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4 years ago

I need help ASAP....

Physics

1 answer:

Svetradugi [14.3K]4 years ago

5 0

Answer:

Meters

Explanation:"How FAR did the athlete run?"

Also it talked about meters

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Someone please help me

Lapatulllka [165]

False... I hope that helps ;)

6 0

3 years ago

A plant dying after being exposed to poison represents a physical change.

Sphinxa [80]

Answer:

False is the correct answer.

It is considered more of a chemical change because all of the cells inside the plant decompose (change chemicaly)

5 0

3 years ago

Read 2 more answers

How to calculate ?Help!!!

Alona [7]

How to calculate what?

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4 years ago

The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the

FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

$B=\frac{2\pi*a }{u*I}$

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

$I=\frac{2\pi*a*B }{u}$ Formula(2)

$B=0.2*10^{-5} T = 0.2*10^{-5} \frac{weber}{m^{2} }$

a =8cm=0.08m

$u=4*\pi *10^{-7} \frac{Weber}{A*m}$

We replace the known information in the formula (2)

$I=\frac{2\pi*0.08*0.2*10^{-5} }{4\pi *10x^{-7} }$

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0

3 years ago

Chegg ) Alice owns 20 grams of a radioactive isotope that has a half-life of ln(4) years. (a) Find an equation for the mass m(t)

stepladder [879]

Answer:

$m(t)=20e^{-0.5t}$

Explanation:

Given:

Initial mass of isotope (m₀) = 20 g

Half life of the isotope $(t_{1/2})$ = (ln 4) years

The general form for the radioactive decay of a radioactive isotope is given as:

$m(t)=m_0e^{-kt}$

Where,

$m(t)\to mass\ after\ 't'\ years\\t\to years\ passed\\k\to rate\ of\ decay\ per\ year$

So, the equation is: $m(t)=20e^{-kt}$

At half-life, the mass is reduced to half of the initial value.

So, at $t=t_{1/2},m(t)=\frac{m_0}{2}$ . Plug in these values and solve for 'k'. This gives,

$\frac{m_0}{2}=m_0e^{-k\times\ln 4}\\\\0.5=e^{-k\times\ln 4}\\\\Taking\ natural\ log\ on\ both\ sides,we\ get:\\\\\ln(0.5)=-k\times \ln 4\\\\k=\frac{\ln 0.5}{-\ln 4}=0.5$

Hence, the equation for the mass remaining is given as:

$m(t)=20e^{-0.5t}$

8 0

3 years ago