A 5kg mass is pushed with a force of 10N for a distance of 2.5 meters. The work done is

A 2400W electric toaster is connected to a standard 120 V wall outlet. A family

jolli1 [7]

Answer:

1) energy = 15.6 kWh, 2) total_cost = $ 2.03

Explanation:

1) The energy dissipated is the product of the power and the time of use In a month it was used t = 6.5 h and the power of the toaster is

P = 2400 W = 2,400 kW

energy = P t

energy = 2,400 6.5

energy = 15.6 kWh

using rounding to a decimal

energy = 15.6 kWh

2) The cost of energy is unit_cost = $ 0.13 / kWh

so the total cost

total_cost = energy unit_cost

total_cost = 15.6 0.13

total_cost = $ 2.028

rounding to two decimal places

total_cost = $ 2.03

5 0

3 years ago

Ask Your Teacher A basketball player shoots toward a basket 5.8 m away and 3.0 m above the floor. If the ball is released 1.7 m

const2013 [10]

Answer:

The answer to your question is vo = 5.43 m/s

Explanation:

Data

distance = d= 5.8 m

height = 3 m

height 2 = 1.7 m

angle = 60°

vo = ?

g = 9.81 m/s²

Formula

hmax = vo²sinФ/ 2g

Solve for vo²

vo² = 2ghmax / sinФ

Substitution

vo² = 2(9.81)(3 - 1.7) / 0.866

Simplification

vo² = 19.62(1.3) / 0.866

vo² = 25.51 / 0.866

vo² = 29.45

Result

vo = 5.43 m/s

5 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

2 years ago

Read 2 more answers

You need to build a prototype with machined parts that withstand a saline corrosive environment and temperatures above 200 degre

kondor19780726 [428]

Answer:

Stainless steel

Explanation:

I will try to order the solutions from the least correct to the most correct.

Since a temperature greater than 200 ° F is required, that is to say approximately 93 ° c, Polycaprolactone is the least indicated. Its melting point is approximately 60 ° C, so it would not serve the required application.

On the other hand we have Untreated aluminum, which although it has a melting point higher than the required one, without a zinc and magnesium treatment it will easily oxidize in a salty environment, so it cannot be used in this choice either.

We have to compare the two steels.

The Mild Steel has a better corrosion resistance than the previous ones, but in a long-term cycle it will end up full of corrosion and therefore its properties will be highly affected.

Finally, we have stainless steel, which, as the name implies, contains in some of its variations chromium, zinc or magnesium in its alloys, which makes it highly resistant to corrosion.

In addition its melting point is above 1500 ° c.

The best choice is stainless steel.

5 0

3 years ago

Not a question, but I need help on the last two questions I posted, preferably no links, please...

Andrei [34K]

Answer:

ok...

Explanation:

7 0

3 years ago

Read 2 more answers

A 5kg mass is pushed with a force of 10N for a distance of 2.5 meters. The work done is​

A 5kg mass is pushed with a force of 10N for a distance of 2.5 meters. The work done is