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neonofarm [45]
3 years ago
8

Which kinds of objects emit visible light in the electromagnetic spectrum?1)all objects2)radioactive objects3)relatively cold ob

jects4)relatively hot objects
Physics
1 answer:
Lostsunrise [7]3 years ago
8 0
Relatively hot objects emit visible light.
Some examples: 
==> the wire coils in the toaster; 
==> the spoon that you stuck in the flame on the stove;
==> the fine wire in the lightbulb when current goes through it.

VERY radioactive objects also do that.  But if you're actually
standing there watching an object that's THAT radioactive,
then you're in big trouble.
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A black, totally absorbing piece of cardboard of area a = 2.1 cm2 intercepts light with an intensity of 8.8 w/m2 from a camera s
PolarNik [594]

Answer:

The value of radiation pressure is 2.933 \times 10^{-8} Pa

Explanation:

Given:

Intensity I = 8.8 \frac{W}{m^{2} }

Area of piece A = 2.1 \times 10^{-4} m^{2}

From the formula of radiation pressure in terms of intensity,

   P = \frac{I}{c}

Where P = radiation pressure, c = speed of light

We know value of speed of light,

 c = 3 \times 10^{8} \frac{m}{s}

Put all values in above equation,

  P = \frac{8.8}{3 \times 10^{8} }

  P = 2.933 \times 10^{-8} Pa

Therefore, the value of radiation pressure is 2.933 \times 10^{-8} Pa

8 0
4 years ago
A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g
mr Goodwill [35]

The distance an object falls from rest through gravity is

                         D  =  (1/2) (g) (t²)

            Distance  =  (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected
if  ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.

                                                      D  =  (1/2) (g) (t²)

Multiply each side by  2 :         2 D  =            g    t² 

Divide each side by ' g ' :      2 D/g =                  t²

Square root each side:        t = √ (2D/g)


Looking at the equation now, we can see what happens
to ' t ' when only ' g ' changes: 

-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
                                             and smaller 'g' ==> longer 't' .

-- They don't change by the same factor, because  1/g  is inside
the square root.  So 't' changes the same amount as  √1/g  does.

Gravity on the surface of the moon is roughly  1/6  the value
of gravity on the surface of the Earth.

So we expect ' t ' to increase by  √6  =  2.45 times.

It would take the same bottle  (2.45 x 4.95) = 12.12 seconds
to roll off the same window sill and fall 120 meters down to the
surface of the Moon.
4 0
3 years ago
Read 2 more answers
How does water get to the tops of the tallest trees against the force of gravity
alexira [117]
Water gets to the leaves in the tops of the tallest trees by something called the cohesion-tension theory. Water has two very unique properties called adhesion and cohesion. Cohesion is the tendency of water molecules to stick together with one another. The water sticks together, leaving no room for air, strengthening the "force" of the water going up the tree. The water also sticks to the sides of the xylem inside the tree. In addition to these properties, there are also the factors of negative and positive water potential. For more information, look up more details of the cohesion-tension theory. 
4 0
3 years ago
Read 2 more answers
A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
3 years ago
You push a cart with mass 15 kg forward , giving it an acceleration of 3 m/s ^ 2 . How much force did you apply ? O A. 0.2N O B.
Alexandra [31]

Answer: force = mass x accelegation

Explanation: 15x3 = 45 N

7 0
3 years ago
Read 2 more answers
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