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3 years ago

10

Manganese-52 has a half-life of 6 days. How many days would a scientist have to wait for the radioactivity to be 12.5% the start

ing amount?

2 answers:

Dahasolnce [82]3 years ago

5 0

Answer:

18 Days

Explanation:

sineoko [7]3 years ago

4 0

Answer:

Let N = N0 where N0 is the number of atoms originally present.

In 6 days N = N0 / 2

In 12 days N = N0 / 4

In 18 days N = N0 / 8 = .125 N0

So it would take 18 days.

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A horizontal force of 140 N is needed to pull a 60kg box across the horizontal floor at constant speed . Find the coefficient of

____ [38]

Explanation:

formula for force is:

force=mass × acceleration

but in case of friction

force =coefficient of friction × Normal Reaction

F. = u × R

U = F/R

but when placed horizontally

R= M×G

M=mass=60kg

G=Gravity(10m/s or 9.8m/s)

F=140N

U=140/60×10

U=140/600

U=0.2333333333

approximately to 3 significant figures

U=0.233

if i am correct rate it 5 star

4 0

1 year ago

How does the thermoflask prevent heat loss

riadik2000 [5.3K]

Answer:

The way that the flask is built it has 3 protective layers.... the inside layer to keep the heat in, the outside layer to reflective the cold, and a vacuum layer, which is an empty layer that limits conduction and convection

Explanation:

3 0

3 years ago

A steel piano string of mass per unit lenght 4×10-⁴kgm-¹ was struck, if the speed of the sound on the string is 300ms-¹. Calcula

attashe74 [19]

Answer:

36 N

Explanation:

Velocity of a standing wave in a stretched string is:

v = √(T/ρ),

where T is the tension and ρ is the mass per unit length.

300 m/s = √(T / 4×10⁻⁴ kg/m)

T = 36 N

3 0

3 years ago

A refrigerator having a power rating of 350W operates for 12 hours a day. calculate the cost of electrical energy to operate it

rodikova [14]

Answer:

See the answers below.

Explanation:

The cost of energy can be calculated by multiplying each given value, a dimensional analysis must be taken into account in order to calculate the total value of the cost in Rs.

$Cost=0.350[kW]*12[\frac{hr}{1day}]*30[days]*4.5[\frac{Rs}{kW*hr} ]=567[Rs]$

The fuse can be calculated by knowing the amperage.

$P=V*I$

where:

P = power = 350 [W]

V = voltage = 240 [V]

I = amperage [amp]

Now clearing I from the equation above:

$I=P/V\\I=350/240\\I=1.458[amp]$

The fuse should be larger than the current of the circuit, i.e. about 2 [amp]

3 0

3 years ago

How would the period of a simple pendulum be affected if it were located on the moon instead of the earth?

OlgaM077 [116]

Answer:

On moon time period will become 2.45 times of the time period on earth

Explanation:

Time period of simple pendulum is equal to $T=2\pi \sqrt{\frac{l}{g}}$ ....eqn 1 here l is length of the pendulum and g is acceleration due to gravity on earth

As when we go to moon, acceleration due to gravity on moon is $\frac{1}{6}$ times os acceleration due to gravity on earth

So time period of pendulum on moon is equal to

$T_{moon}=2\pi \sqrt{\frac{l}{\frac{g}{6}}}=2\pi \sqrt{\frac{6l}{g}}$ --------eqn 2

Dividing eqn 2 by eqn 1

$\frac{T_{moon}}{T}=\sqrt{\frac{6l}{g}\times \frac{g}{l}}$

$T_{moon}=\sqrt{6}T=2.45T$

So on moon time period will become 2.45 times of the time period on earth

5 0

3 years ago