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olasank [31]
2 years ago
13

Calculate the volume occupied by 25.2 g of co2 at 0.84 atm and 25°c. r = 0.08206 latm/kmol.

Physics
1 answer:
Nitella [24]2 years ago
5 0

Answer:

16.68 L

Explanation:

First of all, we need to calculate the number of moles corresponding to 25.2 g of CO2.

The molar mass of CO2 is 44.01 g/mol, therefore the number of moles is:

n=\frac{m}{M_m}=\frac{25.2 g}{44.01 g/mol}=0.573 mol

So now we can use the ideal gas equation:

pV=nRT

where:

p = 0.84 atm is the gas pressure

V = ? is the volume

n = 0.573 mol is the number of moles

R = 0.08206 L atm / K mol is the gas constant

T = 25°c = 298 K is the gas temperature

Substituting into the equation and re-arranging it, we find the volume:

V=\frac{nRT}{p}=\frac{(0.573 mol)(0.08206 Latm/Kmol)(298 K)}{0.84 atm}=16.68 L

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A physics student with a stopwatch drops a rock into a very deep well, and measures the time between when he drops the rocks and
tankabanditka [31]

Answer:

h= 161.06 m

Explanation:

Given that

Speed of the sound ,C= 343 m/s

Total time ,t= 6.2 s

lets take the depth of the well = h

The time taken by stone before striking the water = t₁

we know that

h=\dfrac{1}{2}gt_1^2

t_1=\sqrt{\dfrac{2h}{g}}

The time taken by sound =t₂

t_2=\dfrac{h}{343}

The total time

t = t₁+ t₂

6.2 = \sqrt{\dfrac{2h}{g}}+\dfrac{h}{343}

6.2 = \sqrt{\dfrac{2h}{9.81}}+\dfrac{h}{343}

Now by solving the above equation we get

h= 161.06 m

Therefore the depth of the well will be 161.06 m.

6 0
2 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
A 2 ft x 2 ft x 2 ft box weighs 100 pounds, and the weight is evenly distributed. What is the magnitude of the minimum horizonta
Tems11 [23]

Answer:

Explanation:

Let the force required be F . It is applied at the top of the box . The box is likely to turn about a corner . Torque of this force about this corner

= F x 2

This torque will try to turn the box . On the other hand the weight which is acting at CM will create a torque about the same corner . This torque will try to prevent the box to turn around the corner.

This torque of weight

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7 0
2 years ago
Read 2 more answers
An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's
ivolga24 [154]
1) Focal length

We can find the focal length of the mirror by using the mirror equation:
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d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

In this case, d_o = 18 cm, while d_i=-4 cm (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}
from which we find
f=- \frac{36}{7} cm=-5.1 cm

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
r=2f=2 \cdot 5.1 cm=10.2 cm
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Physics in motion unit 6a the nature of waves
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