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photoshop1234 [79]

3 years ago

8

DISPLACEMENT WITH CONSTANT ACCELEration. Given information: displacement=64m, acceleration=9.81m\s and time interval=3s. calcula

te the initial velocity

1 answer:

balandron [24]3 years ago

3 0

HEYEYEYEYYEYEYDYDYDYD

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A car has wheels which spin forwards. As the wheels spin forwards, they grip the road and push the road backwards. Since forces

alexandr402 [8]

Answer:

the law of motion

Explanation:

because the wheels are moving it means motion i am not sure which number law it is but I believe that it is 2nd but u should look it up to be safe

6 0

2 years ago

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The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?

Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight = 60 .3 (9.81) = 591.54 N

Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift

60.3 ( 9.81 - L ) = 564

solve for L = .457 m/s^2 DOWNWARD

so L = - .457 m/s^2

4 0

1 year ago

if you have a kinetic energy of 1470 J, and you are 60kg mass and 0 m above the ground, what is you velocity?

laiz [17]

Answer:

The 39.

Explanation:

8 0

2 years ago

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A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

7 0

2 years ago

I know how the voltage increasing/not enough/stays the same and how long the battery lasts with series and parallel connections.

kirill [66]

Answer:

It comes out the positive side of the battery and goes in to the negative side of the battery

Explanation:

There are already electrons in wires in a circuit before you add the battery. By adding the battery, you're giving the electrons the energy it needs to move along the circuit.

In a series circuit, the circuit is one continuous loop so there is only one path for the electrons to go - out of the positive side of the battery and around the circuit then goes back into the negative side of the battery.

However, with a parallel circuit, there are two or more ways the electrons can go so they take the path of least resistance. The electrons still go out the positive side of a battery but along the circuit, the electrons will go through the path of least resistance ( I tend to think of it like a net with holes in it - the lower the resistance the bigger the holes for the electrons to go through so more can fit in a set amount of time ) but the electrons still go out of the positive side and in through the negative

3 0

2 years ago