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Dmitry_Shevchenko [17]
3 years ago
9

What happens when infrared hits a shiny surface

Physics
1 answer:
saw5 [17]3 years ago
3 0
Will reflect ........
You might be interested in
Plz help!!! A man drives south for 362 s at 45.8 m/s.  How far does he get?<br><br>​
Aleks [24]

Answer:

16,579.6

Explanation:

i think you multiply 362×45.8= 16,579.6 and then whatever your unit of measurement is

6 0
3 years ago
A sled drops 50 meters in height on a hill. The mass of the rider and sled is 70 kg and the sled is going 10 m/s at the bottom o
navik [9.2K]

Answer:

Efficiency = 10.2 %

Explanation:

Given the following data;

Mass = 70 kg

Height = 50 m

Velocity = 10 m/s

We know that acceleration due to gravity is equal to 9.8 m/s².

To find the efficiency of energy conversion from potential to kinetic;

First of all, we would determine the potential energy;

P.E = mgh

P.E = 70 * 9.8 * 50

P.E = 34300 J

For the kinetic energy;

K.E = ½mv²

K.E = ½ * 70 * 10²

K.E = 35 * 100

K.E = 3500

Therefore, Input energy, I = 34300 J

Output energy, O = 3500 J

Next, we find the efficiency;

Efficiency = O/I * 100

Substituting into the formula, we have;

Efficiency = 3500/34300 * 100

Efficiency = 0.1020 * 100

Efficiency = 10.2 %

4 0
2 years ago
1.Convert 340 cm into m *(answer=0.34m)
Nataly [62]

Answer:

<em>1</em><em>.</em><em>for </em><em>the </em><em>first </em><em>one </em><em>100c</em><em>e</em><em>n</em><em>t</em><em>i</em><em>m</em><em>e</em><em>t</em><em>e</em><em>r</em><em>s</em><em> </em><em>make </em><em>1</em><em> </em><em>meter </em><em>therefore</em>

<em>100c</em><em>m</em><em>-</em><em>1</em><em>m</em>

<em>3</em><em>4</em><em>0</em><em>c</em><em>m</em><em>-</em><em>x</em>

<em>3</em><em>4</em><em>0</em><em>/</em><em>100</em>

<em>=</em><em>3</em><em>.</em><em>4</em>

<em>the </em><em>answer </em><em>is </em><em>supposed</em><em> to</em><em> be</em><em> </em><em>3</em><em>.</em><em>4</em><em>,</em><em> maybe</em><em> </em><em>there's</em><em> </em><em>a </em><em>mistake</em><em> </em><em>with </em><em>the </em><em>question</em><em> </em><em>or </em><em>the </em><em>answer</em>

<em>2</em><em>.</em><em>t</em><em>h</em><em>e</em><em> </em><em>weight</em><em> </em><em>of </em><em>a </em><em>body </em><em>is </em><em>given </em><em>by </em><em>the </em><em>formula</em>

<em>mass×</em><em>g</em><em>r</em><em>a</em><em>v</em><em>i</em><em>t</em><em>y</em><em>,</em><em>in </em><em>this </em><em>case </em><em>the </em><em>mass </em><em>is </em><em>7</em><em>5</em><em>k</em><em>g</em><em> </em><em>and </em><em>the </em><em>gravity </em><em>is </em><em>9</em><em>.</em><em>8</em>

<em>weight</em><em>=</em><em>7</em><em>5</em><em>×</em><em>9</em><em>.</em><em>8</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>7</em><em>3</em><em>5</em><em>N</em>

<em>3</em><em>.</em><em>f</em><em>o</em><em>r</em><em> </em><em>this </em><em>one </em><em>the </em><em>mass </em><em>of </em><em>a </em><em>body </em><em>is </em><em>given</em><em> by</em><em> the</em><em> formula</em>

<em>mass=</em><em>weight/</em><em>gravity</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>4</em><em>2</em><em>0</em><em>/</em><em>9</em><em>.</em><em>8</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>4</em><em>2</em><em>.</em><em>8</em><em>k</em><em>g</em>

<em>I </em><em>hope</em><em> this</em><em> helps</em>

4 0
3 years ago
A point charge of 5.7 µc moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram
Sauron [17]

The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

<h3>What is magnetic force?</h3>

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

\rm F=qvBsin\theta

where

The magnitude of the charge   q=5.7\ \mu C =5.7\times 10^{-6} \ C

The velocity of the charge        v=4.5\times 10^5\ \frac{m}{s}

The magnitude of the magnetic field   3.2\ mT=0.0032\ T

The angle between the directions of v and B  \theta =90^o-37^o=53^o

By substituting the values we will get:

F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)

F=6.6\times 10^{-3}\ N

Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

To know more about Magnetic force follow

brainly.com/question/14411049

3 0
2 years ago
Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109
Oksana_A [137]

Answer:

 A) F = 1.09 10 5 N, b) Yes  

Explanation:

Part A

For this exercise we need the number of free electrons in copper, as the valence of copper +1 there is a free electron for each atom. Let's use the concept of density to find the mass of copper in the sphere

               ρ = m / V

               .m = ρ V = ρ 4/3 π r³

The radius is half the diameter

               r = 1.9 10⁻² / 2 = 0.95 10⁻² m

               ρ = 8960 kg / m3

               m = 8960 4/3 π (0.95 10⁻²)³

               m = 3.2179 10⁻² kg

The molecular weight of copper is 63,546 g / mol which has 6,022 10²³ atoms

With this we can use a rule of proportions to enter the number of atom is this mass

             #_atom = 6.022 10²³ 3.2179 10⁻² / 63.546 10⁻³

             #_atom = 3,049 10²³

Therefore there is the same number of electrons, as they indicate that the charge of the protone and the electon differs by 1/10⁹ the total charge for each spherical is

               q = e / 10⁹    #_atom

               q = e / 10⁹    3,049 1023

               q = 3,049 10⁴  (-1.6 10⁻¹⁹)

               q = -4,878 10-5 C

Electric force is

             F = k q₁q₂ / r²

             F = k q² / r²

             

Let's calculate

             F = 8.99 10⁹ (4.878 10⁻⁵)²2 / (1.4 10⁻²)²

              F = 1.09 10 5 N

This is a force of repulsion.

Part B

 The magnitude of this force is  in very easy to detect

4 0
3 years ago
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