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harkovskaia [24]
3 years ago
6

Газ имеет объем 2 м³ при давлении 10° Па. Найдите обьем этого газа при изотермическом уменьшении давления в два раза.​

Physics
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

sorry i didn't understand

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The third choice is correct
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Two 6 ohm resistors in parallel gives an equivalent resistance of
Mashcka [7]

Answer:

3 ohms

Explanation:

6×6/6+6 =3 .............

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A speeding Thunderbird left skid marks on the road that were 76.7 m long when it came to a stop. If the acceleration of the car
FrozenT [24]

Answer: 39.2 m/s

Explanation:

You can use the kinematic equation:

v_f^2=v_i^2+2a*d

We know the final velocity because it says it came to a stop. So now all we gotta do is plug in.

0^2=v_i^2+2(-10)(76.7)\\v_i^2=1,534\\v_i=\sqrt{1534} \\=39.166 m/s

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3 years ago
How do I find the approximate velocity of "the object", on the graph at 5 seconds?
MakcuM [25]

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7 0
3 years ago
Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part
Fantom [35]

a) Angular acceleration: 17.0 rad/s^2

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

W=mg

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

W_p =mg sin \theta

where \theta is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta

where L is the length of the pencil.

The relationship between torque and angular acceleration \alpha is

\tau = I \alpha (1)

where

I=\frac{1}{3}mL^2

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for \alpha, we find:

\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when \theta=10^{\circ} is

\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude mg

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

\tau = Fd

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

\tau =mg\frac{L}{2} cos \theta

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0
3 years ago
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