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2 years ago

A rubbit gets down from a rump which its /\x=0.85m in 0.5s, The rubbit's mass is 2kg, what is the net Force?

Physics

1 answer:

mestny [16]2 years ago

8 0

Answer: 13.6 N

Explanation:

The equation of motion for the rabbit is:

$\Delta x=V_{ox}t+\frac{1}{2}a_{x}t^{2}$ (1)

Where:

$\Delta x=0.85 m$ is the distance traveled by the rabbit

$V_{ox}=0 m/s$ is the rabbit's initial velocity, assuming it started from rest

$t=0.5 s$ is the time

$a_{x}$ is the acceleration

Isolating $a_{x}$ :

$a_{x}=\frac{2 \Delta x}{t^{2}}$ (2)

$a_{x}=\frac{2 (0.85 m)}{(0.5)^{2}}$ (3)

$a_{x}=6.8 m/s^{2}$ (4)

On the other hand, the force $F_{x}$ is given by:

$F_{x}=m.a_{x}$ (5)

Where $m=2 kg$ is the mass of the rabbit

Substituting (4) in (5):

$F_{x}=(2 kg)(6.8 m/s^{2})$ (6)

Finally:

$F_{x}=13.6 N$

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the resistor of values 6 ohm,6 ohm are connected in series and 12 ohm are connected in parallel. the equivalent resistance of th

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2 years ago

The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the

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3 years ago

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

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Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .