of ice = 38.06g

Explanation:

Since the heat capacity of calorimeter is negligible.

The water is already at 0°C, so the heat loss can no longer reduce the temperature of the water. It is used for fusion and forming more ice.

The equilibrium temperature will be 0°C, because the heat gain by ice is only enough to bring it down to 0°C.

Heat gained by ice = heat loss by water

Heat gained by ice (from -14°C to 0°C) = heat lost to fusion by water (heat of fusion of some amount of the water present in the calorimeter)

mi Ci ∆Ti = mw . L ......1

Where;

mi = mass of ice = 35g = 0.035 kg

Ci = specific heat capacity of ice = 2090 J/kg ∙ K

∆Ti = change in temperature of ice = 0-(-14) = 14 K

mw = the mass of water that have gained enough heat for fusion ( mass of water converted to ice)

L = latent heat of fusion of water = 33.5 × 10^4 J/kg.

From equation 1;

mw = (mi Ci ∆Ti )/L

mw = (0.035×2090×14)/335000

mw = 0.00306 kg

mw = 3.06 g

Therefore, 3.06 g of water has been converted to ice.

When combined with the initial amount of ice initially in the calorimeter (at 0°C)

Total mass of ice = mi + 3.06g = 35g + 3.06g = 38.06g

Total mass of ice = 38.06g

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2 years ago

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Which statement about work and power correctly describes an automobile race?

miv72 [106K]

Well this question looks like it makes some assumptions. So assuming that both cars have the same mass and experience the same wind resistance regardless of speed and same internal frictions, then we could say "The car that finishes last has the lowest power". The reason is that for a given race the cars must overcome losses associated with motion. Since they all travel the same distance, the amount of work will be the same for both. This is because work is force times distance. If the force applied is the same in both cases (identical cars with constant wind resistance) and the distance is the same for both (a fair race track) then W=F·d will be the same.
Power, however, is the work done divided by the time over which it is done. So for a slower car, time t will be larger. The power ratio W/t will be smaller for the longer time (slower car).

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2 years ago

Read 2 more answers