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4 years ago

If the person starts over and moves his hand more quickly the waves will have a

Physics

1 answer:

Mama L [17]4 years ago

6 0

Answer:

larger push force

Explanation:

mark 5 stars and brainliest pls thx I took the test/quiz

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A long, straight wire with a circular cross section of radius R carries a current I. Assume that the current density is not cons

igor_vitrenko [27]

Answer: (a) α = $\frac{3I}{2.\pi.R^{3}}$

(b) For r≤R: B(r) = μ_0. $(\frac{I.r^{2}}{2.\pi.R^{3}})$

For r≥R: B(r) = μ_0. $(\frac{I}{2.\pi.r})$

Explanation:

(a) The current I enclosed in a straight wire with current density not constant is calculated by:

$I_{c} = \int {J} \, dA$

where:

dA is the cross section.

In this case, a circular cross section of radius R, so it translates as:

$I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr$

$I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr$

$I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}$

$\alpha = \frac{3I}{2.\pi.R^{3}}$

For these circunstances, α = $\frac{3I}{2.\pi.R^{3}}$

(b) <u>Ampere's</u> <u>Law</u> to calculate magnetic field B is given by:

$\int\ {B} \, dl =$ μ_0. $I_{c}$

(i) First, first find $I_{c}$ for r ≤ R:

$I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr$

$I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr$

$I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr$

$I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}$

$I_{c} = \frac{I.r^{3}}{R^{3}}$

Calculating B(r), using Ampere's Law:

$\int\ {B} \, dl =$ μ_0. $I_{c}$

$B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} )$ .μ_0

B(r) = $(\frac{Ir^{3}}{R^{3}2.\pi.r})$ .μ_0

B(r) = $(\frac{Ir^{2}}{2.\pi.R^{3}} )$ .μ_0

For r ≤ R, magnetic field is B(r) = $(\frac{Ir^{2}}{2.\pi.R^{3}} )$ .μ_0

(ii) For r ≥ R:

$I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr$

So, as calculated before:

$I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}$

$I_{c} =$ I

Using Ampere:

B.2.π.r = μ_0.I

B(r) = $(\frac{I}{2.\pi.r} )$ .μ_0

For r ≥ R, magnetic field is; B(r) = $(\frac{I}{2.\pi.r} )$ .μ_0.

3 0

3 years ago

An elevator car has a mass of 750 kg, and its three passengers have a combined mass of 135 kg. If the elevator and its passenger

joja [24]

Answer:

The change in gravitational potential energy is -1.80x10⁵ J.

Explanation:

The change in gravitational potential energy is given by:

$\Delta E_{p} = E_{p_{f}} - E_{p_{i}}$

$\Delta E_{p} = mgh_{f} - mgh_{i}$

Where:

"i" is for final and "f" for final

m: is the mass

g: is the gravity = 9.81 m/s²

h: is the height

For the car and the passengers we have:

$\Delta E_{p} = m_{T}g(h_{f} - h_{i}) = (750 kg + 135 kg)9.81 m/s^{2}(0 - 20.7 m) = -1.80 \cdot 10^{5} J$

The minus sign is because when the elevator car and the passengers are up they have a bigger gravitational potential energy than when they are in the ground.

Therefore, the change in gravitational potential energy is -1.80x10⁵ J.

I hope it helps you!

3 0

3 years ago

The weight of a 72.0 kg astronaut on the Moon, where g = 1.63 m/s2 is (5 points) Select one: a. 112 N b. 117 N c. 135 N d. 156 N

Hunter-Best [27]

Answer: The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

Explanation:

Mass of the astronaut on the moon , m= 72 kg

Acceleration due to gravity on moon,g = 1.63 $m/s^2$

According to Newton second law of motion: F = ma

This will changes to = Weight = mass × g

$Weight=72 kg\times 1.63m/s^2=117.36 N$

The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

5 0

4 years ago

Read 2 more answers

THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!! THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THAN

Karo-lina-s [1.5K]

Answer:

450km because this night founder

6 0

3 years ago

Read 2 more answers

3(NH4)2CO3 what does the 3 in front mean?

VladimirAG [237]

Answer:

3 in front means the no.of NH4 atoms

Hope it is correct ^_^

7 0

3 years ago