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Anestetic [448]
3 years ago
7

The half-life of a substance is 10 years. What percent of the substance remains after 2 half-lives.

Physics
1 answer:
lozanna [386]3 years ago
8 0

Answer:

i think its 2.5 as well

Explanation:

You might be interested in
A piece of wire length 30cm and cross sectional area of 0.5mm^2 has a resistance of 5ohms at 20°c. It is then heated to a temper
Grace [21]

Answer:

8.333*10^-6 ohms

Explanation:

Resistivity of a material is expressed as;

p = RA/l

R is the resistance of the material

A is the cross sectional area

l is the length of the material

Given

R = 5 ohms

A = 0.5mm^2

A = 5 * 10^-7m^2

l = 30cm  = 0.3m

Substitute into the formula;

p = (5 *  5 * 10^-7m^2)/0.3

p = 25 * 10^-7/0.3

p = 0.0000025/0.3

p = 8.333*10^-6

Hence its resistivity at 20 degrees Celsius is 8.333*10^-6 ohms

7 0
3 years ago
If you want points for your account just answer yes
jeka94

Answer:

Lol, yes

Explanation:

5 0
3 years ago
An object, initially at rest, is subject to an acceleration of 34 m/s2. How long will it take for that
SIZIF [17.4K]

Answer:

If it is moving 34 m/s it will take 100 seconds, or 1:40 to reach 3400 meters.

Explanation:

I found this answer by dividing 3400 by 34 and converting seconds to minutes

6 0
3 years ago
Read 2 more answers
Coherent microwaves of wavelength 5.00 cm enter a tall, narrow window in a building otherwise essentially opaque to the microwav
zzz [600]
The solution for this problem is:

For 1st minimum, let m be equal to 1. 

d = slit width

D = screen distance. 

Θ = arcsin (m * lambda/ (d))


= 0.13934 rad, 7.9836 deg 

y = D*tan (Θ)

y = 6.50 * tan (7.9836)

= 0.91161 m is the distance from the central maximum to the first-order minimum 
8 0
2 years ago
a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u
n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    a_{c}  = \frac{v^{2} }{r}

   a_{c}  = 23.04/10.3

    a_{c}  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

a_{t} = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

a_{net}  =  \sqrt{a_{c} ^{2} +a_{t} ^{2}   }

a_{net}  =  \sqrt{4.97 + 0.396}

a_{net}  =  2.316 m/s²

So the magnitude of  net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

#SPJ4

8 0
1 year ago
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