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sdas [7]
4 years ago
14

Which of these actions will increase friction? Check all that apply.

Physics
1 answer:
Lapatulllka [165]4 years ago
3 0

Answer:

scratching a surface to make it rougher

increasing the size of a flying object

adding extra weight to an object

Explanation:

Friction exists between two surfaces when they are or tend to be in relative motion. This happens due to the interlocking of the microscopic hills and valleys structure of the two surfaces. Friction depends on the roughness of the surface. If a surface is scratched, it can be made more rough and thus, friction would increase. Polishing or oiling the surfaces can reduce friction between them. Friction can also be increased by increasing the weight of the object. It is difficult to move a heavier body. also, with increase in the contact surface, friction increases. Therefore, if the size of a flying object is increased, air drag increases.

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Animals are classifield as two main groups in the animal kingdom : Vertebrates and invertebrates .

Explanation:

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3 years ago
How can you increase the potential energy of a bouncing ball
ad-work [718]

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When you lift the ball, you are doing work to increase its gravitational potential energy. When you then release the ball, gravitational energy is transformed into kinetic energy as the ball falls. When the ball hits the floor, the ball's shape changes as it flattens against the floor.

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3 years ago
Is it better to wire a house using a series circuit or a parallel circuit?
Svetach [21]

Answer: its better to use parallel because, in parallel connection there will be more advantages than a series connection. and also the electronic devices are wired in series so thats why you should use parralel in house wiring

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Explanation:

5 0
3 years ago
a car travels east with a velocity of 80 km/h for 90 minutes how far did the car travel (show your work)
olga55 [171]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Given terms :

  • velocity (v) = 80 km/h

  • time (t) = 90 minutes = 1.5 hour

As we know, Distance covered is equal to :

  • velocity \times time

  • 80 \times 1.5

  • 120 \:  \: km
3 0
3 years ago
A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge
zheka24 [161]

Answer:

The acceleration of an electron is 1.2\times10^{20}\ m/s^2

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

F_{1}=\dfrac{kq_{1}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}

F_{1}=8\times10^{-11}j\ \ N

We need to calculate the force on electron due to q₂

Using formula of force

F_{2}=\dfrac{kq_{2}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}

F_{2}=6.93\times10^{-11}i\ \ N

We need to calculate the net force

Using formula of net force

F=F_{1}+F_{2}

Put the value into the formula

F=8\times10^{-11}j+6.93\times10^{-11}i

The magnitude of the net force

F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}

F=1.058\times10^{-10}\ N

We need to calculate the acceleration of an electron

Using formula of force

F = ma

a=\dfrac{F}{m}

Put the value into the formula

a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}

a=1.2\times10^{20}\ m/s^2

Hence, The acceleration of an electron is 1.2\times10^{20}\ m/s^2

3 0
4 years ago
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