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2 years ago

12

A red card is illuminated by red light. Part A What color will the card appear? What color will the card appear? a. Red b. Black

c. White d. Green

1 answer:

Wewaii [24]2 years ago

4 0

Red light reflects off the card into your eyes and you see the red card as red. The light will just make the card brighter. So A

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All mutations are favorable and increase the ability for the

masya89 [10]

False sometimes you get a bad mutation

3 0

2 years ago

What would happen if you touched an object that is the same temperature as your hand? Would any heat transfer happen? Explain yo

SashulF [63]

Answer:

Explanation:

When two objects are in thermal equilibrium they are said to have the same temperature. During the process of reaching thermal equilibrium, heat, which is a form of energy, is transferred between the object

which means that it refers to transfer through a selectively permeable partition, the contact path.[1] For the relation of thermal equilibrium, the contact path is permeable only to heat; it does not permit the passage of matter or work; it is called a diathermal connection. According to Lieb and Yngvason, the essential meaning of the relation of thermal equilibrium includes that it is reflexive and symmetric. It is not included in the essential meaning whether it is or is not transitive. After discussing the semantics of the definition, they postulate a substantial physical axiom, that they call the "zeroth law of thermodynamics", that thermal equilibrium is a transitive relation. They comment that the equivalence classes of systems so established are called isotherms

plz dont be mad that i coppied it sounded so good so i wanted veryone to see it when they look bc i am dumb

4 0

2 years ago

Shawn uses 45 N of force to stop the cart 27 meter from running his foot over. How much work does he do?

Ede4ka [16]

Answer:

W=f×d

w=45×27

w=1215 j.............

3 0

3 years ago

If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.

AleksAgata [21]

Answer:

a) $f_1=5.587Hz$

b) $f_{n+1}-f_n=5.587Hz$

Explanation:

The frequency of the $n^{th}$ harmonic of a vibrating string of length <em>L, </em>linear density $\mu$ under a tension <em>T</em> is given by the formula:

$f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}$

a) So for the <em>fundamental mode</em> (n=1) we have, substituting our values:

$f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz$

b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:

$f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz$

3 0

2 years ago

An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplan

MArishka [77]

Answer:

2560m or 2.56km (rounded to 3 significant figures)

Explanation:

First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):

$Vertical \ velocity \ (\frac{m}{s} ) = u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) = u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) = a_{v} = 9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) = a_{h} = 0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}$

The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;

First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;

Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);

What we have is the horizontal velocity but we don't have the time taken;

One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;

What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:

800 = (0)t + ¹/₂·(9.8)t²

800 = 4.9t²

t² = 163.26...

t = 12.77...

We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:

$u_{h} = \frac{s_{h} }{t} \\\\ 200 = \frac{s_{h} }{12.77...} \\\\ s_{h} = 200(12.77...) \\\\ s_{h} = 2555.5...$

7 0

2 years ago