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3 years ago

How do you work out the potential diffrance

Physics

1 answer:

timurjin [86]3 years ago

6 0

Answer:

The potential difference is the drop in voltage that occurs across a resistor as current flows through it in a circuit, potential difference or voltage(V) = current (I) *resistance (R), or to abbrevate V = I*R. In this case, I = 5amps and R = 10 ohms, so V = 5 * 10 = 50volts

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A particle travels along the x-axis in such a way that its acceleration at time t is a(t) = t + t2. if it starts at the origin w

Olegator [25]

The acceleration of the particle as a function of time t is
$a(t) = t + t^2$
The velocity of the particle at time t is the integral of the acceleration:
$v(t) = \int {a(t)} \, dt = \frac{t^2}{2} + \frac{t^3}{3} + C$
where the constant C can be found by requiring that the velocity at time t=0 is v=3:
$v(0) = 3$
and we find $C=v_0=3$
so the velocity is
$v(t)=3+ \frac{t^2}{2}+ \frac{t^3}{3}$

The position of the particle at time t is the integral of the velocity:
$x(t)=\int {v(t) } \,dt = 3t + \frac{t^3}{6}+ \frac{t^4}{12} +D$
where D can be found by requiring that the initial position at time t=0 is zero:
x(0)=0
from which we find D=0, so
$x(t)=3t + \frac{t^3}{6}+ \frac{t^4}{12}$

To solve the problem, now we just have to substitute t=5 into x(t) and v(t) to find the position and the velocity of the particle at t=5.

The position is:
$x(5)=3(5) + \frac{5^3}{6}+ \frac{5^4}{12}=87.92$
and the velocity is:
$v(5) = 3+ \frac{5^2}{2}+ \frac{5^3}{3}=57.17$

6 0

3 years ago

If you get scared when swimming,panicking just makes you breathe faster and uses up more precious energy.

Nonamiya [84]

The answer: True :))

8 0

3 years ago

The picture on the left represents the reactants in the equation and the picture on the right represents the products. Does this

V125BC [204]

The answer is , n<span>o, because there are not the same number of molecules in the product side as in the reactant side.

The law of conservation of matter says that the system remains the same throughout. The number of atoms, mass and energy must stay constant. If you look at your figure there are more white atoms on the left side than the right, this means that there was a loss and this should not be the case.
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7 0

3 years ago

Question 11 of 11 | Page 11 of 11

KiRa [710]

Answer:

Decreases the time period of revolution

Explanation:

The time period of Cygnus X-1 orbiting a massive star is 5.6 days.

The orbital velocity of a planet is given by the formula,

v = √[GM/(R + h)]

In the case of rotational motion, v = (R +h)ω

ω = √[GM/(R + h)] /(R +h)

Where 'ω' is the angular velocity of the planet

The time period of rotational motion is,

T = 2π/ω

By substitution,

Hence, from the above equation, if the mass of the star is greater, the gravitational force between them is greater. This would reduce the time period of revolution of the planet.

3 0

3 years ago

An iron block of mass 10 kg rests on a wooden plane inclined at 30° to the horizontal. It is found

Kaylis [27]

I assume the 100 N force is a pulling force directed up the incline.

The net forces on the block acting parallel and perpendicular to the incline are

∑ F[para] = 100 N - F[friction] = 0

∑ F[perp] = F[normal] - mg cos(30°) = 0

The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.

Then

F[friction] = 100 N

F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N

If µ is the coefficient of static friction, then

F[friction] = µ F[normal]

⇒ µ = (100 N) / (84.9 N) ≈ 1.2

5 0

2 years ago