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hram777 [196]
2 years ago
15

Calculate the velocity of an apple that falls freely from rest and drops for 3.5 seconds.

Physics
1 answer:
padilas [110]2 years ago
6 0

Answer:

initial velocity(u)=0

time(t)=3.5sec

acceleration (a)=9.8m/s²

final velocity (v)=?

Explanation:

we have

v=u+at=0+9.8×3.5=34.3m/s

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A force of 8.0 N is along x direction, another force of 6.0 N is along +y direction. If both forces are acting on a point object
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Answer:

Resultant force, R = 10 N

Explanation:

It is given that,

Force acting along +x direction, F_x=8\ N

Force acting along +y direction, F_y=6\ N

Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,

R=\sqrt{F_x^2+F_y^2+F_xF_y\ cos\theta}

Here \theta=90^{\circ}

R=\sqrt{F_x^2+F_y^2}

R=\sqrt{8^2+6^2}

R = 10 N

So, the resultant force on the object is 10 N. Hence, this is the required solution.

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Answer: (b) direct object

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A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .
Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}

Taking out constants from the integral:
B =\frac{\mu_0 i}{4\pi R^2}  \int ds

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

B =\frac{\mu_0 i}{4\pi R^2}  \int\limits^{2\pi R}_0 \, ds

Evaluate:
B =\frac{\mu_0 i}{4\pi R^2}  (2\pi R- 0) = \frac{\mu_0 i}{2R}

Plugging in our givens to solve for the magnetic field strength of one loop:

B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T

Multiply by the number of loops to find the total magnetic field:
B_T = N B = 0.00631 = \boxed{6.318 mT}

b)

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.  

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}

Using the diagram, if 'z' is the point's height from the center:

r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}

Substituting this into our expression:
dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds

Evaluate:
B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}

Multiplying by the number of loops:
B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}

Plug in the given values:
B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ =  0.00006795 = \boxed{67.952 \mu T}

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