For each picture below identify the material and closing the sandwich as transparent translucent or opaque

The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p

NISA [10]

Answer:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be $v_{s}$ = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

$t = \frac{d}{v_{s} }$

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be $v_{r}$ = 2.5 m/s

$S = v_{r} t$

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

$cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}$

d) Downstream velocity of the swimmer, $v_{y} = v_{s} sin \theta\\$

$v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s$

The vertical displacement is given by, $y = v_{y} t$

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

$v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s$

The downstream horizontal distance of the swimmer, $x = v_{x} t$

x = 1.6 * 66.67

x = 106.67 m

7 0

2 years ago

In a lever, the effort arm is two times as long as the load arm. The resultant force will be

jasenka [17]

Answer is B.

In a lever, the effort arm is 2 times as a long as the load arm. The resultant force will be twice the applied force.

Hope it helped you.

-Charlie

5 0

2 years ago

Read 2 more answers

Differentiate scalar & vector quantity?

Keith_Richards [23]

$\textbf{Hello Friend}$

Scalar Quantity :-

→ These are the quantities with magnitude only . These quantities doesn't have to be mentioned with direction

eg.)=> Mass , Temprature .

Vector Quantity :-

→ These quantities are described with both Magnitude and Direction . These quantities follow special type of algebra called Vector algebra .

eg.)=> Force , Displacement

_______________________________

Hope It Helps You. ☺

5 0

3 years ago

If you placed a dot on a world map to represent the location of every earthquake in the last 100 years, where would you see eart

Vedmedyk [2.9K]

If we are to place dots to teh places that have been struck by an earthquake these past 100 years, the dots would be concentrated in the east and southeast Asia region. This is because of the presence of the Pacific ring of fire. This is a major area in the Pacific Ocean where most of the earthquakes are likely to occur.

6 0

2 years ago

Imagine an alternate universe where the value of the Planck constant is . In that universe, which of the following objects would

HACTEHA [7]

Question: The planck constant was not given. In this calculation, planck constant of 6.62607*10^-9 Js is used for the calculation.

Answer:

(a) A virus -------------Classical

(b) A buckyball -----Classical

(c) A mosquito ------ Quantum

(d) A turtle ------------Quantum

Explanation:

Calculating the wavelength using the formula;

λ= h/(mv)

where

λ= Wavelength

h = Planck Constant = 6.62607*10^-9 Js

m = mass in kg

v = velocity in m/s

Virus size = 280. nm = 2.80*10⁻⁷ m

a)

A Virus:

m = 9.4 x 10-17 g 9.4*10⁻²⁰ kg

v = 0.50 µm/s = 5 *10⁻⁷ m/s

h = 6.62607*10^-9 Js

Virus size = 280 nm = 2.80*10⁻⁷ m

Substituting into the formula; we have

λ= h/(mv)

λ= 6.62607*10^-9/ (9.4*10⁻²⁰* 5 *10⁻⁷)

= 6.62607*10^-9/4.7*10^-26

= 1.4*10^17 m

Classical : Wavelength is bigger than it's size

(b)

A buckyball

m = 1.2 x 10-21 g = 1.2 *10⁻²⁴ kg

V = 37 m/s

Size = 0.7 nm = 7*10⁻¹⁰ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 1.2 *10⁻²⁴* 37)

= 6.62607*10^-9/4.44*10^-23

= 1.49 *10^14 m

Classical : Wavelength is bigger than it's size

(c)

A mosquito

Mass = 1.0 mg = 1*10⁻⁶ kg

v = 1.1 m/s

Size = 6.3 mm = 6.3*10⁻³ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 1*10⁻⁶* 1.1)

= 6.62607*10^-9/1.1*10^-6

= 6.02*10^-3 m

Quantum Approach: The wavelength and the size are comparable

(d)

A turtle

Mass = 710. g = 0.71 kg

Size = 22. cm = 0.22 m

V = 2.8 cm/s. = 0.028 m/s

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 0.71* 0.028)

= 6.62607*10^-9/0.01988

= 3.33*10^-7 m

Quantum Approach: The wavelength and the size are comparable

8 0

3 years ago

For each picture below identify the material and closing the sandwich as transparent translucent or opaque￼

For each picture below identify the material and closing the sandwich as transparent translucent or opaque