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Montano1993 [528]
2 years ago
10

vector ????⃗ has a magnitude of 17.9 and its direction is 80∘ counter‑clockwise from the x- axis. what are the x- and y- compone

nts of the vector?
Physics
1 answer:
Reika [66]2 years ago
5 0

We have vector (17.9*cos80^{0},17.9*sin80^{0})

Therefore,

x component = 17.9 * cos80 degree = 3.108

y component = 17.9 * sin80 degrees = 17.628

<h3>What is a vector?</h3>

An object with both magnitude and direction is referred to be a vector. A vector can be visualized geometrically as a directed line segment, with an arrow pointing in the direction and a length equal to the magnitude of the vector. The vector points in a direction from its tail to its head.

If the magnitude and direction of two vectors match, they are the same vector. This shows that if we move a vector to a different location without rotating it, the final vector will be the same as the initial vector. The vectors that denote force and velocity are two examples. The direction of force and velocity are both fixed. The size of the vector would represent the force's strength or the velocity's corresponding speed.

To know more about vectors, visit:

brainly.com/question/12937011

#SPJ4

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How many orbitals are allowed in a subshell if the angular momentum quantum number for electrons in that subshell is 3?
svp [43]

Answer:

7 orbitals are allowed in a sub shell if the angular momentum quantum number for electrons in that sub shell is 3.

Explanation:

We that different values of m for a given l provide the total number of ways in which a given s, p,d and f sub shells in presence of magnetic field can be arranged in space along x, y ,z- axis or total number of orbitals into which a given subshell can be divided.

    Range for given l lies between -l to +l .

The possible values of m are -3 , -2 , -1 , 0 , 1 ,2 , 3 .

    Total number of orbitals are 7.

4 0
3 years ago
A thin spherical shell with radius R1 = 4.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm . Both
kakasveta [241]

Answer:

The right response will be "450 volts".

Explanation:

The given values are:

R1 = 4.00 cm

R2 = 6.00 cm

q1 = +6.00 nC

q2 = −9.00 nC

As we know,

The potential difference between the two shell's difference will be:

⇒  \Delta V=K[(\frac{q1}{R1}+\frac{q2}{R2})-(\frac{q1}{R1} +(\frac{q2}{R2}))]

           =K[\frac{q1}{R2}-\frac{q1}{R1} ]

On substituting the values, we get

           =(9\times 10^9)[\frac{6\times 10^{-9}}{0.04}-\frac{6\times 10^{-9}}{0.06}]  

       Δ =450 \ volts

3 0
3 years ago
4
Zigmanuir [339]

(a) The object moves with uniform velocity from A to B.

(b) The object moves with constant velocity from B to C.

(c) The object moves with increasing velocity from C to D.

<h3>Velocity of the object from point A to B</h3>

V(A to B) = (6 - 0)/(4 - 0) = 1.5 m/s

<h3>Velocity of the object from point B to C</h3>

V(B to C) = (6 - 6)/(11 - 4) = 0 m/s

<h3>Velocity of the object from point C to D</h3>

V(C to D) = (7 - 6)/(12 - 11) = 1 m/s

final velocity = 1 + 1.5 m/s = 2.5 m/s

Thus, we can conclude the following;

The object moves with uniform velocity from A to B.

The object moves with constant velocity from B to C.

The object moves with increasing velocity from C to D.

Learn more about velocity here: brainly.com/question/6504879

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4 0
2 years ago
The electric potential 1.34 m from a charge is 580 V. What is the value of the charge? Include the sign, + or -. (The answer is
blagie [28]

Answer: 8.6

Explanation:

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In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 1.00 mm away. part a
mihalych1998 [28]
<span>We can use Coulomb's law to find the force F acting on the proton that is released. F = k x Q1 x Q2 / r^2 k = 9 x 10^9 Q1 is the charge on one proton which is 1.6 x 10^{-19} C Q2 is the same charge on the other proton r is the distance between the protons F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2 F = 2.304 x 10^{-22} N We can use the force to find the acceleration. F = ma a = F / m a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg) a = 1.38 x 10^5 m/s^2 The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
8 0
3 years ago
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