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2 years ago

You are riding a bicycle. You apply a forward force of 150 N, and you and the

Physics

2 answers:

yawa3891 [41]2 years ago

7 0

Answer:

1.67 m/s2

Explanation:

other guys is a f moron

ella [17]2 years ago

3 0

Answer:

B. 0.60 m/s2

- Hope this helped!

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Current that moves in one direction from negative to positive. May be created by a battery. Is generally NOT found in U.S. elect

Alla [95]

That's "DC" . . . Direct Current .

8 0

2 years ago

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If 2 objects had the same momentum, what must be true about the mass of the object that traveled the fastest?

julsineya [31]

Yes, the above-given statement is true

Explanation:

The product of the mass x the velocity will be the same for both. Momentum is the action of a body with a particular mass through space and there is the conservation of momentum.
Momentum is described as the mass of the object multiplied by its velocity.
Momentum (p) = Mass (M) * Velocity (v)
Therefore for two objects with many masses to have a similar momentum, then the lighter one has to be moving quicker than the heavier object.

4 0

2 years ago

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:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

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

5 0

1 year ago

Read 2 more answers

A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car

dybincka [34]

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is option B.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

$Normal force = Gravitational force - centripetal force$

At the foot of a round hill

$Normal Force = centripetal force + Gravitational force$

4 0

3 years ago

A ball dropped from a window strikes the ground 2.76 seconds later. How high is the window above the ground

sweet [91]

Answer:

37.33m

Explanation:

Using the equation of motion

S = ut + 1/2gt^2

Time t = 2.76secs

g = 9.8m/s^2

S = 0 + 1/2(9.8)(2.76)^2

S = 4.9*7.6176

S = 37.33

Hence the window is 37.33m above the ground

7 0

2 years ago