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3 years ago

What is the chemical in groundwater called that causes cancer?

Physics

1 answer:

melisa1 [442]3 years ago

3 0

The chemical in groundwater that causes cancer is called Methadone :)

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What is the change in momentum of a ball if a force of 5.0 N acts on it for a brief time of 4 ms?

Gnoma [55]

Answer:

0.02 Ns

Explanation:

From the question given above, the following data were obtained:

Force (F) = 5 N

Time (t) = 4 ms

Change in momentum =?

Next, we shall convert 4 ms to s. This can be obtained as follow:

1 ms = 1×10¯³ s

Therefore,

4 ms = 4 ms × 1×10¯³ s / 1 ms

4 ms = 4×10¯³ s

Thus, 4 ms is equivalent to 4×10¯³ s

Change in momentum = Impulse

Impulse (I) = Force (F) × time (t)

Change in momentum = Force (F) × time (t)

Force (F) = 5 N

Time (t) = 4×10¯³ s

Change in momentum =?

Change in momentum = 5 × 4×10¯³

Change in momentum = 0.02 Ns

Therefore, the change in the momentum of the ball is 0.02 Ns.

4 0

3 years ago

A spring is used to launch a coffee mug. The 20cm long spring can be compressed by a maximum of 8cm. The mug has a mass of 500 g

Otrada [13]

Answer:

7664.06249 N/m

Explanation:

m = Mass of mug = 0.5 kg

x = Compression of spring = 0.08 m

h = Height of fall = 5 m

g = Acceleration due to gravity = 9.81 m/s²

The kinetic energy of the spring and potential energy of the fall are conserved

$\frac{1}{2}kx^2=mgh\\\Rightarrow k=\frac{2mgh}{x^2}\\\Rightarrow k=\frac{2\times 0.5\times 9.81\times 5}{0.08^2}\\\Rightarrow k=7664.06249\ N/m$

The spring constant of the spring is 7664.06249 N/m

8 0

3 years ago

The red lines on this

tino4ka555 [31]

The circular lines you see on the chart are isobars, which join areas of the same barometric pressure.

8 0

3 years ago

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) Ampere's Law 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

A maglev train can “float” because magnets in the bottom of the train and on the guideway...

miss Akunina [59]

C. Have like poles that repel each other

4 0

3 years ago