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1 year ago

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Often what one expects to see influences what is perceived in the surrounding environment. Please select the best answer from th

e choices provided T F

2 answers:

Stella [2.4K]1 year ago

8 0

Hello <span>Andijwiltbank
</span>

Question: <span>Often what one expects to see influences what is perceived in the surrounding environment. True or False?

Answer: True

Reason: What we observe about the environment decides what we believe about it and how we react.

Hope This Helps :-)
-Chris</span>

prohojiy [21]1 year ago

5 0

This statement is in-doubtingly true

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A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x

mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

Vf^2 = Vi^2 + 2*a*x

2*a*x = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

Let x = Xf -Xi

Where Xf is the final position of the cart and Xi the initial position of the cart.

x = 12.5 - 0

x = 12.5

The cart comes to a stop before changing direction

Vf = 0 m/s

a = (0^2 - 5^2)/ 2*12.5

a = - 1 m/s^2

The cart is decelerating

Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.

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1 year ago

Levi and Clara are trying to move a very heavy box. Levi is pushing the box with a force of 30 N, and Clara is pulling the box w

Komok [63]

There is a ner force of 15 N allowing Levi and Clara to mobe the box.

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1 year ago

Read 2 more answers

To support a tree damaged in a storm, a 12-foot wire is secured from the ground to the tree at a point 10 feet off the ground. T

podryga [215]

Answer:

The wire meet the ground at an angle of 56.4 degrees

Explanation:

It is given that,

To support a tree damaged in a storm, a 12-foot wire is secured from the ground to the tree at a point 10 feet off the ground.

The hypotenuse is, H = 12 foot

The perpendicular distance is, P = 10 feet

The angle between the tree and the ground is 90 degrees

Using Pythagoras theorem as :

$sin\theta=\dfrac{P}{H}$

$sin\theta=\dfrac{10}{12}$

$\theta=56.4^{\circ}$

So, the wire meet the ground at an angle of 56.4 degrees. Hence, the correct option is (d).

6 0

1 year ago

Coherent red light of wavelength λ = 700 nm is incident on two very narrow slits. The light has the same phase at both slits. (a

oksano4ka [1.4K]

(a) 0.028 rad

The angular separation of the nth-maximum from the central maximum in a diffraction from two slits is given by

$d sin \theta = n \lambda$

where

d is the distance between the two slits

$\theta$ is the angular separation

n is the order of the maximum

$\lambda$ is the wavelength

In this problem,

$\lambda=700 nm=7\cdot 10^{-7} m$

$d=0.025 mm=2.5\cdot 10^{-5} m$

The maximum adjacent to the central maximum is the one with n=1, so substituting into the formula we find

$sin \theta = \frac{n \lambda}{d}=\frac{(1)(7\cdot 10^{-7} m)}{2.5\cdot 10^{-5} m}=0.028$

So the angular separation in radians is

$\theta= sin^{-1} (0.028) = 0.028 rad$

(b) 0.028 m

The screen is located 1 m from the slits:

D = 1 m

The distance of the screen from the slits, D, and the separation between the two adjacent maxima on the screen (let's call it y) form a right triangle, so we can write the following relationship:

$\frac{y}{D}=tan \theta$

And so we can find y:

$y=D tan \theta = (1 m) tan (0.028 rad)=0.028 m$

(c) $2.8\cdot 10^{-4} rad$

In this case, we can apply again the formula used in part a), but this time the separation between the slits is

$d=2.5 mm = 0.0025 m$

so we find

$sin \theta = \frac{n \lambda}{d}=\frac{(1)(7\cdot 10^{-7} m)}{0.0025 m}=2.8\cdot 10^{-4}$

And so we find

$\theta= sin^{-1} (2.8\cdot 10^{-4}) = 2.8\cdot 10^{-4} rad$

(d) $7.0\cdot 10^{-6} m = 7.0 \mu m$

This part can be solved exactly as part b), but this time the distance of the screen from the slits is

$D=25 mm=0.025 m$

So we find

$y=D tan \theta = (0.025 m) tan (2.8\cdot 10^{-4} rad)=7.0\cdot 10^{-6} m = 7.0 \mu m$

(e) The maxima will be shifted, but the separation would remain the same

In this situation, the waves emitted by one of the slits are shifted by $\pi$ (which corresponds to half a cycle, so half wavelength) with respect to the waves emitted by the other slit.

This means that the points where previously there was constructive interference (the maxima on the screen) will now be points of destructive interference (dark fringes); on the contrary, the points where there was destructive interference before (dark fringes) will now be points of maxima (bright fringes). Therefore, all the maxima will be shifted.

However, the separation between two adjacent maxima will not change. In fact, tall the maxima will change location exactly by the same amount; therefore, their relative distance will remain the same.

8 0

1 year ago

If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100

SCORPION-xisa [38]

Answer:

Charge enter a 0.100 mm length of the axon is $8.98\times 10^{-12} C$

Explanation:

Electric field E at a point due to a point charge is given by

$E=k \frac{q}{r^2}$

where $k$ is the constant = $9.0 \times 10^9 Nm^2 / C^2$

$q$ is the magnitude of point charge and $r$ is the distance from the point charge

Charges entering one meter of axon is $5.\times 10^{11} \times (+e)$

Charges entering 0.100 mm of axon is $5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}$

substituting the value of $+e=1.6\times 10^{-19} C$ in above equation, we get charge enter a 0.100 mm length of the axon is

$q=5.\times 10^{11} \times1.6\times 10^{-19} \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C$

3 0

1 year ago