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4 years ago

Warm air has greater pressure then cool air. True or False

1 answer:

6 0

Your description of the experiment is very sketchy.

If you have a jar full of air, sealed so that no air can get in or out,
then the pressure is lower when you cool the jar, and higher when
you heat the jar.

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You are standing on a bathroom scale in an elevator in a tall building. Your mass is

Oksanka [162]

Answer:

94 kg or 921.6 N

Explanation:

The velocity function is $v(t) = 3t + 0.2t^2$ . The acceleration function, a(t), is the time derivative of the velocity function.

$a(t) = \dfrac{d}{dt}v(t) = 3 + 0.4t$

At $t=4$ ,

$a(4) = 3 + 0.4\times4 = 3+1.6 =4.6\text{ m/s}{}^2$

Because the elevator is going upwards, the net acceleration, $a_N = g + a(4)$ where g us acceleration of gravity.

$a_N = 9.8 + 4.6 = 14.4\text{ m/s}{}^2$

The resultant weight is

$W =m a_N =64\times14.4=921.6 \text{ N}$

Since the bathroom scale is graduated in kg, it's reading is

$m =\dfrac{W}{g}=\dfrac{921.6}{9.8}=94\text{ kg}$

3 0

3 years ago

HELLLPPPPPP!!!!!!

Bond [772]

DIVIDE THEN MUTPLY THEN YOU AEMPLY SUCK

3 0

3 years ago

A ball is dropped from a cliff and has an acceleration of 9.8 m/s^2 how long will it take the ball to reach a speed of 24.5 m/s

Fiesta28 [93]

<span>Acceleration of ball (a) = 9.8 m/s^2 Speed (v) = 24.5 m/s Time (t) = Acceleration / Time = v / a = 24.5/9.8 = 2.5 s So it will take 2.5 seconds to reach that speed.</span>

5 0

3 years ago

A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.46 m, and is initially unc

KonstantinChe [14]

(a) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can solve the different part of the problem by using Gauss theorem.

Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:

$E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

where q is the charge contained in the spherical surface, so

$q=5.00 C$

Solving for E(r), we find the expression of the field for r<a:

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

(b) 0

The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.

So, if we use Gauss theorem for the region a < r < b, we get

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$

however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so

q' = + q - q = 0

And so we find

E(r) = 0

(c) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can use again Gauss theorem:

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$ (1)

where this time r > b (outside the shell), so the gaussian surface this time contained:

- the charge +q at the centre

- the inner surface, with a charge of -q

- the outer surface, with a charge of +q

So the net charge is

q' = +q -q +q = +q

And so solving (1) we find

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

which is identical to the expression of the field inside the shell.

(d) $-12.3 C/m^2$

We said that at r = a, a charge of -q is induced. The induced charge density will be

$\sigma_a = \frac{-q}{4\pi a^2}$

where $4 \pi a^2$ is the area of the inner surface of the shell. Substituting

q = 5.00 C

a = 0.18 m

We find the induced charge density:

$\sigma_a = \frac{-5.00 C}{4\pi (0.18 m)^2}=-12.3 C/m^2$

(e) $-1.9 C/m^2$

We said that at r = b, a charge of +q is induced. The induced charge density will be

$\sigma_b = \frac{+q}{4\pi b^2}$

where $4 \pi b^2$ is the area of the outer surface of the shell. Substituting

q = 5.00 C

b = 0.46 m

We find the induced charge density:

$\sigma_b = \frac{+5.00 C}{4\pi (0.46 m)^2}=-1.9 C/m^2$

3 0

3 years ago

The position of an object is given by the equation x = 4.0t2 - 2.0 t - 4.5where x is in meters and t is in seconds. What is the

s344n2d4d5 [400]

Answer:

Explanation:

x = 4 t^2 - 2 t - 4.5

Position at t = 3 s

x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m

Velocity at t = 3 s

v = dx / dt = 8 t - 2

v ( t = 3 s) = 8 x 3 - 2 = 22 m/s

Acceleration at t = 3 s

a = dv / dt = 8

a ( t = 3 s ) = 8 m/s^2

When is the velocity = 0

v = 0

8 t - 2 = 0

t = 0.25 second

When is the position = 0

x = 0

4 t^2 - 2 t - 4.5 = 0

$t = \frac{2 \pm \sqrt{4 + 72}}{8}$

t = 1.4 second

8 0

3 years ago