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3 years ago

How important are volcanoes in adding carbon dioxide to the atmosphere

Physics

1 answer:

leonid [27]3 years ago

5 0

Answer:

Important enough for someone to ask you to do it.

Explanation:

If someone asked you this question, then it must be important and stuff. This is proven by scientific stuff that no one cares about.

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A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.

Arturiano [62]

Answer:

a) $\alpha=0.5117\ rad.s^{-2}$

b) $\theta=8.4\ rad$

Explanation:

Given:

initial rotational speed of phonograph, $\omega_i=0\ rad.s^{-1}$

final rotational speed of phonograph, $N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}$

time taken for the acceleration, $t=5.73\ s$

Now angular acceleration:

$\alpha=\frac{\omega_f-\omega_i}{t}$

$\alpha=\frac{2.932-0}{5.73}$

$\alpha=0.5117\ rad.s^{-2}$

Using eq. of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2$

$\theta=8.4\ rad$

5 0

3 years ago

Given a volume of 1000. Cm^3 of an ideal gas at 300 k, what volume would iy occupy at a temperature of 600 k

LUCKY_DIMON [66]

Answer:2000 cm³

Here, pressure remains constant.

So, b the gas law

V/V' = T/ T'

1000 / V' = 300 / 600

V' = 2000 cm³

Explanation:also pls mark brainliest

8 0

3 years ago

A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket m

slamgirl [31]

Answer: Acceleration of the car at time = 10 sec is 108 $m/s^{2}$ and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

$M\frac{dv}{dt} = u\frac{dM}{dt}$

$\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt$

= $u\int_{M_{o}}^{M_{f}} \frac{dM}{M}$

$v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}$

$v_{o} = 0$

As, $v_{f} = u ln (\frac{M_{f}}{M_{o}})$

u = -2900 m/s

$M_{f} = M_{o} - m \times t_{f}$

= $2500 kg + 1000 kg - 95 kg \times t_{f}s$

= $(3500 - 95t_{f})s$

$v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s$

Also, we know that

a = $\frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}$

= $\frac{u}{3500 - 95 t} \times (-95) m/s^{2}$

= $\frac{95 \times 2900}{3500 - 95t} m/s^{2}$

At t = 10 sec,

$v_{f}$ = 918.34 m/s

and, a = 108 $m/s^{2}$

3 0

3 years ago

At time=0s, the object is at the 21.0-meter position along the roadway. Where is the object at time = 10 s?

Jet001 [13]

B.) 70m is your answer!

4 0

3 years ago

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

Elan Coil [88]

Answer: 230.50 m

Explanation:

We have the following information:

$h_{Hg-TOP}=675mmHg=0.675m$ the barometric reading at the top of the building

$h_{Hg-BOT}=695mmHg=0.695m$ the barometric reading at the bottom of the building

$\rho _{air}=1.18 kg/m^{3}$ density of air

$\rho _{Hg}=13600 kg/m^{3}$ density of mercury

$g=9.8/m^{2}$ gravity

And we need to find the height of the building.

In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building $P_{TOP}$ and the perssure at the bottom $P_{BOT}$ :

$P_{TOP}=\rho _{Hg} g h_{Hg-TOP}$ (1)

$P_{BOT}=\rho _{Hg} g h_{Hg-BOT}$ (2)

From (1): $P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa$ (3)

From (2): $P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa$ (4)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure $\Delta P$ :

$\Delta P=P_{BOT} - P_{TOP}$ (5)

$\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa$ (6)

On the other hand, we have a column of air with a cross-section area $A$ and the same height of the building, lets name it $h_{air}$ .

As pressure is defined as the force $F$ exerted on a specific area $A$ , we can write:

$\Delta P=\frac{F}{A}$ (7)

If we isolate $F$ we have:

$F= A \Delta P$ (8)

Also, the force gravity exerts on this column of air (its weight) is:

$F=m_{air} g$ (9)

Knowing the density of air is: $\rho_{air}=\frac{m_{air}}{V_{air}}$ (10)

where the volume of air can be written as: $V_{air}=(A)(h_{air})$ (11)

Substituting (1) in (10):

$\rho_{air}=\frac{m_{air}}{(A)(h_{air}}$ (12)

Isolating $m_{air}$ :

$m_{air}=(\rho_{air}) (A) (h_{air})$ (13)

Substituting (13) in (9):

$F=(\rho_{air}) (A) (h_{air}) (g)$ (14)

Matching (8) and (14)

$A \Delta P=(\rho_{air}) (A) (h_{air}) (g)$ (15)

Isolating $h_{air}$ :

$h_{air}=\frac{\Delta P}{g \rho_{air}}$ (16)

Substituting the known and calculated values:

$h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})}$ (17)

Finally:

$h_{air}=230.50 m$ This is the height of the building

8 0

3 years ago