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2 years ago

5

Physics

2 answers:

KiRa [710]2 years ago

7 0

D, the wheels
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NNADVOKAT [17]2 years ago

4 0

Answer:

Explanation:

moving car experience the most is the Tyre

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A force can give a certain mass a certain acceleration. if i double the force and double the mass, the acceleration of this obje

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The acceleration of this item will remain constant if I double the force and double the mass.

<h3>What does acceleration mean ?</h3><h3 />

The rate at which a moving object's speed and direction change over time. When something moves faster or slower, it is considered to be accelerating. Motion on a circle accelerates even while the speed is constant because the direction is always changing.

If an object is accelerating and moving in the appropriate direction, it has positive acceleration. The speeding automobile in the first scenario showed positive acceleration. The car is moving forward and accelerating in the same direction as the acceleration.

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4 months ago

How can you find the nuetrons in youre element ??

miskamm [114]

Answer: you subtract the number of protons from the mass number, on the periodic table your atomic number is your protons and your atomic mass is the mass number

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2 years ago

The picture below shows a wheelbarrow. Using a wheelbarrow can make it easier to lift a heavy object by

timofeeve [1]

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2 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

2 years ago

how do you find work when only given the angle a sled is pulled, the mass, the coefficent of kinetic friction and distance

Sergio039 [100]

Answer:

W = F * s

Work done equals applied force * distance traveled

Apparent weight = M g (1 - sin θ) since some of applied force will lighten sled

μ = coefficient of kinetic friction

F cos θ = force applied to motion of sled

s = distance traveled

[μ M g (1 - sin θ)] cos θ * s = work done in moving sled

Note that F = μ M g if applied force is in the horizontal direction

8 0

1 year ago