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Zigmanuir [339]
3 years ago
7

For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vec

tors? For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vectors? The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector. The velocity vector is parallel to the acceleration vector; the acceleration vector is parallel to the net force vector. The velocity vector is perpendicular to the acceleration vector; the acceleration vector is perpendicular to the net force vector. The velocity vector is parallel to the acceleration vector; the acceleration vector is perpendicular to the net force v
Physics
1 answer:
tigry1 [53]3 years ago
7 0

Answer: The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector.

Explanation:

In a uniform circular motion, the meaning of "uniform"is the same as for an uniform straight motion, i.e., the module of the velocity vector (its speed) is constant.

Now, if the object were not describing a circular trajectory, it should move at constant speed, in a straight line, provided no external forces acted upon it.

If there is an external force acting on it, making it to follow a circular trajectory, this force doesn't change the instantaneous value of the velocity, but it changes his direction instead.

While the direction is changing, it always keep tangential to the trajectory, due to if at any moment the force disappears, the body must continue in a straight line at constant speed, following a line tangent to the circle.

It can be showed, that the acceleration vector, defined as the change in velocity over time, always aims towards the center of the circle, and is perpendicular to the velocity vector.

As the only net force acting on the object (assuming a horizontal trajectory), is the one that causes the acceleration, the acceleration vector has the same direction as the net force.

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Determine the angle between the directions of vector a = 3.00i 1.00j and vector b = 1.00i 3.00j .
Katena32 [7]

The angle between the two vectors is 126° 52' 11".

The given parameters;

vector A = 3.00i + 1.00j

vector B = 1.00i + 3.00j

The angle between the two vectors is calculated as follows;

cos  \theta = \frac{A.B}{|A|.|B|}

The dot product of vectors A and B is calculated as;

A.B = ( 3i + 1j ) . ( 1i +3j )

      = ( 3 × 1 ) + ( 1  × 3 )

      = 3 + 3

      = 6

The magnitude of vectors A and B is calculated as;

|A|  = \sqrt[]{3^2 + 1^2} = \sqrt[]{10} \\

|B| = \sqrt[]{1^2 + 3^2} = \sqrt[]{10} \\

The angle between in two vectors is calculated as;

Cos \theta = \frac{6}{\sqrt[]{10}\sqrt[]{10}  } \\\\Cos \theta = \frac{6}{\ 10 }\\\\Cos \theta = 0.6\\\\\theta = cos^-^1 (0.6)\\\\\theta = 126\\

Therefore, the angle between the two vectors is 126° 52' 11".

Learn more about vectors here:

brainly.com/question/25705666

#SPJ4

6 0
2 years ago
while doing a lab , a student found the density of a piece of pure aluminum to be 2.80 g/cm . The accepted value for the density
postnew [5]

Look at the attachment or go online and search the question to get the answers you need


4 0
4 years ago
A stiff wire 44.0 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the l
Helga [31]

The magnitude of force acting on wire will be 1.90 N and the direction of the force acting on the wire will be 41.9 degrees below the negative y axis.

Explanation:

It is known that the force acting on a current carrying conductor placed in a magnetic field is

F=BIL sin theta

Here B is the magnetic field, I is the current flowing through the wire and L is the length of the wire which is given as 44 cm.

Since the wire is bended in the middle at right angle so the length of the two sides of the wire will be 22 cm each. Also one part is lying over z axis and another part lies in the plane of xy in the equation of line y = 2x. So the slope of this wire will be

\frac{y}{x} =2

This will be equal to tan θ.

So θ = tan⁻¹ (2) =63.4°

Then, the length of the wire will be written as components of i, j and k.

L = (-22)k+(22) cos ( 63.4) i+(22) sin (63.4)j

L = 0.098 i+0.197 j-0.22k

Then,

F = I (L × B)

F = 20.5 ((0.0985 i + 0.197 j -0.22k) * (0.316 i))

F = 20.5 (\left[\begin{array}{ccc}i&j&k\\0.098&0.197&-0.22\\0.316&0&0\end{array}\right] )

F = 20.5(i(0)-j(0-(-0.22*0.316))+k(0-(0.316*0.197))) = 20.5(-0.069 j-0.062 k)

F = -1.415 j-1.271 k

The magnitude of force on the wire will be

F = \sqrt{(-1.415)^{2}+(-1.27)^{2}  } = \sqrt{3.615}=1.90 N

And the direction can be found by the tan inverse of the ratio of k component to j component of the force.

theta = tan-1(\frac{-1.271}{-1.415})= 41.9 degrees

So the magnitude of force acting on wire will be 1.90 N and the direction of the force acting on the wire will be 41.9 degrees below the negative y axis.

5 0
4 years ago
Light shines through a single slit whose width is 5.7 x 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a
lilavasa [31]

Answer:

\lambda = 570\ nm

Explanation:

Given,

Width of slit, W = 5.7 x 10⁻⁴ m

Distance between central bright fringe, L = 4 m

distance between central bright fringe and first dark fringe, y = 4 mm

Diffraction angle

tan \theta = \dfrac{y}{L}

tan \theta = \dfrac{4}{4\times 10^3}

\theta = 0.0572

Now.

W sin \theta = m \lambda

m = 1

5.7 \times 10^{-4} \times sin (0.0572) = 1 \times \lambda

\lambda = 569.99 \times 10^{-9}\ m

\lambda = 570\ nm

4 0
4 years ago
The motor on a helicopter turns at an angular speed of 6.2 x 102 revolutions per minute. (a) Express this angular speed in radia
arsen [322]

Answer:

  • 64.93 rad/s
  • 38.956 km

Explanation:

(a)

  \dfrac{6.2\cdot 10^2\,\text{rev}}{60\,\text{s}}\times\dfrac{2\pi\,\text{rad}}{\text{rev}}=\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\approx\boxed{64.93\,\text{rad/s}}

__

(b)

  d=r\theta=(3.0\,\text{m})(2.0\cdot 10^2\,\text{s})\left(\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\right)=124\pi\cdot 10^2\,\text{m}=\boxed{38\,956\,\text{m}}

6 0
3 years ago
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