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svet-max [94.6K]
2 years ago
11

Unit Test Review

Physics
1 answer:
PolarNik [594]2 years ago
4 0

Answer:

option B...

they represent different concept...

i hope this will answer your question

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Two particles, each with charge Q, and a third charge q, are placed at the vertices of an equilateral triangle as shown. The tot
Gelneren [198K]

Answer:

<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>

Explanation:

The image is shown below.

The force on the particle with charge q due to each charge Q = \frac{kQq}{r^{2} }

we designate this force as N

Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.

Resolving the forces on the particle, we have

for the x-component

N_{x} = N cosine 60° + (-N cosine 60°) = 0

for the y-component

N_{y} = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N

The above indicates that there is no resultant force in the x-axis, since it is equal to zero (N_{x} = 0).

The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.

<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>

5 0
4 years ago
magine two carts, one with twice the mass of the other, that are going to have a head-on collision. In order for the two carts t
scoray [572]

Answer:

Twice as fast

Explanation:

Solution:-

- The mass of less massive cart = m

- The mass of Massive cart = 2m

- The velocity of less massive cart = u

- The velocity of massive cart = v

- We will consider the system of two carts to be isolated and there is no external applied force on the system. This conditions validates the conservation of linear momentum to be applied on the isolated system.

- Each cart with its respective velocity are directed at each other. And meet up with head on collision and comes to rest immediately after the collision.

- The conservation of linear momentum states that the momentum of the system before ( P_i ) and after the collision ( P_f ) remains the same.

                             P_i = P_f

- Since the carts comes to a stop after collision then the linear momentum after the collision ( P_f = 0 ). Therefore, we have:

                             P_i = P_f = 0

- The linear momentum of a particle ( cart ) is the product of its mass and velocity as follows:

                             m*u - 2*m*v = 0

Where,

                 ( u ) and ( v ) are opposing velocity vectors in 1-dimension.

- Evaluate the velcoity ( u ) of the less massive cart in terms of the speed ( v ) of more massive cart as follows:

                          m*u = 2*m*v

                              u = 2*v

Answer: The velocity of less massive cart must be twice the speed of more massive cart for the system conditions to hold true i.e ( they both come to a stop after collision ).

8 0
3 years ago
What is the frequency for a beam of electrons orbiting in a field of 4.62 x 10^-3 T? Let the mass of an electrons m = 9.31 x 10^
melisa1 [442]

Frequency: 1.27\cdot 10^8 Hz

Explanation:

The force experienced by an electron in a magnetic field is

F=qvB

where

q=1.6\cdot 10^{-19}C is the electron charge

v is the speed of the electron

B is the strength of the magnetic field

Since the force is perpendicular to the direction of motion of the electron, the force acts as centripetal force, so we can write:

qvB=m\frac{v^2}{r}

where

r is the radius of the orbit

m=9.31\cdot 10^{-31}kg is the mass of the electron

Re-arranging the equation,

\frac{v}{r}=\frac{qB}{m} (1)

We also know that in a circular motion, the speed is equal to the ratio between circumference of the orbit and orbital period (T):

v=\frac{2\pi r}{T}

Substituting into (1),

\frac{2\pi}{T}=\frac{qB}{m}

We also know that 1/T is equal to the frequency f, so

f=\frac{qB}{2\pi m}

In this problem,

B=4.62\cdot 10^{-3}T

Therefore, the frequency of the electrons is

f=\frac{(1.6\cdot 10^{-19})(4.62\cdot 10^{-3})}{2\pi(9.31\cdot 10^{-31})}=1.27\cdot 10^8 Hz

4 0
3 years ago
Two point masses are held in place a distance d apart. Another point mass M is midway between them. M is then displaced a small
sasho [114]
THAT LINK IS A VIRUS NEVER GO TO A LINK and if you go to “goggle” you should see a camera icon and take a picture of the question and get the answer there
5 0
3 years ago
What Current is drawn by a vacuum cleaner from 120 V circuit having a resistance of 28
Ostrovityanka [42]

Answer:

4.3A

Explanation:

V=IR

120=I×28

I=120/28

=4.3

hope this helps

please like and Mark as brainliest

5 0
3 years ago
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