Which types of light are absorbed by genetic material?

A bicyclist starts at point P and travels around a triangular path that takes her through points Q and R before returning to the

REY [17]

Answer:

Displacement by cyclist is zero.

Explanation:

In the given question bicyclist is travelling in a rectangular track having P , Q and R edges.

The bicyclist starts from P and travel through Q and R and returned to P again.

We need to find its displacement.

We know displacement of a body is its difference between its initial position to final position.

Here in the given question the bicyclist returns to P again.

Therefore, total displacement by bicyclist is zero.

Hence, this is the required solution.

4 0

3 years ago

You are driving to the grocery store at 20 m/s. You are 110 m from an intersection when the light turns red. You have a reaction

Lynna [10]

Answer:

a)

reaction time = 0.70 s

distance travelled in reaction time = v*t

= 20 m/s * 0.70 s

= 14 m

So, when brake is applied, distance remaining= 110 m - 14 m = 96 m

Answer: 96 m

b)

vf = 0 m/s

d = 96 m

vi = 20 m/s

use:

vf^2 = vi^2 + 2*a*d

0 = 20^2 + 2*a*96

-400 = 2*a*96

a = -2.08 m/s^2

Answer: -2.08 m/s^2

c)

use:

vf = vi + a*t

0 = 20 - 2.08*t

t = 9.6 s

Answer: 9.6 s

Explanation:

8 0

2 years ago

A car is initially moving at 20 m/s east and a little while later it is moving at 10 m/s north. Which of the following best desc

Nina [5.8K]

Answer:d

Explanation:

Given

First car is moving towards east with velocity 20 m/s

$\vec{v_1}=20\hat{i}$

then it turns towards north then velocity is

$\vec{v_2}=10\hat{j}$

suppose car takes t sec to change its path so average acceleration is given by

$a=\frac{v_2-v_1}{t}$

$a=\frac{1}{t}(10\hat{j}-20\hat{i})$

So average acceleration is towards North of west.

5 0

3 years ago

At a certain elevation, the ________ , the air becomes saturated and water-vapor molecules ________.

andriy [413]

At certain altitude, the temperature of air decrease, The air becomes saturated and water vapour molecules starts condensing.

As the altitude of air increase, the atmospheric pressure decrease due to which the temperature of the air decrease. The water molecules in the atmosphere start condensing, which saturate the air (that is air can no hold water molecules), due to which the water vapour molecules starts condensing and falls on the earth in the form of rain.

4 0

3 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

3 years ago