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LUCKY_DIMON [66]

3 years ago

5

A difference In density between gases creates buoyancy which is the ability of an object to float. 0°C helium has a density of .

18 kg/m³. 0°C regular air has a density Of 1.29 kg/m^3.
In kilograms how much mask in a cubic meter of helium float in regular air at 0°C?

1 answer:

rjkz [21]3 years ago

6 0

Answer:

mass = 0.18 [kg]

Explanation:

This is a classic problem where we can apply the definition of density which is equal to mass over volume.

$density = \frac{mass}{volume} \\\\where:\\volume = 1 [m^3]\\density = 0.18[kg/m^3]$

mass = 0.18*1

mass = 0.18 [kg]

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A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

vitfil [10]

The acceleration of the sled will be 1.30 m/s². Force is defined as the product of mass and acceleration.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Given data;

m(mass of sled)=8 kg

Θ is the inclination of force= 50°

Force of friction,f=2.4 N.

The applied force at the given angle is resolved into the two-component as;

$\rm F_h=F cos \theta \\\\ F_h= 20 cos 50 ^0 \\\\ F_h= 12.85 \ N$

$\rm F_v=F sin \theta \\\\ F_v=20 sin 50^0 \\\\ F_v=15.32 \ N$

The net vertical force is zero;

$\rm F_N=mg-Fsin50^0 \\\\ \rm F_N=8 \times 9.81 -15.32 \\\\ F_N=63.1 \ N \\\\$

From Newton's second law the net force as;

$\rm \sum F_{net}=ma \\\\ Fcos 60^0-f =ma \\\\ a=\frac{12.855-2.4}{8} \\\\a = 1.30 \ m/s^2$

Hence, the acceleration of the sled will be 1.30 m/s².

To learn more about the force refer to the link;

brainly.com/question/26115859

#SPJ1

5 0

2 years ago

Read 2 more answers

Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr

nirvana33 [79]

Answer:

t = 444.125 sec

Explanation:

Given data:

V = 24 volt

I = 0.1 ampere

mass of water mw = 51 gm

cr = 4.18 J/gm degree K^-1

mass of resistor = 8 gm

cr = 3.7 J/gm degree K^-1

we know that power is given as

Power P = VI

But P =E/t

so equating both side we have

$\frac{E}{t} = VI$

solving for t

$t = \frac{E}{VI}$

$t = \frac{m_w C_w \Delta T}{VI}$

$t = \frac{51 \times 4.18 \times (5 -0)}{24\times 0.1}$

t = 444.125 sec

5 0

4 years ago

Read 2 more answers

A transport plane takes off from a level landing field with two gliders in tow, one behind the other. The mass of each glider is

velikii [3]

Answer:

(a) Length =136.58 m

(b) T=5995 N

Explanation:

for the glider in the back

T - 1900 = 700 a

for the glider in front

12000-T -1900 = 700a

add equations

12000-3800 = 1400 a

a=5.85 m/s^2

v^2 = v0^2 + 2 a x

40^2 = 2*5.85*x

Length =136.58 m

b) plug the a back into one of the previous formula

T - 1900 = 700*5.85

T=5995 N

8 0

3 years ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

So

$mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=> $w_f = 1.531 \ rad/ s$

3 0

3 years ago

A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.1-m-long rope. The ball is pulled to one s

AlladinOne [14]

Answer:

The tension in the rope is 262.88 N

Explanation:

Given:

Weight $W = 150$ N

Length of rope $r = 4.1$ m

Initial speed of ball $v = 5.5 \frac{m}{s}$

For finding the tension in the rope,

First find the mass of rod,

$mg = 150$ ( $g = 9.8 \frac{m}{s^{2} }$ )

$m = \frac{150}{9.8}$

$m = 15.3$ kg

Tension in the rope is,

$T = mg + \frac{mv^{2} }{r}$

$T = 150 + \frac{15.3 \times (5.5)^{2} }{4.1}$

$T = 262.88$ N

Therefore, the tension in the rope is 262.88 N

7 0

3 years ago