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Sophie [7]
3 years ago
8

If A > B, under what condition is |A-BI=|A|- IB|? a. Vectors A and B are in opposite directions b. Vectors A and B are in the

same direction. c. The statement is never true. d. Vectors A and B are in perpendicular directions. e. The statement is always true.

Physics
1 answer:
kupik [55]3 years ago
5 0

Answer:

b) Vectors A and B are in the same direction.

Explanation:

To understand this problem we will say that vector A has a magnitude of 5 units and vector B a magnitude of 3 units. In the subtraction of vectors the initial parts of vectors always bind together. And the vector resulting from the subtraction is traced from the end of the second vector (B) to the end of the first vector (A).

The length of the resultant vector will be 5 - 3 = 2

In the attached image, we analyze case a), b), and d)

For a)

As we can see in the attached image the resultant vector has a length of 8 units.

For d)

As we can see in the attached image the resultant vector has a length of 5.83 units.

For b)

The resultant vector has a length of 2 units.

Therefore the case given in b) is true

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6 0
3 years ago
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A light wave passes through an aperture (that is, a narrow slit). When it does so, the degree to which the wave spreads out will
crimeas [40]

Explanation:

Single slit diffraction

Diffraction is the phenomenon of spreading out of waves as they pass through an aperture or around objects. Diffraction occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave. For very small aperture sizes, the vast majority of the wave is blocked. in case of  large apertures the wave passes by or through the obstacle without any significant diffraction.

7 0
3 years ago
Which surface exhibits the highest albedo measurement
k0ka [10]
Fresh snow is the surface that exhibits the highest albedo measurement. 
8 0
3 years ago
Real bodies emit and absorb more radiation than a blackbody at the same temperature. True or False
monitta

Answer: False

Explanation:

Relative to the concept of radiations, a black body is an object capable of absorbing any form of electromagnetic radiation irrespective of its frequency or angle of incidence when incident on such object.

However, the same cannot be said about real bodies as real bodies are those which reflect all rays incident on them completely and uniformly in all directions.

One very important characteristic of black bodies is that they are ideal emmiters.

The concept of emmisivity is brought about by the existence of real bodies .

This is due to the fact that they are only able to emit radiation at a fraction of the black body energy levels.

Please note that by convention, the emmisivity of a real body is always less thaan 1.

As such they are not able to emit as much radiation as a black body at the same temperature.

3 0
3 years ago
A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.
gtnhenbr [62]

Answer:

(a)  vmax = 0.34m/s

(b)  v = 0.13m/s

(c)  v = 0.31m/s

(d)  x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

v_{max}=\omega A       (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

\omega=\sqrt{\frac{k}{m}}           (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

x=Acos(\omega t)

You solve the previous equation for t:

t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s

With this value of t, you can calculate the speed of the object with the following formula:

v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s

The position will be:

x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0
4 years ago
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