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-BARSIC- [3]
2 years ago
12

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 n frictional force. he pushes in a dire

ction 25.0? below the horizontal. what is the work done on the cart by the gravitational force?
Physics
1 answer:
alexandr1967 [171]2 years ago
5 0

Work is calculated as the product of Force, Distance, and angular motion. In this case, the work done by gravity is perpendicular to the motion of the cart, so θ = 90°

 and W=Fdcosθ

W=35.0 N x 20.0 m x cos90

W=0 J

This means that work done perpendicular to the direction of the motion is always zero.

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The forces acting on this object are _______ and the net force is ________ Balanced, 0 N Balanced, 10 N Unbalanced, 10 N Unbalan
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1st Blank Answer: Balanced   
2nd Blank Answer: 0n 

I took the test and got it right
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2 years ago
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Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On
8090 [49]

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks.  So if Fork-A is 256 Hz and the beat is      6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz.  But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz.  That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz.  While it was loaded with wax, it was 261 Hz.

4 0
3 years ago
An 100 kg object traveling at 50 m/s collides (perfectly inelastic) with a 50 kg object initially at rest.
qaws [65]

Answer:

Option C. 5,000 kg m/s

Explanation:

<u>Linear Momentum on a System of Particles </u>

Is defined as the sum of the momenta of each particles in a determined moment. The individual momentum is the product of the mass of the particle by its speed

P=mv

The question refers to an 100 kg object traveling at 50 m/s who collides with another object of 50 kg object initially at rest. We compute the moments of each object

m_1=(100\ kg)(50\ m/s)=5,000\ kg\ m/s

m_2=(50\ kg)(0\ m/s) = 0

The sum of the momenta of both objects prior to the collision is

P=5,000\ kg\ m/s+0\ kg\ m/s

\boxed{ P=5,000\ kg\ m/s}

7 0
3 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}

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}

5 0
1 year ago
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Why is an element considered a pure substance?
ElenaW [278]

Answer:

Explanation:

right

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2 years ago
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