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4 years ago

15

The specific heat of a liquid Y is 4.2 cal/g degrees Celsius. A sample grams of this liquid at 137 K is heated to 265K. The liqu

id absorbs. 7.82 kcal. What is the sample of liquid in grams?

1 answer:

sergiy2304 [10]4 years ago

8 0

Answer:

54

Explanation:

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which of the following statements best explains how consumers determine growth in technological areas

Mrac [35]

If these were the missing choices:

a) Consumers fill out questionnaires concerning their need for new products.

b) Consumers vote for politicians who decide which kind of research to support

c) Consumers decide what to buy and what not to buy

d) Consumers influence the decisions of private foundations by deciding where to donate money.

My answer would be: c) <span>Consumers decide what to buy and what not to buy</span>

Every growth is based on the demand of the people. If a good or service is needed then its demand will increase. If a good or service is not needed then its demand will decrease until such time that said good or service will be eliminated.

6 0

3 years ago

If the only forces acting on a 2.0kg mass are F1 = (3i-8j)N and F2 = (5i+3j)N, what is the magnitude of the acceleration of the

belka [17]

Answer: 4.7m/s²

Explanation:

According to newton's first law,

Force = mass × acceleration

Since we are given more the one force, we will take the resultant of the two vectors.

Mass = 2.0kg

F1+F2 = (3i-8j)+(5i+3j)

Adding component wise, we have;

F1+F2 = 3i+5i-8j+3j

F1+F2 = 8i-5j

Resultant of the sum of the forces will be;

R² = (8i)²+(-5j)²

Since i.i = j.j = 1

R² = 8²+5²

R² = 64+25

R² = 89

R = √89

R = 9.4N

Our resultant force = 9.4N

Substituting in the formula

F = ma

9.4 = 2a

a = 9.4/2

a = 4.7m/s²

Therefore, magnitude of the acceleration of the particle is 4.7m/s²

3 0

3 years ago

The layer of earth that has the lightest elements is the

Sonja [21]

If I remember correctly (from my studies long time ago) the layers are from the outer to the center:
SiAl : Silicon-Aluminum
SiMa : Silicon-Magnesium (although should be Mg)
NiFe : Nickel-Iron

The SiMa layer should have the lightest elements (Magnesium is lighter than Aluminum)

8 0

4 years ago

If two charged objects in a laboratory are brought to a distance of 0.22 meters away from each other. What is

zysi [14]

Answer:

$q_2=2.47\times 10^{-4}\ C$

Explanation:

The charge on one object, $q_1=9.9\times 10^{-5}\ C$

The distance between the charges, r = 0.22 m

The force between the charges, F = 4,550 N

Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

$F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C$

So, the charge on the other sphere is $2.47\times 10^{-4}\ C$ .

7 0

3 years ago

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y

Gnesinka [82]

The representation of this problem is shown in Figure 1. So our goal is to find the vector $\overrightarrow{R}$ . From the figure we know that:

$\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}$

From geometry, we know that:

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Then using vector decomposition into components:

$For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85$

Therefore:

$R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m$

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector $\overrightarrow{R}$ , that is:
</span>
$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span> $\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}$

6 0

3 years ago