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3 years ago

a radioactive tracer has a half-life of 2 hours. how much of a 2500g sample will be avalible after 18 hours

Physics

1 answer:

zhannawk [14.2K]3 years ago

8 0

Answer:

i believe the answer is 2.44140625 grams remaining

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A Person whose weight is 5.20 x 10^2 N is being pulledup

Gnoma [55]

Answer:

Explanation:

Given

Weight of Person $W=5.20\times 10^{2} N$

Cave is $h=35.1 m$ deep

Breaking stress $T=569 N$

Net Force on Person

$F_{net}=569-520=49 N$

$a_{net}=\frac{F_{net}}{\frac{W}{g}}$

$a_{net}=\frac{49}{\frac{520}{9.8}}$

$a_{net}=0.923 m/s^2$

The shortest time such that the person can be taken out of cave

$h=ut+\frac{1}{2}at^2$

where

h=distance moved

t=time

a=acceleration

$35.1=0+\frac{1}{2}(0.923)(t)^2$

$t^2=76.05$

$t=\sqrt{76.05}$

$t=8.72\ s$

6 0

3 years ago

1) mechanical

Dafna1 [17]

I am pretty sure about these answers.Thermal goes in the 4th blank.
Mechanical goes in the 2nd blank.
Electrical goes in the 3rd blank.
I think chemical goes in the 1st blank and light goes in the 5th blank
Hope this helps

7 0

3 years ago

A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is

Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with uniform circular motion, her centripetal acceleration $a_{c}$ is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$a_{c}=17 m/s^{2}$ is the centripetal acceleration

$V$ is the tangential speed

$r=12 m$ is the radius of the circle

Isolating $V$ from (1):

$V=\sqrt{a_{c}r}$ (2)

$V=\sqrt{(17 m/s^{2})(12 m)}$

Finally:

$V=14.28 m/s$ This is the girl's tangential speed

3 0

3 years ago

How do you calculate force?

DIA [1.3K]

Force (N) = mass (kg) × acceleration (m/s²)

6 0

3 years ago

Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106V/m ) if

xxMikexx [17]

(a) The electric field strength between two parallel conducting plates does not exceed the breakdown strength for air ( $3 \times 10^{6} V / m$ )

(b) The plates can be close together to 1.7 mm with this applied voltage

Explanation:

Given data:

Dielectric strength of air = $3 \times 10^{6} V / m$

Distance between the plates = 2.00 mm = $2.00 \times 10^{-3} \mathrm{m}$

Potential difference, V = $5.0 \times 10^{3} V$

We need to find

a) whether the electric field strength between two parallel conducting plates exceed the breakdown strength for air or not

b) the minimum distance at which the plates can be close together with this applied voltage.

The voltage difference (V) between two points would be equal to the product of electric field (E) and distance separation (d). The equation form is and apply all given value,

$E=\frac{V}{d}=\frac{5.0 \times 10^{3}}{2.00 \times 10^{-3}}=2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$

From the above, concluding that The electric field strength between two parallel conducting plates ( $2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$ ) does not exceed the breakdown strength for air ( $3 \times 10^{6} V / m$ )

b) To find how close together can the plates be with this applied voltage:

The formula would be,

$d_{\min }=\frac{V}{E_{\max }}$

Apply all known values, we get

$d_{\min }=\frac{5.0 \times 10^{3}}{3 \times 10^{6}}=1.7 \times 10^{-3} \mathrm{m}=1.7 \mathrm{mm}$

3 0

3 years ago