PHYSICS COLLEGE

An electron has a velocity of 3.2 x 10^6 m/s. What is its’ momentum? (b) What is its’ wavelength? (c) What other objects/materials have this space/size? (d) Assuming that we can measure the velocity to an accuracy of 10%. Use the Heisenberg uncertainty principle to calculate the uncertainty in the position.

Answers

Answer 1

Answer:

Answer:

P = 2.91*10^{-24} kg m/s

$\lambda = 2.73 *10^(-10) m$

size of atom hat lie in range of 1 to 5 Angstrom

$\Delta x = 0.2272$ Angstrom

Explanation:

A) MOMENTUM

p = mv

where m is mass of electron

so momentum p can be calculated as

p = 9.11*10^{-31} *3.2*10^{6}

P = 2.91*10^{-24} kg m/s

b) wavelength

$\lambda = (h)/(mv)$

where h is plank constant

so $\lambda = (6.626*10^(-34))/(2.91*10^(-24))$

$\lambda = 2.73 *10^(-10) m$

c) size of atom hat lie in range of 1 to 5 Angstrom

d) from the information given in the question we have

$(\Delta v)/(v) = 0.1$

$\Delta v = 0.1 v$

we know that

$\Delta p *\Delta x = (h)/(4\pi)$

$m \Delta v \Delta x =(h)/(4\pi)$

$\Delta x = (h)/(m \Delta v)$

$\Delta x = (2.272)/(0.1)$ [ $\Delta v = 0.1 v$ ]

$\Delta x = 0.2272$ Angstrom

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An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)

A wooden block with mass 1.05 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 35.0 degrees (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 4.90m up the incline from A, the block is moving up the incline at a speed of 5.10 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is 0.55. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

Answers

The amount of potential energy that was initially stored in the spring due to the wooden block is 65.3 joules.

What is potential energy?

Potential energy is the energy which body posses because of its position.

The potential energy of a body is given as,

$PE=mgh$

Here, (m) is the mass of the body, (g) is the gravitational force and (h) is the height of the body.

The energy stored in the spring is the sum of all the potential energy, kinetic energy and the energy dissipated due to friction. Therefore, it can be given as,

$E=mgh+(1)/(2)mv^2+\mu mgd\cos\theta$

Here, the mass of the wooden block is 1.05 kg . Angle of inclination is 35.0 degrees (point A). The distance from point B is 4.90m up the incline from A.

The speed of the block is 5.10 m/s and the coefficient of kinetic friction between the block and incline is 0.55. Therefore, put the values in the above formula as,

$E=1.05(9.81)(4.9\sin(35))+(1)/(2)(1.05)(5.1)^2+(0.55)(1.05)(9.8)(4.9)\cos(35)\nE=65.3\rm J$

Hence, the amount of potential energy that was initially stored in the spring due to the wooden block is 65.3 joules.

Learn more about the potential energy here;

brainly.com/question/15896499

Answer:

Explanation:

energy stored in spring initially

= kinetic + potential energy of block + energy dissipated by friction

= 1/2 mv² + mgh + μ mgcosθ x d

m is mass , v is velocity at top position , h is vertical height , μ is coefficient of friction ,θ is angle of inclination of plane

= m (1/2 v² + gh + μ gcosθ x d )

= 1.05 ( .5 x 5.1² + 9.8 x 4.9 sin35 + .55 x 9.8 cos35 x 4.9 )

= 1.05 ( 13.005 + 27.543 + 21.635)

= 65.3 J .

Two objects of equal mass are a distance of 5.0 m apart and attract each other with a gravitational force of 3.0 x 10^-7 N find their mass.A) 150 kg
B) 9.8 kg
C) 11.000 kg
D) 340 kg

Answers

Answer

I Think Its 150

If the population of penguins increased, then this would have a direct effect on the populations of?

Answers

Answer:

Globel warming

Explanation:

hope this helpex

For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is p1. If the length of the pipe is then doubled, what is the relation of the new pressure drop p2 to the original pressure drop p1 at the original mass flow rate?

Answers

Answer: ∆p2 = 2* ∆p1

Explanation:

Given that all other factors remain constant. The pressure drop across the pipeline is directly proportional to the length.

i.e ∆p ~ L

Therefore,

∆p2/L2 = ∆p1/L1

Since L2 = 2 * L1

∆p2/2*L1 = ∆p1/L1

Eliminating L1 we have,

∆p2/2 = ∆p1

Multiplying both sides by 2

∆p2 = 2 * ∆p1

Which of these objects are constantly in motion? Select all that apply.A.
Earth

B.
Planes

C.
Trains

D.
Blood

E.
Sun

F.
Cars

Answers

The object Earth,Sun, and Blood are constantly in motion. The correct option is A, D, and E.

What is motion?

if a body changes its position with respect to its surroundings in a given interval of time, Then the body is said to be in motion

Motion is generally classified as follows.:

i) Rectilinear motion.

ii) Circular motion.

iii) Rotational motion.

iv) Periodic motion.

The Earth is continuously in motion because it continuously revolves around The Sun in an elliptical orbit, due to which a year is 365 days. and also The earth rotates about its own axis once a day.

The Sun also revolves around the galactic center of our Milkyway galaxy. and it also rotates about its own axis continuously. so that the sun is also continuously in motion.

The Human blood is continuously in motion Because our blood is continuously circulating whole over the body with the help of our heart. The heart continuously pumps our blood and circulates it inside the human body.

Hence the Earth, Sun, and blood are continuously in motion.

To learn more about Motion click:

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#SPJ3

earth, blood, and the sun

Q18: A cube of aluminum has an edge length of 20 cm. Aluminum has adensity of 2.7 g/cm and a specific heat of 0.217 cal/ g.°С. When
the internal energy of the cube increases by 47000 cal its temperature
increases by:
A
B
C
D
E
5 °C
10 °C
20 °C
100 °C
200 °C

Answers

The change in temperature of this cube of aluminum is equal to: B. 10°C

Given the following data:

Edge length, L = 20 cm.

Density of Aluminum = 2.7 g/cm

Specific heat capacity (C) of aluminum = 0.217 Cal/g°С

Internal energy = 47000 calories.

To find the change in temperature of this cube of aluminum:

First of all, we would determine the volume of this cube of aluminum.

$Volume \;of \;a \;cube = L^3\n\nVolume \;of \;a \;cube = 20^3\n\nVolume \;of \;a \;cube = 8000\; cm^3$

Next, we calculate the mass of this cube of aluminum:

$Mass = Density * Volume\n\nMass = 2.7 * 8000$

Mass = 21,600 grams.

Now, we can find the change in temperature of this cube of aluminum:

Mathematically, the quantity of heat energy is given by the formula;

$Q = mc\theta$

Where:

Q represents the quantity of heat energy.

m represents the mass of an object.

c is the specific heat capacity.

∅ is the change in temperature.

Substituting the parameters into the formula, we have;

$47000 = 21600 * 0.217 * \theta\n\n47000 = 4687.2 \theta\n\n \theta =(47000)/(4687.2) \n\n \theta = 10.03$

Change in temperature = 10°C

Answer:

10 °C

Explanation:

From the question given above, the following data were obtained:

Egde length (L) of aluminum = 20 cm

Density of Aluminum = 2.7 g/cm³

Specific heat capacity (C) of aluminum = 0.217 cal/ g°С

Heat (Q) energy = 47000 cal

Change in Temperature (ΔT) =?

Next, we shall determine the volume of the aluminum. This can be obtained as follow:

Egde length (L) of aluminum = 20 cm

Volume (V) of aluminum =?

V = L³

V = 20³

V = 8000 cm³

Thus, the volume of the aluminum is 8000 cm³

Next, we shall determine the mass of the aluminum. This can be obtained as follow:

Density of Aluminum = 2.7 g/cm³

Volume of Aluminum = 8000 cm³

Mass of aluminum =.?

Density = mass/volume

2.7 = mass /8000

Cross multiply

Mass of aluminum = 2.7 × 8000

Mass of Aluminum = 21600 g

Finally, we shall determine the change in temperature of the aluminum as follow:

Specific heat capacity (C) of aluminum = 0.217 Cal/g°С

Heat (Q) energy = 47000 Cal

Mass (M) of Aluminum = 21600 g

Change in Temperature (ΔT) =?

Q = MCΔT

47000 = 21600 × 0.217 × ΔT

47000 = 4687.2 × ΔT

Divide both side by 4687.2

ΔT = 47000 / 4687.2

ΔT = 10 °C

Therefore, the increase in the temperature of the aluminum is 10 °C.

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