PHYSICS HIGH SCHOOL

Friction always acts in a direction ______ to the direction of motion.a. equal
c. perpendicular
b. opposite

Answers

Answer 1
Answer:

Answer:

b.Opposite

Explanation:

Friction: It is that force which oppose the motion of any object.It is rubbing force.

We have to fill given blank space

Friction always act in opposite direction of motion.

Friction force is applied to overcome the motion of object.

If equal friction force applied in opposite direction  to   applied force on the object due to which an object in motion  then the object will come to in rest.

When applied force is not equal to friction force then motion will not come in rest and continue move.

Hence, friction force always acts in a direction opposite to the direction of motion.

Answer : b.Opposite

Answer 2
Answer: Friction always acts in a direction opposite to the direction of motion

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Which is not a vector quantity? a. displacement
b. velocity
c. time
d. acceleration

Answers

Time is not a vector quantity

What kind of image is formed by an object that is placed beyond the center of curvature on the principal axis of a concave mirror?A.
real, inverted, the same size as the object, and at the same distance as the object
B.
real, inverted, larger than the object, and farther from the mirror than the object
C.
virtual, upright, left-right reversal, the same size as the object, and at the same distance as the object
D.
real, inverted, smaller than the object, and closer to the mirror than the object
E.
virtual, upright, larger than the object, and farther from the mirror than the object

Answers

Answer:D.

Real, inverted, smaller than the object, and closer to the mirror than the object

Explanation: Real images will be inverted. As it is beyond the centre of curvature it is also beyond 2F which means that the image is inside the centre of curvature ( between F and 2F from the mirror ) As the image is closer to the mirror than the object it must be diminished in size

An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uniform magnetic field ???? = 0.030T ???? − 0.15T ???? . (a) Find the force on the electron due to the magnetic field (b) Repeat your calculation for a proton having the same velocity.

Answers

Explanation:

It is given that,

Velocity of the electron, v=(2* 10^6i+3* 10^6j)\ m/s

Magnetic field, B=(0.030i-0.15j)\ T

Charge of electron, q_e=-1.6* 10^(-19)\ C

(a) Let F_e is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

F_e=q_e(v* B)

F_e=1.6* 10^(-19)* [(2* 10^6i+3* 10^6j)* (0.030i-0.15j)]

F_e=-1.6* 10^(-19)* (-390000)(k)

F_e=6.24* 10^(-14)k\ N

(b) The charge of electron, q_p=1.6* 10^(-19)\ C

The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force of static friction f on the block?A) f > mg
B) f = mgsin(?)
C) f > mgcos(?)
D) f = mgcos(?)
E) f > mgsin(?)

Answers

When the block is at rest, the static frictional force is equal to the horizontal component of the block's weight (F = mgsin(θ)).

The static frictionalforce on the body at rests is determined by applying Newton's second law of motion.

F = ma

where;

  • F is applied force on a body
  • m is the mass of the body
  • a is the acceleration of the body

If the block is at rest, then the net horizontal force on the block is zero.

\Sigma F_x = 0\n\nmg sin(\theta ) - F_s = 0\n\nF_s = mg sin(\theta)

Thus, when the block is at rest, the static frictional force is equal to the horizontal component of the block's weight.

Learn more here:brainly.com/question/13758352

Answer:

Option B

Explanation:

For a system of block on inclined ramp shown in the attached image. From the attached image, the Normal force N, weight mg and frictional force f act on the block.  The sum of vertical forces should be zero just as sum of vertical forces should be zero when the system is in equilibrium condition.

Taking sum of forces along the inclined plane we deduce that  

[tex]f=mgsin \theta[tex]

Therefore, option B is the correct option.

A 0.4-kg toy train car moving forward at 3 m/s collides with and sticks to a 0.8–kg toy car that is traveling in the opposite direction at –2 m/s. What is the size and direction of the final velocity of the two cars

Answers

Hey there!

Seems like you're looking for the size and direction to the final velocity of the two cars. To find it, you must solve it like this.

0.4 kg(3 m/s) + 0.8kg(–2 m/s) = 1.2 kg m/s -1.6 kg m/s = –0.4 kg m/s

–0.4 kg m/s = 1.2 kg(v) = (–0.4 kg m/s)/(1.2 kg) = v = –0.33 m/s


So, the cars are traveling at -0.33 m/s in the direction of the second car.


Hope this helps


Tobey

Anyone knows what this is?

Answers

I think it is D.) “Cycles around” but I am not positive.
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