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3 years ago

8

If you wake up in the morning, go to school, go to the grocery store, and then return to the same place where you woke up, what

is your displacement for the day

1 answer:

Gekata [30.6K]3 years ago

8 0

Answer:

0 I think

Explanation:

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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 gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assumin

alekssr [168]

Answer:

The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.

Solution:

As per the question:

Radius of an alveolus, R = $200\mu m = 200\times 10^{- 6}\ m$

Gauge Pressure inside, $P_{in} = 25\ mmHg$

Blood Pressure outside, $P_{o} = 10\ mmHg$

Now,

Change in pressure, $\Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa$

Since the alveolus is considered to be a spherical shell

The surface tension can be calculated as:

$\Delta P = \frac{4\pi T}{R}$

$T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m$

And we know that the surface tension of water is 72.8 mN/m

Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.

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3 years ago

6. A 145-g baseball moving 30.5 m/s strikes a stationary 5.75-kg brick resting on small rollers so it moves without significant

Sindrei [870]

Answer:

Explanation:

a )

momentum of baseball before collision

mass x velocity

= .145 x 30.5

= 4.4225 kg m /s

momentum of brick after collision

= 5.75 x 1.1

= 6.325 kg m/s

Applying conservation of momentum

4.4225 + 0 = .145 x v + 6.325 , v is velocity of baseball after collision.

v = - 13.12 m / s

b )

kinetic energy of baseball before collision = 1/2 mv²

= .5 x .145 x 30.5²

= 67.44 J

Total kinetic energy before collision = 67.44 J

c )

kinetic energy of baseball after collision = 1/2 x .145 x 13.12²

= 12.48 J .

kinetic energy of brick after collision

= .5 x 5.75 x 1.1²

= 3.48 J

Total kinetic energy after collision

= 15.96 J

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3 years ago

Why do the passengers in high-altitude jet planes feel the sensation of weight while passengers in an orbiting space vehicle, su

kykrilka [37]

Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.

In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.

Remember that the Force of Gravity is given under the principle

$F_g = \frac{GMm}{r^2}$

Where,

G = Gravitational Universal constant

M = Mass of the planet

m = mass of the object

r = Distance from center of the planet

When the radius grows considerably the gravitational force begins to decrease.

7 0

2 years ago

How can you find the reading of main scale and vernier scale

anygoal [31]

Answer:

this pdf should help you out

Explanation:

7 0

2 years ago