Is crest in pitch or volume

When attempting to determine the coefficient of kinetic friction, why is it necessary to move the block with constant velocity

Readme [11.4K]

Answer:

Explanation:

In order to measure the coefficient of friction , we apply external force to move the body . When external force comes in motion , we adjust the external force so that it moves with zero acceleration or uniform velocity . In this case external force becomes equal to kinetic frictional force and then net force becomes zero because

net force = mass x acceleration = m x 0 = 0

Now frictional force = μ mg where μ is coefficient of kinetic friction

so F = μ mg where F is external force applied

μ = F / mg

Hence , to make external force equal to frictional force , it is necessary to make acceleration of body zero .

4 0

2 years ago

What is the magnitude of velocity for a 2000 kg car possessing 3000 kgm/s of momentum

Alex

For this case we have that by definition, the momentum is given by:

$p = mv$

Where,

<em>m: mass </em>
<em>v: speed </em>

Therefore, replacing values we have:

$3000 = (2000) v$

From here, we clear the value of the speed:

$v = \frac {3000} {2000}\\v = 1.5 \frac {m} {s}$

Answer:

The magnitude of velocity is:

$v = 1.5 \frac {m} {s}$

4 0

3 years ago

Read 2 more answers

A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$ .

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ

According to the question, we have -

Entering Velocity (v) = 20 i m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $$\frac{2\pi r}{4}$ = $$\frac{\pi R}{2}$

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $$\frac{\pi R}{2v} = \frac{\pi R}{40}$

Hence, the time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

#SPJ4

6 0

1 year ago

Find Vxl and Vyl of a pumpkin launched at a velocity of 55 m/s at an angle of 20 degrees

Vinvika [58]

Answer:

Explanation:

is A projectile is any object on which the only force acting is gravity and air resistance (drag).

Examples of projectiles are:

baseballs and softballs in the air after being hit by the bat

golf balls hit by a club

objects dropped from aircraft, such as people (skydivers), bombs, crates of food being dropped to refugees

objects launched by cannons, such as cannonballs, shells, and circus performers

Once the baseball, softball, golf ball, skydiver, bomb, crate, cannonball, shell, or clown are no longer touching the bat, club, aircraft, or cannon, and are in the air with only gravity and slight air resistance acting on it, then it is a projectile.

Here is an online projectile motion applets to play with, just for fun.

Unless otherwise stated in a particular problem or discussion, we will be ignoring the effects of air resistance.

The key to understanding the motion of projectiles is that the horizontal motion and the vertical motion of the projectile are independent of each other. So we can write separate equations for the displacement of the projectile in the horizontal (x) and vertical (y) directions.

The only common variable between these two equations is t, the time. Because in projectile problems there is usually no acceleration (i.e. we ignore air resistance) in the horizontal direction, we can write

The velocity components follow the same equations we used for one-dimensional motion.

Because there is usually no acceleration in the x direction, the x-velocity is constant.

3 0

2 years ago

The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?

Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight = 60 .3 (9.81) = 591.54 N

Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift

60.3 ( 9.81 - L ) = 564

solve for L = .457 m/s^2 DOWNWARD

so L = - .457 m/s^2

4 0

2 years ago