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3 years ago

X-component of length 5 and a y-component of length 4. What is the angle of the vector?

1 answer:

4 0

The direction of the vector with respect to the x-axis
is the angle whose tangent is (4/5) .

Arctan(4/5) = 38.7° counterclockwise from the x-axis.

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Any help guys? I am stuck on two problems.

Juli2301 [7.4K]

Ax=A*Cos56

= 20*(0.559)

= 11.18 N

8 0

3 years ago

A camera has a single converging lens with a fixed focal length f. (a) How far should the lens be from the film (or in a present

Varvara68 [4.7K]

Answer:

a) Due to the characteristic that a converging lens focuses light rays from infinity and parallel to its main axis. Therefore, the lens should be placed at a distance "f" from the film, in this way it will form the image of the object placed at infinity in said film.

b) Since the converging lens produces an image of an object placed at a distance of 2f, the lens must be placed at the same distance (2f), so that this object that is placed at a distance of 2f is focused.

Explanation:

4 0

2 years ago

A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. ten second later, he is moving at 15 m/s. what is his

pantera1 [17]

$\Large {{ \sf {Question :}}}$

<h3>A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. Ten second later, he is moving at 15 m/s. What is his acceleration?</h3>

$\Large {{ \sf {Given :}}}$

<h3>Initial Velocity (<em>u</em>) - 5 m/s</h3><h3>Final Velocity (<em>v</em>) - 15 m/s</h3><h3>Time (<em>t</em>) - 10 sec</h3>

$\Large {{ \sf {Formulae :}}}$

<h3>If the velocity of an object changes from an initial value <em>u </em>to the final value <em>v </em>in time <em>t,</em><em> </em>the acceleration <em>a</em> is, </h3><h3> $a \: = \frac{v - u}{t}$ </h3><h3> $\Large {{ \sf {Step-by-step explanation :}}}$ </h3>

$a \: = \frac{v - u}{t} \\ or \: \: a = \frac{(15 - 5)}{10} m \: s^{ - 2} \\ or \: \: a \: = \frac{10}{10}m \: s^{ - 2} \\ or \: \: a = 1m \: s^{ - 2}$

$\Large {{ \sf {Answer :}}}$

<h3>His acceleration is </h3><h3> $1m \: s^{ - 2}$ </h3><h3 /><h3 />

5 0

3 years ago

Question: Self-test 3.12 Calculate the change in G for ice at -10°C, with density 917 kg mº, when the pressure is increased from

Akimi4 [234]

The change in the Gibb's free energy per mole (G) is 1.96 J.

The given parameters:

Density of the ice, ρ = 917 kg/m³
Initial pressure, P₁ = 1.0 bar
Final pressure, P₂ = 2.0 bar
Temperature, T = - 10 C
Mass of water = 18 g

The change in the Gibb's free energy per mole (G) is calculated as follows;

$\Delta G = V(P_2-P_1) \\\\$

where;

V is the volume of the ice

$Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density} \\\\Volume = \frac{18 \times 10^{-3} \ kg}{917 \ m^3} \\\\Volume = 1.96 \times 10^{-5} \ m^3\\\\Volume = 1.96 \times 10^{-5} \ m^3 \times \frac{1000 \ L}{m^3} \\\\Volume = 0.0196 \ L$

Change in pressure;

$P_2 - P_1 = 2.0 \ bar \ - \ 1.0 \ bar = 1.0 \ bar = 0.987 \ atm$

The change in the Gibb's free energy per mole (G);

$\Delta G= V(P_2-P_1)\\\\\Delta G = 0.0196\ L \times 0.987\ atm \\\\\Delta G = 0.0193 \ L.atm\\\\1 \ L.atm = 101.325 \ J\\\\\Delta G = 0.0193 \ L.atm \times \frac{101.325 \ J}{1 \ L.atm} \\\\\Delta G = 1.96 \ J$

Thus, the change in the Gibb's free energy per mole (G) is 1.96 J.

Learn more about Gibb's free energy here: brainly.com/question/10012881

3 0

2 years ago

Which element would have the lowest electronegativity? (1 point)O an element with a large number of valence electrons and a larg

nekit [7.7K]

Answer:

an electron with element with large number of valence electron and large atomic <u>radius</u>

8 0

2 years ago