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3 years ago

10

A very powerful vacuum cleaner has a hose 2.86 cm in diameter. with the end of the hose placed perpendicularly on the flat face

of a brick, what is the weight of the heaviest brick that the cleaner can lift?

1 answer:

prisoha [69]3 years ago

8 0

Answer:

F = 75.49 N

Explanation:

given,

atmospheric pressure Po = 1 atm

Po = 101325 Pa

diameter =3.08 cm,

radius = 1.54 cm

1 cm = 0.01 m

1.54 cm = 0.0154 m

Force = Pressure x area

area = π r²

Force = 101325 Pa x π x 0.0154 x 0.0154

F = 75.49 N

Weight of the heaviest brick that the cleaner can lift is F = 75.49 N

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3 years ago

The free-fall acceleration at the surface of planet 1 is 22 m/s^2. The radius and the mass of planet 2 are twice those of planet

algol13

Answer:

g₂ = 11 m/s²

Explanation:

The value of free-fall acceleration on the surface of a planet is given by the following formula:

$g = \frac{Gm}{r^2}$

where,

g = free-fall acceleration

G = Universal Gravitational Constant

m = mass of the planet

r = radius of planet

FOR PLANET 1:

$g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2$ --------------------- equation (1)

FOR PLANET 2:

$g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\$

using equation (1):

$g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}$

<u>g₂ = 11 m/s²</u>

8 0

3 years ago

A cube is 4.4 cm on a side, with one corner at the origin. Part 1 (a) What is the unit vector pointing from the origin to the di

Sidana [21]

Answer:

(a) $\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) $\theta = 85.44^{\circ}$

Solution:

As per the question:

Side of the cube, a = 4.4 cm

Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm

Now,

(a) To calculate the unit vector:

$\hat{A} = \frac{\vec{A}}{|A|}$

$\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}$

$\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}$

$\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) To calculate the angle between the two vectors say A and A' is given by:

$\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta$

$\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})$ (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

$\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}$

Now,

Using equation (1) :

$\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})$

$|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261$

Thus

$\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})$

$\theta = cos^{- 1}(0.07946) = 85.44^{\circ}$

4 0

3 years ago

The Bohr radius a0 is the most probable distance between the proton and the electron in the Hydrogen atom, when the Hydrogen ato

katen-ka-za [31]

Answer:

The electric force is $F = 11.9 *10^{-9} \ N$

Explanation:

From the question we are told that

The Bohr radius at ground state is $a_o = 0.529 A = 0.529 ^10^{-10} \ m$

The values of the distance between the proton and an electron $z = 2.63a_o$

The electric force is mathematically represented as

$F = \frac{k * n * p }{r^2}$

Where n and p are charges on a single electron and on a single proton which is mathematically represented as

$n = p = 1.60 * 10^{-19} \ C$

and k is the coulomb's constant with a value

$k =9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F = \frac{9*10^{9} * [(1.60*10^{-19} ]^2)}{(2.63 * 0.529 * 10^{-10})^2}$

$F = 11.9 *10^{-9} \ N$

3 0

3 years ago

A football player throws and accelerates a ball (m = 0.150 kg) from rest to a velocity of 50 m/s in 0.0121 sec. Determine the ac

Rasek [7]

Answer:

Acceleration=4m/s²

Force applied =619.8N

Explanation:

Using equation of motion

V=u+at we have: u=o, v=50m/s

50= 0 + a×0.0121

a = 50/0.0121

a= 4m/s²

Neglecting resistance forces

F= ma, where a = v-u/t

F=m×(v-u)/t

F= 0.150 ×(50-0)/0.0121

F=7.5/0.0121

F= 619.8N

5 0

3 years ago