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brilliants [131]
3 years ago
14

Suppose you increase your walking speed from 6 m/s to 13 m/s in a period of 1 s. What is your acceleration? m/s2

Physics
1 answer:
ziro4ka [17]3 years ago
5 0
Average acceleration is
Change in Velocity/change in time
So you could then do Vf-Vi/Tf-Ti
Which would look like 13m/s-6m/s / 1s-0s
Which then is 7m/s/1s which means the acceleration is 7m/s^2
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Two pith balls hang side by side close to each other without touching as shown in the figure below. They are both neutral to beg
Mila [183]

a)

When a plastic rod is rubbed with wool, charging by friction occurs.

This is one of the three mechanisms of charging an object: in this particular case, as the rod is rubbed with the wool, electrons jump from the wool to the rod.

As a result, the plastic rod will remain with an excess of electrons (negative charge), while the wool will remain with a lack of electrons (positive charge).

However, the number of electrons gained by the rod is equal to the number of electrons given off by the wool: therefore, the total net charge of the system rod + cloth has not changed.

b)

When the rod touches sphere A, some electrons are transferred from the rod to the sphere (this method of charging is called charging by conduction).

Then, the sphere A will also remain negatively charged, as well as the plastic rod.

Later, the rod is brought close to sphere A without touching it: here, the electrons in the rod repel the electrons in the sphere (like charges repel each other), so the rod and the sphere will repel each other.

c)

Before part b), the negative charge on the plastic rod was equal to the positive charge on the wool.

After part b), the plastic rod has given off part of its electrons to sphere A. As a result, the negative charge on the plastic rod will be now less than the positive charge on the wool.

Therefore, the net charge of the rod + wool system will now be positive.

d)

The electrostatic force between two charged object is:

- Attractive if the charges on the objects have opposite sign

- Repulsive if the charges on the objects have same sign

Since sphere A is charged negatively, when it is brought close to sphere B, the electrons in sphere B will be attracted on the opposite side relative to sphere A, while the positive charges will migrate close to sphere A. Therefore, the two spheres will attract each other, since opposite charges attract each other.

e)

When the plastic rod (which is negatively charged) touches sphere B, some electrons are transferred to sphere B (charging by induction). As a result, charge B will also be negatively charged now, as it has an excess of electrons.

At the same time, sphere A is also negatively charged. Therefore, as like charges repel each other, this means that the two spheres will now repel each other.

f)

The net charge of the rod and the two spheres is negative.

In fact, the electrons given off by the rod to the two spheres were initially on the rod itself: this means that the total net charge of the rod + 2 spheres is equal to the net charge on the rod before the process.

Since the net charge on the rod before the process was negative, then the net charge of the system consisting of the rod + 2 spheres must be negative as well.

g)

When a glass rod is rubbed with silk, charging by friction occurs, similar to part a). However in this case, the glass rod gives off electrons to the silk: therefore, the glass rod remains positively charged.

Later, the glass rod touches sphere B: as a result, positive charge is transferred from the rod to sphere B, so sphere B will remain positively charged.

In the meantime, sphere A is negatively charged: since opposite charges attract each other, this means that now sphere A and sphere B will attract each other.

h)

When the two spheres are moved a bit further apart, the attractive force between them decreases a bit. In fact, the electrostatic force between two charged objects is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the  charges on the two objects

r is the separation between the objects

As we can see, the force is inversely proportional to the square of the distance: therefore, if the separation between the spheres increases, the force will decrease.

4 0
2 years ago
Find a unit vector normal to the plane containing Bold u equals 2 Bold i minus Bold j minus 3 Bold k and Bold v equals negative
Nastasia [14]

Answer:

\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}

Explanation:

It is given that,

\vec{u}=2i-j-3k

\vec{v}=-3i+j-2k

Taking the cross product of v and v such that,

\vec{w}=u\times v

\vec{w}=(2i-j-3k)\times (-3i+j-2k)

\vec{w}=5i+13j-k

|w|=\sqrt{5^2+13^2(-1)^2}

|w| = 13.92

Let \hat{w} is the unit vector normal to the plane containing u and v. So,

\hat{w}=\dfrac{\vec{w}}{|w|}

\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}

Hence, this is the required solution.

7 0
3 years ago
A small sphere with mass 5.00.×10−7kg and charge +3.00μC is released from rest a distance of 0.400 m above a large horizontal in
julia-pushkina [17]
I will assume you are asking what the initial acceleration of the sphere is since the information provided seems to indicate that.
First we need to know Newton's Law
F=ma.  
We know the mass of the sphere and we want a so we solve to get
a=F/m.
Now we need the force on the charged sphere.  This is given by the electric field, E and the charge, Q.  The relationship is F=Q×E.  (Recall that the electric field units can be expressed in Newtons/Coulomb).
Now the electric field above a large (~infinite) sheet of charge with a known charge density σ, is given by
E = σ/(2ε0)
Plug in your values of σ, to get E, then the sphere charge Q to get F, the the mass into a = F/m to get the acceleration

6 0
3 years ago
Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to anothe
levacccp [35]

Answer:

Obviously Lengthen...   T = 2\pi \sqrt{L/g}   or   g = 4\pi ^{2} L/g

Explanation:

As we can observe from the equation, time period of a simple pendulum depends upon the length directly. When the gravitational acceleration increases the time period of the pendulum decreases and vice versa. So, by increasing the length, the time period can be adjusted...

4 0
2 years ago
A ball of radius R and mass m is magically put inside a thin shell of the same mass and radius 2R. The system is at rest on a ho
dlinn [17]

Answer:

x =\frac{-R}{2}

Explanation:

From the question we are told that mass

Thin layer radius = 2R

Generally the expression for ths solution is given as

Xcm =(m*0 =m(-2R))/2m =-mR/(2m)=-R/2

the center of mass will not move at initial state  

Considering the center of mass of both bodies

xcm=\frac{m*x+m*x)}{2m} =x

x =\frac{-R}{2}

Therefore the enclosing layer moves x =\frac{-R}{2}                          

7 0
3 years ago
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