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3 years ago

The mass of an object was measure on a balance and reported as 65.35g and it displaced 3.4ml. What is the density of the object

Physics

1 answer:

Sindrei [870]3 years ago

5 0

Answer:

your [email protected] father

Explanation:

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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.

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3 years ago

Which of the following is a mood disorder?

gulaghasi [49]

Bipolar disorder is the answer

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3 years ago

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HELP FAST!!!!!!!

Alex17521 [72]

<u>Answer</u>

A. Metals A and metals B

<u>Explanation</u>

Heat transfer takes place whenever there is temperature difference. When two bodies of different temperatures are brought together, heat energy will move from one body to the other until equilibrium temperature is reached.

In our case, heat transfer will take place in all four metals.

Metal A will transfer heat to the water since it's temperature is higher than that of water.

Metal B will also transfer heat to the water since it's temperature is higher than that of water.

Metal C will get heat from the water since it's colder than the water.

Metal D will also get heat from the water since it is colder than water.

7 0

3 years ago

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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 physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Oiled Steel fry

Alinara [238K]

Answer:

N = 6.67 N

Explanation:

The frictional or frictional force is a force that arises from the contact of two bodies and opposes movement.

The friction is due to imperfections and roughness, mainly microscopic, that exist on the surfaces of the bodies. Upon contact, these roughnesses engage with each other making movement difficult. To minimize the effect of friction, either the surfaces are polished or lubricated, since the oil fills the imperfections, preventing them from snagging.

As the frictional force depends on the materials and the force exerted on one another, its magnitude is obtained by the following expression:

f = μ*N Formula (1)

where:

f is the friction force (N)

μ is the coefficient of friction

N is the normal force (N)

Data

f = 0.2 N : frictional force between the steel spatula and the Oiled Steel frying pan

μ = 0.03 :coefficient of kinetic friction between the two materials

Calculating of normal force

We replace data in the formula (1)

f = μ*N

0.2 = 0.03*N

N = 0.2 / 0.03

N = 6.67 N

5 0

3 years ago