Why do the four outer planets have lower density than the four inner planets?

Section of two thermometers are shown. The thermometer on the left is labeled A and the thermometer on the right is labeled B. T

jolli1 [7]

Answer : The correct option is, (b) Thermometer B, because it measures temperature more accurately than A.

Explanation :

Accuracy : It is defined as the closeness of a measured value to a standard or known value.

For Example: If the mass of a substance is 50 kg and one person weighed 48 kg and another person weighed 55 kg. Then, the weight measured by first person is more accurate.

Precision : It is defined as the closeness of two or more measurements to each other.

For Example: If you weigh a given substance five times and you get 1.8 kg each time. Then the measurement is said to be precise.

Level of precision is determined by the maximum number of decimal places.

As per question, the thermometer B has markings 40°, 45° and 50° at equal intervals and there are four small markings equally spaced in between the numbers and thermometer B has markings 40°, 45° and 50° at equal intervals but there is no small markings equally spaced in between the numbers. That means, thermometer B is measures temperature more accurately than thermometer A.

Hence, correct option is, (b) Thermometer B, because it measures temperature more accurately than A.

8 0

3 years ago

Read 2 more answers

Hey guys i need help (potential energy)

Vlad1618 [11]

Explanation:

Michael should put the vase at the bottom of the shelf to reduce the potential energy because the height of the vase to the floor is nearly zero.

8 0

2 years ago

Sound travels around 340 m/s the function d(t =340 gives the distance d(t in meters that sound travels in t seconds how far does

MAXImum [283]

Sound would travel 10,540 meters

7 0

2 years ago

The force P is applied to the 45-kg block when it is at rest. Determine the magnitude and direction of the friction force exerte

cluponka [151]

Answer:

Check attachment for free body diagram of the question.

I used the free body diagram and the angles given in the diagram missing in the question above, but I used the data given in the above question.

Explanation:

Let frictional force be Fr acting down the plane

Let analyze the structure before inserting values

Using Newton's second law along the y-axis

ΣFy = Fnet = m•ay

Since the body is not moving in the y-direction, then ay = 0

N+PSinβ — WCosθ = 0

N+PSin20—441.45Cos15 = 0

N+PSin20—426.41 = 0

N = 426.41 — PSin20 , equation 1

The maximum Frictional force to be overcome is given as

Fr(max) = μsN

Fr(max) = 0.25(426.41 — PSin20)

Fr(max)= 106.6 —0.25•PSin20

Fr(max) = 106.6 — 0.08551P, equation 2

This is the maximum force that must be overcome before the body starts to move

Using Newton's law of motion in the x direction

Note, we took the upward direction up the plane as the direction of motion since the force want to move the block upward

Fnetx = ΣFx

Fnetx = P•Cosβ —W•Sinθ — Fr

Fnetx = P•Cos20—441.45•Sin15—Fr

Fnetx = 0.9397P — 114.256 — Fr

Equation 3

When Fnetx is positive, then, the body is moving up the plane, if Fnetx is negative, then, the maximum frictional force has not yet being overcome and the object is still i.e. not moving

a. When P = 0

From equation 2

Fr(max) = 106.6 — 0.08551P

Fr(max) = 106.6 — 0.08551(0)

Fr(max)= 106.6 N

So, 106.6N is the maximum force to be overcome

So, here the only force acting on the body is the weight and it acting down the plane, trying to move the body downward.

Wx = WSinθ

Wx = 441.45× Sin15

Wx = 114.256 N.

Since the force trying to move the body downward is greater than the maximum static frictional force, then the body is not in equilibrium, it is moving downward.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — PSin20 , equation 1

N = 426.41 N, since P=0

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 426.41

Fr = 93.81 N.

b. Now, when P = 190N

From equation 2

Fr(max) = 106.6 — 0.08551(190)

Fr(max) = 106.6 —16.2469

Fr(max)= 90.353 N

So, 90.353 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 190Cos20 — 441.45Sin15

Fnetx = 64.29N

So the force moving the body up the incline plane is 64.29N

Fnetx < Fr(max)

Then, the frictional force has not being overcome yet.

Then, the body is in equilibrium.

Then, applying equation 3.

Fnetx = 0.9397P — 114.256 — Fr

Fnetx = 0, since the body is not moving

0 = 0.9397(190) —114.246 — Fr

Fr = 64.297 N

Fr ≈ 64.3N

c. When, P = 268N

From equation 2

Fr(max) = 106.6 — 0.08551(268)

Fr(max) = 106.6 —16.2469

Fr(max)= 83.68 N

So, 83.68 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 268Cos20 — 441.45Sin15

Fnetx = 137.58 N

So the force moving the body up the incline plane is 137.58 N

Fnetx > Fr(max)

Then, the frictional force has being overcome.

Then, the body is not equilibrium.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — 268Sin20 , equation 1

N = 334.75 N, since P=268N

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 334.75

Fr = 73.64 N

d. The required force to initiate motion is the force when the block want to overcome maximum frictional force.

So, Fnetx = Fr(max)

Px — Wx = Fr(max)

From equation 1

Fr(max) = 106.6 — 0.08551P,

P•Cosβ-W•Sinθ = 106.6 — 0.08551P

P•Cos20 — 441.45•Sin15 = 106.6 — 0.08551P

P•Cos20—114.256=106.6 - 0.08551P

PCos20+0.08551P =106.6 + 114.256

1.025P=220.856

P = 220.856/1.025

P = 215.43 N

3 0

2 years ago

A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge

zheka24 [161]

Answer:

The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

$F_{1}=\dfrac{kq_{1}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}$

$F_{1}=8\times10^{-11}j\ \ N$

We need to calculate the force on electron due to q₂

Using formula of force

$F_{2}=\dfrac{kq_{2}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}$

$F_{2}=6.93\times10^{-11}i\ \ N$

We need to calculate the net force

Using formula of net force

$F=F_{1}+F_{2}$

Put the value into the formula

$F=8\times10^{-11}j+6.93\times10^{-11}i$

The magnitude of the net force

$F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}$

$F=1.058\times10^{-10}\ N$

We need to calculate the acceleration of an electron

Using formula of force

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}$

$a=1.2\times10^{20}\ m/s^2$

Hence, The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

3 0

3 years ago