Calculate the average speed of a gazelle that runs 140 meters in 5.0 seconds

Two rigid tanks of equal size and shape are filled with different gases. The tank on the left contains oxygen, and the tank on t

fredd [130]

Answer:

The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.

Explanation:

Given:

Molar mass of oxygen, $M_O=32$

Molar mass of hydrogen, $M_H=2$

We know ideal gas law as:

$PV=nRT$

where:

P = pressure of the gas

V = volume of the gas

n= no. of moles of the gas molecules

R = universal gs constant

T = temperature of the gas

∵ $n=\frac{m}{M}$

where:

m = mass of gas in grams

M = molecular mass of the gas

∴Eq. (1) can be written as:

$PV=\frac{m}{M}.RT$

$P=\frac{m}{V}.\frac{RT}{M}$

as: $\frac{m}{V}=\rho\ (\rm density)$

So,

$P=\rho.\frac{RT}{M}$

Now, according to given we have T,P,R same for both the gases.

$P_O=P_H$

$\rho_O.\frac{RT}{M_O}=\rho_H.\frac{RT}{M_H}$

$\Rightarrow \frac{\rho_O}{32}=\frac{\rho_H}{2}$

$\rho_O=16\rho_H$

∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>

5 0

3 years ago

SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the

Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0

2 years ago

A girl throws a rock horizontally at 10 m/s from the top of a building, 22 m above street level. Assuming free fall conditions a

MAXImum [283]

Answer:21.18 m

Explanation:

Given

initial speed u=10 m/s

height of building h=22 m

time taken to complete 22 m

$h=ut+\frac{1}{2}at^2$

initial vertical velocity =0

$22=\frac{1}{2}gt^2$

$t=\sqrt{\frac{22\times 2}{g}}$

$t=2.11 s$

Horizontal Distance moved

$R=u_x\times t$

$R=10\times 2.11$

$R=21.18 m$

6 0

3 years ago

A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang

olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: $s_c = \omega_ct = 0.233t$