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3 years ago

What is the answer to this question?

Physics

2 answers:

sashaice [31]3 years ago

8 0

Answer: The answer is C

Explanation:

Hope it helps :)

Charra [1.4K]3 years ago

3 0

Answer:

Explanation:

The temperature of an object does not change when it is going through a phase change, since there is a single specific point on the temperature gradient where that change occurs. Once the phase change has occurred, the water's temperature can continue to change in that direction, but during the phase change it does not. Hope this helps!

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Two thermometers are calibrated, one in de

zlopas [31]

-40 c = -40 f but k would be 233.15

6 0

3 years ago

1. Applied research observational evidence 2. Basic research the experimental factor that changes in response to a change in the

lesya [120]

Answer:

This question is about matching each definition with its correct term. Please find the term matched with their appropriate definition below.

Explanation:

1. Empirical evidence: An empirical evidence is an observational evidence i.e an evidence gathered by observation or use of senses.

2. Dependent variable: Dependent variable is an experimental factor that changes in response to a change in the independent variable. In other words, it is dependent on the independent variable.

3. Applied research: Applied research is a type of research oriented at solving a present problem or need. It encompasses the production of products that can be sold for profit.

4. Hypothesis: A hypothesis in an experiment is a proposed explanation for a scientific problem that itself can be tested by experimentation. A hypothesis aims at providing a testable explanation to an observed problem.

5. Control: A control is a quantity in an experiment that remains unchanged or constant. It is kept the same by the experimenter for all groups in the experiment in order not to influence the outcome.

6. Basic research: Basic research is the research that expands knowledge in a particular area. It is the kind of research that aims at filling a knowledge void or satiating curiosity.

7. Independent variable: The independent variable is the experimental factor that is changed or manipulated deliberately by the scientist.

8 0

4 years ago

When an atom that has no charge looses two electrons it becomes a

alex41 [277]

Answer: it becomes a positive ion

Explanation:

So, when an atom loses 2 electrons there will be no change in the number of neutrons. Therefore, an isotope will not form. Thus, it is concluded that when an atom with no charge loses two electrons, it becomes a positive ion.

8 0

3 years ago

Read 2 more answers

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

A sample of helium behaves as an ideal gas as energy is

alisha [4.7K]

Answer:

0.0321 g

Explanation:

Let helium specific heat $c_h = 5.193 J/g K$

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

$E_h = m_hc_h \Delta T = 20 J$

where $m_h$ is the mass of helium, which we are looking for:

$m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g$

4 0

3 years ago