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2 years ago

You walk to the north, then turn 90° to your left and walk another How far are you from where you originally started?

Physics

1 answer:

rodikova [14]2 years ago

3 0

Whatever distance north and then west you walked, you are then

(1.41 x that distance)

northwest of where you started.

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3. A cat walks 0.220km North, then 0. 120 km South in a time of 400 seconds. whats the displacement and average velocity?

Andru [333]

Answer:

The rate at which velocity changes with respect to a change in time is called. acceleration.

Explanation:

4 0

3 years ago

Weddell seals make holes in sea ice so that they can swim down to forage on the ocean floor below. Measurements for one seal sho

Mariulka [41]

Answer:

B. 1 m/s

Explanation:

Metric unit conversions:

0.3 km = 300m

5 minutes = 5*60 = 300 seconds

So if a seal can reach a depth of 300m in a time of 300 seconds, its diving speed is the distance divided by time duration

v = s/t = 300/300 = 1m/s

So B is the correct answer

3 0

2 years ago

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed,

KATRIN_1 [288]

Answer:

a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

V = i (R₁ + R₂)

with R₁ connected

V = (i + 0.5) R₁

with R₂ connected

V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

V = i ( $\frac{V}{ i+0.5} + \frac{V}{i+0.25}$ )

1 = i ( $\frac{1}{i+0.5} + \frac{1}{i+0.25}$ )

1 = i ( $\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }$ ) = $\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}$

i² + 0.75 i + 0.125 = 2i² + 0.75 i

i² - 0.125 = 0

i = √0.125

i = 0.35355 A

with the second equation we look for R1

R₁ = $\frac{V}{i+0.5}$

R₁ = 12 /( 0.35355 +0.5)

R₁ = 14.1 Ω

with the third equation we look for R2

R₂ = $\frac{V}{i+0.25}$

R₂ = $\frac{12}{0.35355+0.25}$

R₂ = 19.9 Ω

3 0

2 years ago

A blue-green photon (λ = 488 nm ) is absorbed by a free hydrogen atom, initially at rest. What is the recoil speed of the hydrog

Natalka [10]

Answer:

The recoil speed is $2.207\times 10^{4} m/s$

Solution:

Wavelength of a blue-green photon, $\lambda_{BG} = 488 nm = 488\times 10^{- 9} m$

Now, the energy associated with the blue-green photon:

$E_{BG} = \frac{hc}{\lambda_{BG}}$

where

h = Planck's constant

C = speed of light ion vacuum

$E_{BG} = \frac{6.626\times 10^{- 34}\times 3\times 10^{8}}{488\times 10^{- 9}}$

$E_{BG} = 4.07\times 10^{- 19} J$

Also, we know that the recoil speed can be calculated by the KInetic energy which is equal to the Energy of the blue-green photon:

$KE_{H} =\frac{1}{2}m_{p}v_{H}$

where

$v_{H}$ = velocity of Hydrogen atom

$m_{p} = 1.67\times 10^{- 27} kg$ = mass of H-atom

Now,

$KE_{H} =\frac{1}{2}m_{p}(v_{H})^{2}$

$4.07\times 10^{- 19} =\frac{1}{2}\times 1.67\times 10^{- 27}\times (v_{H})^{2}$

$v_{H} = \sqrt(4.87\times 10^{8}) = 2.207\times 10^{4} m/s$

7 0

3 years ago

Which of the following is of particular interest to scientists in their observations because of potential threats to the Earth?

nordsb [41]

The answer is Near Earth Objects (NEOs).

Near Earth Objects (NEOs) have gained particular interest to scientists in their observations because of its potential threats to the Earth. It can be comets or asteroids. It is called Near Earth since their orbits are too close to cross the orbit of the Earth thus NEO can reach the space near-Earth that more likely give potential threats to the Earth.

8 0

3 years ago