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3 years ago

Consider a block on frictionless ice. Starting from rest, the block travels a distance din

Physics

1 answer:

sweet [91]3 years ago

3 0

Answer:

The distance is now 4d

Explanation:

Mechanical Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

$\displaystyle a=\frac{F}{m}$

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

$\displaystyle d = vo.t+\frac{at^2}{2}$

If the block starts from rest, vo=0:

$\displaystyle d = \frac{at^2}{2}$

Substituting the value of the acceleration:

$\displaystyle d = \frac{\frac{F}{m}t^2}{2}$

Simplifying:

$\displaystyle d = \frac{Ft^2}{2m}$

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

$\displaystyle a'=\frac{4F}{m}$

And the distance is now:

$\displaystyle d' = \frac{4Ft^2}{2m}$

Dividing d'/d:

$\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}$

Simplifying:

$\displaystyle \frac{d' }{d}=4$

Thus:

d' = 4d

The distance is now 4d

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The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo

Lina20 [59]

I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J.
At point A, all this energy has converted into kinetic energy, which is:
$K= \frac{1}{2}mv^2$
And since K=7.35 J, we can find the velocity, v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s$

3 0

3 years ago

Hello,help me with this out please i need it hurry but please ensure your answer is correct..I attach here with my question.

lilavasa [31]

Answer:

.409 N

Explanation:

For this to balance, the moments around the fulcrum must sum to zero.

On the left you have .21 ( is that down? I will assume it is)

Counterclockwise moments :

.21 * 40 + 1.0 * 20

Clockwise moments :

.5 * 20 + F * 45

these moments must equal each other

.21*40 + 1 *20 = .5 * 20 + F * 45

F = .409 N

7 0

2 years ago

Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis

solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

$v^{2}=u^2+2as$

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

$v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s$

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

$v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds$

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equals $T=2\times 30.51\approx61seconds$

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals $39.66m/s$

4 0

4 years ago

Physics B 2020 Unit 3 Test

weqwewe [10]

Answer:

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

$F=qvB sin \theta$

where here:

For the proton in this problem:

$q=1.602\cdot 10^{-19}C$ is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

$\theta=65^{\circ}$ is the angle between the directions of v and B

So the force is

$F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N$

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

The right hand rule gives the direction of the force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

The radius of a particle moving in a magnetic field is given by:

$r=\frac{mv}{qB}$

where here we have:

$m=6.64\cdot 10^{-22} kg$ is the mass of the alpha particle

$v=2155 m/s$ is the speed of the alpha particle

$q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C$ is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

$r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m$

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

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

where here:

$q=1.602\cdot 10^{-19}C$ is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

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

Substituting, we find:

$f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz$

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.

The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:

$\epsilon=-\frac{N\Delta \Phi}{\Delta t}$

where

N is the number of turns in the loop

$\Delta \Phi$ is the change in magnetic flux through the loop

$\Delta t$ is the time elapsed

From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if $\Delta \Phi \neq 0$ , which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.