A car weighs 14500 N. What is its mass?

A small, positively charged ball is moved close to a large, positively charged ball. which describes how the small ball likely r

NNADVOKAT [17]

Answer;

-it will move away from the large ball because like charges repel.

Explanation;

-Electric force is the force that pushes apart two like charges, or that pulls together two unlike charges. The basic law of electrostatics Like charges of electricity repel each other, whereas unlike charges attract each other.

When small, positively charged ball is moved close to a large, positively charged ball it would be pushed away from the large positively charged ball since they are both positively charged. One has to put in energy to try to move the small ball closer to the large ball. The closer one try to move it to the large ball, the more energy one has to put in, so the more electrical potential energy the small ball would have.

6 0

3 years ago

Read 2 more answers

If a plane is flying level at 910 km/h and the banking angle is not to exceed 50 ∘, what's the minimum curvature radius for the

hoa [83]

Answer:

5.5 km

Explanation:

First, we convert the distance from km/h to m/s

910 * 1000/3600

= 252.78 m/s

Now, we use the formula v²/r = gtanθ to get our needed radius

making r the subject of the formula, we have

r = v²/gtanθ, where

r = radius of curvature needed

g = acceleration due to gravity

θ = angle of banking

r = 252.78² / (9.8 * tan 50)

r = 63897.73 / (9.8 * 1.19)

r = 63897.73 / 11.662

r = 5479 m or 5.5 km

Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km

4 0

3 years ago

A motorcycle accelerated from 10 metre per second to 30 metre per second in 6 seconds. How far did it travel in this time ?

Ber [7]

U=10 m/s
v=30 m/s
t=6 sec

therefore, a=(v-u)/t
=(30-10)/6
=(10/3) ms^-2

now, displacement=ut+0.5*a*t^2
=60+ 0.5*(10/3)*36
=120 m
And you can solve it in another way:

v^2=u^2+2as
or, s=(v^2-u^2)/2a
=(900-100)/6.6666666.......
=120 m

7 0

3 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago

How much does coast to coast membership cost?

CaHeK987 [17]

The price of coast to coast membership in united states could lie anywhere between $2,000 to $ 5,000
Unless you're a frequent user of this type of event, i think it would be economically more efficient if you pay the resort on one-day price

8 0

3 years ago