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3 years ago

A person jogs 10 meters in 5 seconds calculate her avarage speed

Physics

2 answers:

Bas_tet [7]3 years ago

8 0

Answer:

So first determine how many times does 5 go into 100.

100/5=20

Thus in 100 seconds the boy would have traveled a distance of 10 meters 20 times.

10*20=200

If the boy travels 10meters in 5 seconds, he travels 200m in 100seconds.

Colt1911 [192]3 years ago

5 0

Here, we know that the the total distance is 10m and the total time is 5s . so the avg. speed is 10m5s=2m/s .

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A constant 10.0-N horizontal force is applied to a 20.0-kg cart at rest on a level floor. If friction is negligible, what is the

Otrada [13]

Answer:

2.83 m/s

Explanation:

Given:

Mass of the cart (m) = 20.0 kg

Initial velocity of the cart (u) = 0 m/s

Final velocity (v) = ? m/s

Displacement of the cart (S) = 8.0 m

Horizontal force acting on the cart (F) = 10.0 N

Surface is frictionless. So, only horizontal force is the force acting on the cart.

Now, as per work-energy theorem, the work done by the net force acting on a body is equal to the change in the kinetic energy of the body.

Here, the work done by the horizontal force is given as:

$W_{net}=FS=(10\ N)(8.0\ m)=80\ Nm$

The change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}(20.0\ kg)(v^2-0)\\\\\Delta K=10v^2$

Now, from work-energy theorem:

$\Delta K=W_{net}\\\\10v^2=80\\\\v^2=\frac{80}{10}\\\\v=\sqrt{8}=2.83\ m/s$

Therefore, the speed of the cart when it has been pushed 8.0 m is 2.83 m/s.

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2 years ago

What change will most likely increase the strength of a magnetic field produced by an electromagnet? a. Reduce the number of tur

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Answer:

Option D

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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

kkurt [141]

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, $\displaystyle \theta = 90^{\circ}$ .