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Paladinen [302]

3 years ago

8

If the v – t graph of a particle is parallel to the t – axis the body has (a) uniform velocity (b) uniform acceleration (c) zero

velocity (d) infinite velocity

1 answer:

maxonik [38]3 years ago

5 0

Answer:

zero velocity

Explanation:

if the v-t graph is parallel to the time axis its mean body has covered no path in other words the body is at rest so the velocity of the body should be zero

hope my answer will helps u plz mark me brainlist

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Blocks A and B of unknown masses m1 and m2, respectively, are set up on an inclined plane as shown. Block A is attached to block

Korvikt [17]

Newton's second law we can find that the correct answer is:

E) It cannot be determiner whick block has more masses from the information provided

Newton's second law establishes the relationship between force, mass, and acceleration of a body. Since force and acceleration are vector quantities, their components must be added on each axis

For this problem we have two bodies, let's write Newton's second law for the body B, we assume that the body B descends

W_b - T = m_b a

W_b = m_b g

m_b - T = m_b a

Where W_b is the weight of block B, T the tension of the string, mb the mass of block b and the acceleration

Now let's find the relation for block A

let's set a datum with the x axis parallel to the ramp

T - Wₓ = mₐ a

sin θ = Wₓ / W

Wₓ = Wₐ sin θ

Wₐ = mₐ g

Where Wₓ is the component of the weight, Wₐ the weight of the body A and θ the angle of the plane

Let's write our system of equations

m_b g - T = m_b a

T - mₐ g sin θ = mₐ a

let's add the equations

g (m_b - mₐ sin θ) = (m_b + mₐ) a

a = $\frac{m_b - m_a \ sin \ \theta}{m_b+m_a} \ g$

Let's analyze this expression

The numerator is positive the body B descends, this occurs when

m_b - mₐ sin θ > 0

The numerator is negative, body B rises

m_b - mₐ sin θ <0

We can observe that the acceleration is positive or negative depending on the relation of the masses and the angle of the plane.

In conclusion using Newton's second law we find that the correct answer is

E ) It cannot be determiner whick block has more masses from the information provided

learn more about Newton's second law here:

brainly.com/question/9099891

8 0

3 years ago

Show that the Mass spring system executes simple harmonic motion(SHM)?

murzikaleks [220]

Explanation:

Show that the motion of a mass attached to the end of a spring is SHM

Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed

at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.

If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.

According to "Hook's Law

F = - Kx ---- (1)

Negative sign indicates that the elastic restoring force is opposite to the displacement.

Where K= Spring Constant

If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion

from a to b and then b to a.

According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by

F = ma ---- (2)

Comparing equation (1) & (2)

ma = -kx

Here k/m is constant term, therefore ,

a = - (Constant)x

or

a a -x

This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.

5 0

3 years ago

What is an example of a stable system

oksian1 [2.3K]

Some examples of stable system are:

1) functions of sine

2) DC

3) signum

4) unit step

5) cosine.

Happy Studying! ^^

6 0

3 years ago

Read 2 more answers

Two objects are moving at equal speed along a level, frictionless surface. the second object has twice the mass of the first obj

lbvjy [14]

Answer:

They both rises to same height.

Explanation:

When an object is sliding up in friction less surface than according to conservation of energy its potential energy will be converted into kinetic energy.

$mgH=\frac{1}{2}mv^{2}\\ v=\sqrt{2gH}$

Here, m is the mass, v is the velocity, g is the acceleration due to gravity and H is the height.

Here the height is independent on the mass of an object and its only depend on velocity.

Now according to the question, two objects have same velocity but they have different masses.

Therefore, they rises to the same height because height will not change with mass.

8 0

3 years ago

Read 2 more answers

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

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

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

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

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

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

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago