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2 years ago

Half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value after

Physics

1 answer:

makkiz [27]2 years ago

3 0

Answer:

44 years

Explanation:

Use half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.25 A₀ = A₀ (½)^(t / 22)

0.25 = (½)^(t / 22)

t / 22 = 2

t = 44

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A 2500-N net force acting on a 880-kg car accelerates it at a rate of ______ m/s/s

Bond [772]

Answer:

a = 2.84 m/s²

Explanation:

Given that,

Net force, F = 2500 N

Mass of the car, m = 880 kg

We need to find the acceleration of the car. Net force is given by :

F = ma

$a=\dfrac{F}{m}\\\\a=\dfrac{2500\ N}{880\ m/s^2}\\\\a=2.84\ m/s^2$

So, the acceleration of the car is 2.84 m/s².

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2 years ago

What part of this transverse wave is the green arrow identifying?<br><br> Help plz

cupoosta [38]

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2 years ago

A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg

yulyashka [42]

Answer:

$T=245.76N$

Explanation:

We know that the frequency of the nth harmonic is given by $f_n=nf$ , where $f$ is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:

$f_{n+1}-f_n=(n+1)f-nf=nf+f-nf=f$

Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

$f=f_{n+1}-f_n=480Hz-400Hz=80Hz$

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

$f=\frac{1}{2L} \sqrt{\frac{T}{\mu}}$

So the tension is:

$T=\mu(2Lf)^2$

Which for our values is:

$T=(0.0006kg/m)(2(4m)(80Hz))^2=245.76N$

6 0

2 years ago

A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw

meriva

Answer:

As 28m/s = 28m/s

Explanation:

r = the radius of the curve

m = the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

$a = \frac{v^{2}}{r}$

therefore

$\mu mg = m \frac{v^{2}}{r} \\\\ \mu = \frac{v^{2}}{rg}$

$v = \sqrt{\mu rg}$

$\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s$

As 28m/s = 28m/s

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3 years ago

Two scientists are discussing their beliefs about something they cannot observe

svetoff [14.1K]

Could it be hypothesis?

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2 years ago

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