Does potential energy increase,kinetic energy decrease when a book is placed on a shelf

The usefulness of blotting techniques in molecular biology is that

densk [106]

Answer:

Transferred material is in the same relative position on the disk as on the original sample

Explanation:

The usefulness of blotting techniques in molecular biology is that transferred material is in the same relative position on the disk as on the original sample

3 0

3 years ago

A rocket initially at rest accelerates at a rate of 99. 0 meters/second2. Calculate the distance covered by the rocket if it att

creativ13 [48]

The rocket will cover $1 \times 10^3\rm \ m$ distance in 4. 5 s. Acceleration can be defined as the change in velocity.

<h3>What is acceleration?</h3>

Acceleration can be defined as the change in speed or the direction of the object.

From kinamatic equation:

$D = v_{t} +\dfrac 12at^2$

Where,

$D$ - final velocity = 445 m/s

$v_0$ - initial valocity = 0 m/s

$a$ - acceleration = 99. 0 m/s²

$t$ - time = 4. 50 s

Put the values in the formula,

$D = 0\times ( 4.5) + \dfrac12\times (99)(4.5)^2\\\\D = 1002 {\rm \ m}\\\\D = 1 \times 10^3\rm \ m$

Therefore, the rocket will cover $1 \times 10^3\rm \ m$ distance in 4. 5 s.

Learn more about Acceleration :

brainly.com/question/2697545

7 0

2 years ago

What is 9414 divided by 18

Levart [38]

9414/18 is equal to 523

3 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work

aksik [14]

It is required an infinite work. The additional electron will never reach the origin.

In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
$F=k_e \frac{q_e q_e}{d^2}$
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.

5 0

3 years ago