If you wanted the exact answer, you would need the distance from the window to the ground; and than divide by 3.0. Otherwise, 9.8 meters a second would be your best bet.

6 0

2 years ago

Read 2 more answers

Iodine 131 half life is 8.0 days. Ten percent of the original sample o his isotope remains after (a) 22.7 days (b) 24.9 days (c)

Artyom0805 [142]

Answer:

option (c) is correct

Explanation:

Half life of a substance is the time in which the element becomes half of is initial value.

half life, T = 8 days

Amount remaining, N = 10 % of original value

Let the original value is No.

N = 10% of No

N = 0.1 No

Let the time taken is t and the decay constant is λ.

The relation between the decay constant and the half life is given by

$\lambda =\frac{0.6931}{T}=\frac{0.6931}{8}=0.08664 per day$

Us the equation of radioactivity

$N=N_{0}e^{-\lambda t}$

$0.1N_{0}=N_{0}e^{-0.08664 t}$

$e^{0.08664 t}=10$

Taking natural log on both the sides, we get

0.08664 t = 2.303

t = 26.6 days

4 0

2 years ago

The blackbody radation emmitted from a furnace peaks at a wavelength of 1.9 x 10^-6 m (0.0000019 m). what is the temperature ins

krek1111 [17]

Answer:

Temperature, T = 1542.10 K

Explanation:

It is given that,

The black body radiation emitted from a furnace peaks at a wavelength of, $\lambda=1.9\times 10^{-6}\ m$

We need to find the temperature inside the furnace. The relationship between the temperature and the wavelength is given by Wein's law i.e.

$\lambda\propto \dfrac{1}{T}$

$\lambda=\dfrac{b}{T}$

b = Wein's displacement constant

$\lambda=\dfrac{2.93\times 10^{-3}}{T}$

$T=\dfrac{2.93\times 10^{-3}}{\lambda}$

$T=\dfrac{2.93\times 10^{-3}}{1.9\times 10^{-6}\ m}$

T = 1542.10 K

So, the temperature inside the furnace is 1542.10 K. Hence, this is the required solution.

3 0

2 years ago

Assuming motion is on a straight path what would result in two positive components of a vector

polet [3.4K]

If the object being represented is going both up and to the right.

5 0

2 years ago