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3 years ago

12

In a certain chemical process, a lab technician supplies 292 J of heat to a system. At the same time, 68.0 J of work are done on

the system by its surroundings. What is the increase in the internal energy of the system?

1 answer:

vekshin13 years ago

3 0

Answer:

The increase in the internal energy of the system is 360 Joules.

Explanation:

Given that,

Heat supplied to a system, Q = 292 J

Work done on the system by its surroundings, W = 68 J

We need to find the increase in the internal energy of the system. It can be given by first law of thermodynamics. It is given by :

$dE=dQ+dW\\\\dE=292\ J+68\ J\\\\dE=360\ J$

So, the increase in the internal energy of the system is 360 Joules. Hence, this is the required solution.

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What is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?

Art [367]

CORRECT ANSWER:

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

STEP-BY-STEP EXPLANATION:

The complete question from book is

According to Figure 9.6, what is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

b- Signaling molecules that bind to cell-surface receptors lead to cellular responses restricted to the cytoplasm; signaling molecules that bind to intracellular receptors lead to cellular responses restricted to the nucleus.

c- Cell-surface receptors bind to specific signaling molecules; intracellular receptors bind any signaling molecule.

d- Cell-surface receptors typically bind to signaling molecules that are smaller than those bound by intracellular receptors.

e- None of the other answer options is correct.

3 0

3 years ago

Course hero N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches

aleksandrvk [35]

Answer:

Therefore it is save to carry a 62kg adult

Explanation:

From the question we are told that:

Mass $m=380kg$

Height of supporting Rock $X=85cm$

Length of Board $L_r=4.5m$

Mass of board $M_b=22kg$

Mass of adult $M_a=62$

Generally the moment of balance about wedge part about is mathematically given by

$N -Q + R = Mg + mg$

$0.85*N - Mg*2.25 - mg*(2.25 + x) = 0$

$0.85*N = + Mg*2.25 + mg*(2.25 + x)$

where

$N+R=4547$

therefore

$N = 570.70588 + 1608.3529 + 714.823 x$

if N=0 at fallen person

$x=3.04m$

Therefore it is save to carry a 62kg adult

8 0

2 years ago

two electrons are an angstrom (1x10^-10m) apart. What electrostatic force do they exert on one another?

Irina-Kira [14]

Answer:

2.30 × 10⁻⁸ N if the two electrons are in a vacuum.

Explanation:

The Coulomb's Law gives the size of the electrostatic force $F$ between two charged objects:

$\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}}$ ,

where

$k$ is coulomb's constant. $k = 8.99\times 10^{8}\;\text{N}\cdot\text{m}^{2}\cdot\text{C}^{-2}$ in vacuum.
$q_1$ and $q_2$ are the signed charge of the objects.
$r$ is the distance between the two objects.

For the two electrons:

$q_1 = q_2 = 1.60\times 10^{-19}\;\text{C}$ .
$r = 1\times 10^{-10}\;\text{m}$ .

$\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}} = -\frac{8.99\times 10^{8}\times (1.60\times 10^{-19})^{2}}{(1\times 10^{-10})^{2}} = 2.30\times 10^{-8}\;\text{N}$ .

The sign of $F$ is negative. In other words, the two electrons repel each other since the signs of their charges are the same.

8 0

3 years ago

rocks move trough the rock cycle through uplift,folding,and faulting cased by movements of the earth´s________

garik1379 [7]

Tectonic plates and convection cells move tectonic plates

4 0

3 years ago

The Jurassic Park ride at Universal Studios theme park drops 25.6 m straight down essentially from rest. Find the time for the d

ankoles [38]

Answer:

V=22.4m/s;T=2.29s

Explanation:

We will use two formulas in order to solve this problem. To determine the velocity at the bottom we can use potential and kinetic energy to solve for the velocity and use the uniformly accelerated displacement formula:

$mgh=\frac{1}{2}mv^{2}\\\\X= V_{0}t-\frac{gt^{2}}{2}$

Solving for velocity using equation 1:

$mgh=\frac{1}{2}mv^{2} \\\\gh=\frac{v^{2}}{2}\\\\\sqrt{2gh}=v\\\\v=\sqrt{2*9.8\frac{m}{s^2}*25.6m}=22.4\frac{m}{s}$

Solving for time in equation 2:

$-25.6m = 0\frac{m}{s}t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\-51.2m=-9.8\frac{m}{s^{2}}t^{2}\\\\t=\sqrt{\frac{51.2m}{9.8\frac{m}{s^{2}}}}=2.29s$

7 0

3 years ago