Question 1 of 10

Match the following

Fofino [41]

Answer:

1. b

2. d

3. e

4. c

5. a

Explanation:

These are just basic definitions. Let me know if you need further clarification.

7 0

3 years ago

A sample of radium-226 will decay to ¼ of its original amount after 3200 years. What is the half-life of radium-226?

lesantik [10]

<h3><u>Answer</u>;</h3>

1600 years

<h3><u>Explanation</u>;</h3>

Half life is the time taken for a radioactive isotope to decay by half of its original amount.
We can use the formula; N = O × (1/2)^n ; where N is the new mass, O is the original amount and n is the number of half lives.
A sample of radium-226 takes 3200 years to decay to 1/4 of its original amount.

Therefore;

Thus; <em>3200 years is equivalent to 2 half lives.</em>

<em>Hence, the half life of radium-226 is 1600 years</em>

3 0

3 years ago

The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 sec

marin [14]

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

$\epsilon = \frac{\sigma}{\sqrt{n} }$

To reduce the standard error to 0.03 s, let the additional number of trials = x

$0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials$

Therefore, the additional trials needed is 48 trials.

6 0

3 years ago

What is scientific learning

bixtya [17]

Hello here

It means that a person has the ability to describe, explain, and predict natural phenomena.

4 0

2 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$