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likoan [24]
2 years ago
15

This means that the speed at which the bullet travels across Earth's surface (its magnitude of horizontal velocity) does not aff

ect _____.
the speed at which it falls toward the Earth

the rate at which it slows down

the distance it will travel

friction from the air
Physics
1 answer:
Dmitry_Shevchenko [17]2 years ago
6 0

Answer: the speed at which it falls toward the Earth.


Explanation:


A bullet travelling across Earth's surface with some horizontal velocity is classical example of projectile motion.


Projectile motion is an idealization of the motion under the action of gravity neglecting the influence of the air (no drag force nor friction).


This  kind of motion is the result of two independent motions: vertical motion and horizontal motion.


The observed net velocity is the vectorial sum of the vertical and horizontal velocities.


The horizontal velocity is constant, since there is not any force acting in the horizontal axis. Thi is, the object, following the first Law of Newton (inertia law) tends to continue in uniform rectilinear movement (with zero acceleration).


The vertical velocity, this is the velocity at which the bullet falls toward the Earth, is influenced (accelerated) by the action of the gravity of the Earth. So, the vertical velocity is accelerated by the pull of the Earth.


Vertical and horizontal velocities are independent of each other, which means that the speed or the magnitude of the horizontal velocity does not affect the speed at which an object (the bullet) falls toward the Earth.

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there is no questions associated with this what are the questions that go with this senario


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2 years ago
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Chromatic aberration comes from the fact that different wavelengths of light travel at different speeds through the material of
gtnhenbr [62]

Answer:

 y_red / y_blue = 1.11

Explanation:

Let's use the constructor equation to find the image for each wavelength

         1 /f = 1 /o + 1 /i

Where f is the focal length, or the distance to the object and i the distance to the image

Red light

           1 / i = 1 / f - 1 / o

           1 / i_red = 1 / f_red - 1 / o

           1 / i_red = 1 / 19.57 - 1/30

           1 / i_red = 1,776 10-2

           i_red = 56.29 cm

Blue light

            1 / i_blue = 1 / f_blue - 1 / o

            1 / i_blue = 1 / 18.87 - 1/30

            1 / i_blue = 1,966 10-2

            i_blue = 50.863 cm

Now let's use the magnification ratio

             m = y ’/ h = - i / o

             y ’= - h i / o

Red Light

            y_red ’= - 5 56.29 / 30

            y_red ’= - 9.3816 cm

Light blue

            y_blue ’= 5 50,863 / 30

            y_blue ’= - 8.47716 cm

The ratio of the height of the two images is

            y_red ’/ y_blue’ = 9.3816 / 8.47716

            y_red / y_blue = 1,107

            y_red / y_blue = 1.11

5 0
2 years ago
80 foot-pounds of work is needed to move the sofa in Tyler's apartment. Which of the following statements is true?
erastova [34]
D is the correct answer
hop it helped.
3 0
2 years ago
An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst
sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

T_1 = 500°C

T_2 = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to T_{avg} = 535.5K \approx 550K

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

P_r = 0.63

A)

Number for the first strips is equal to

R_e_x = \frac{u_o.L}{v}

R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4

Calculating heat transfer coefficient from the first strip

h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3

h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2

The rate of convection heat transfer from the first strip is

q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W

The rate of convection heat transfer from the fifth trip is equal to

q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)

h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2

Calculating h_o_-_4

h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2

The rate of convection heat transfer from the tenth strip is

q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)

h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2

Calculating

h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2

Calculating the rate of convection heat transfer from the tenth strip

q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W

The rate of convection heat transfer from 25th strip is equal to

q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)

Calculating h_o_-_2_5

h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2

Calculating h_o_-_2_4

h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2

Calculating the rate of convection heat transfer from the tenth strip

q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W

6 0
2 years ago
Which of the following can be reduced to a single number in standard form?
raketka [301]

Complete question is;

Which of the following can be reduced to a single number in standard form?

A) 3√3 + 5√8

B) 2√5 + 5√45

C) √7 + √9

D) 4√2 + 3√6

Answer:

Only option B) 2√5 + 5√45 can be reduced to a single number

Explanation:

A) For 3√3 + 5√8;

Let's simplify it to get;

3√3 + 5√(4 × 2)

From this, we get;

3√3 + (5 × 2)√2 = 3√3 + 10√2

This is 2 numbers and not a single number. Thus it can't be reduced to a single number in standard form.

B) 2√5 + 5√45

Simplifying to get;

2√5 + 5√(9 × 5)

This gives;

2√5 + (5 × 3)√5 = 2√5 + 15√5

Adding the surds gives;

17√5.

This is a single number and thus can be reduced to a single number

C) For √7 + √9

Simplifying, to get;

√7 + 3.

This is 2 numbers and not a single number. Thus it can't be reduced to a single number in standard form.

D) 4√2 + 3√6

Thus can't be simplified further because both numbers inside the square root don't have factors that are perfect squares.

Thus, it remains 2 numbers and not a single number and can't be reduced to a single number in standard form.

6 0
2 years ago
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