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ycow [4]
3 years ago
8

If a baseball has a zero velocity at some instant, is the acceleration of the baseball necessarily zero at that time? Explain -

making sure to give an example of an object that has a zero velocity but non-zero acceleration.
Physics
1 answer:
ipn [44]3 years ago
3 0

Answer:

No, not necessarily

Explanation:

If an object is moving with an acceleration that causes its speed to be reduced, there will be a moment in which it reaches v = 0, but this doesn't necessarily mean that the acceleration isn't acting anymore. If the object continues its movement with the same acceleration, it's velocity will become negative.

An example of an object that has zero velocity but non-zero acceleration:

If you throw an object in the air with a certain velocity, it will move vertically, reducing its velocity in a 9,8 m/s^{2} rate (which is the acceleration caused by gravity). At a certain point, the object will reach its maximum height, and will start to fall. In the exact moment that it reaches the maximum height, before it starts falling, its velocity is zero, but gravity is still acting on the object (this is the reason why it starts falling instead of just being stopped at that point). Therefore, at that point, the object has zero velocity but an acceleration of 9,8 m/s^{2}.

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An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2
pishuonlain [190]

Answer:

E_{total}=4.82*10^6N/C

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

E_{total}=E_1+E_2

<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:

E=\sigma/(2\epsilon_o)

Then:

E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

7 0
3 years ago
Read 2 more answers
Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

PE = -\frac{GMm}{R}

As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

v = 1.83*10^7m/s

Therefore the velocity when they are about to collide is 1.83*10^7m/s

7 0
3 years ago
The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x
Diano4ka-milaya [45]

Answer:

Ff=m\times \dfrac{V_o^2}{2X_1}

Explanation:

Given that

At X=0 V=Vo

At X=X1  V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

V^2=U^2-2aS

0=V_o^2-2a X_1

a=\dfrac{V_o^2}{2X_1}

So the friction force on the box

Ff= m x a

Ff=m\times \dfrac{V_o^2}{2X_1}

Where m is the mass of the box.

4 0
3 years ago
You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the
lana [24]

Answer:

<em>The fifth option is the correct answer: mv; 2 mv</em>

Explanation:

<u>Change of Momentum</u>

Assume an object has a momentum p1 and after some interaction it now has a momentum p2, the change of momentum is

\Delta p=p_2-p_1

The momentum is computed as

p=mv

Where m is the mass of the object and v its speed. Now let's analyze the situation of both the ball and the clay.

The clay has an initial speed v and a mass m, thus its initial momentum is

p_1=mv

When it hits the wall, it sticks, thus its final speed is 0 and

p_2=0

The change of momentum is

\Delta p=0-mv=-mv

The absolute change is mv

Now for the ball, the initial condition is the same as it was for the clay, but the ball hits back at the same speed, thus its final momentum is

p_2=-mv

The change of momentum is

\Delta p=-mv-mv=-2mv

The absolute change is 2mv

The fifth option is the correct answer: mv; 2 mv

3 0
3 years ago
A2 kg object moves at a constant 5 m/s across a level surface. Determine its kinetic energy.
AURORKA [14]
KE = 1/2mv^2
KE= 1/2(2)(5)^2
KE= 25 J
6 0
2 years ago
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