Would you please help me with this? I can't figure it out, please! I need to know what the E means!

When the accuracy in measuring the velocity of a particle increases, that of its position decreases. The correct option is d. decrease.

The Heisenberg's uncertainty principle states that there is always an uncertainty in any attempt to measure accurately the value of any two complementary variables simultaneously. This implies that measuring of position and momentum of a particle simultaneously would lead to uncertainty in their values.

So that;

Δx.ΔP ≥ [tex]\frac{h}{2\pi }[/tex]

Where Δx is the uncertainty in the value of its position, ΔP is the uncertainty in the value of momentum and h is the Planck's constant.

This principle simply explains and provides the required answer to the given question. So that according to Heisenberg's uncertainty principle, the accuracy in measuring its position will definitely decrease.

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Answer:

The magnitude of the magnetic field is 1.83 x [tex]10^{-5}[/tex] T.

Explanation:

The flow of an electric current in a straight wire induces magnetic field around the wire. When current is flowing through two wires in the same direction, a force of attraction exists between the wires. But if the current flows in opposite directions, the force of repulsion is felt by the wires.

In the given question, the direction of flow of current through the wires is opposite, thus both wires applies the same field on each other. The result to repulsion between them.

The magnetic field (B) between the given wires can be determined by:

B = [tex]\frac{U_{o}I }{2\pi r}[/tex]

where: I is the current, r is the distance between the wires and [tex]U_{0}[/tex] is the magnetic field constant.

But, I = 11 A, r = 0.12 m and [tex]U_{0}[/tex] = 4[tex]\pi[/tex] x [tex]10^{-7}[/tex] Tm/A

So that;

B = [tex]\frac{4\pi *10^{-7}*11 }{2\pi *0.12}[/tex]

   = 1.8333 x [tex]10^{-5}[/tex]

B = 1.83 x [tex]10^{-5}[/tex] T

A  5

v=  dt/ dx  =−5+12t

Initial velocity means at t=0, which is −5+0=−5.

Thus, −v=5n

Answer:

You should multiply 60 kg*9.8 and answer will come.

Hope this will help you.

Answer:

yes she is right you should multiple 60*9.8

have a great day God bless you

The price of Pesos is increased, and it takes more to buy a given amount of Pesos

According to the demand and supply economic theory, made popular by Adam Smith in 1776, which provides a price determination model based on the relationship between the amount of a good producers of the good are willing to sell at a given price and the amount of that good that the end users wish to purchase, there is an inverse relationship between price and supply, therefore as the supply decreases, the price increases, and therefore, there is a direct relationship between the demand and price, such that as the demand increases, the price of the goods increases because the ratio of supply to demand is decreased

Given that the arrival of thousands of tourists to the remote Los Cabos, will result in the tourists requiring Pesos, the demand of Pesos will rise, which according to the demand and supply theory, will increase the price of Pesos

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Answer:

233.33°F

Explanation:

(385K - 273.15) * 9/5 + 32 = 233.33°F

Answer:

Following are the complete solution to the given question:

Explanation:

In this question, before tug they are at rest, thus their initial momentum is 0 [tex]\pi=0[/tex], according to the law of conservation [tex]p_f=0[/tex] thus both should move in the opposite direction whose sum of the momentums is equaled to zero as there is no friction they will continue to move with their velocities until their collision. Its collision doesn't happen it will continue up to an infinite distance.

The motion of both you and your partner depends on the tension in the rope.

We must note that frictional force is the force that acts between two surfaces in contact with each other. The frictional force depends on the nature of the surfaces in contact.

Now, since there is no friction between your leg and the ice. The motion of both you and your partner depends on the tension in the rope.

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Answer:

2bsnsnsnns181991oiwiw

Answer:

the distance moved by the puck after 2.5 s is 7.8 m

Explanation:

Given;

mass of the puck, m = 2 kg

initial velocity of the puck, u = 0

applied force, F = 5 N

time of motion, t = 1.5 s

Acceleration of the puck is calculated from Newton's second law of motion;

F = ma

a = F/m

a = 5/2

a = 2.5 m/s²

The distance moved by the puck after 2.5 s is calculated as;

s = ut + ¹/₂at

s = 0 + ¹/₂at²

s = ¹/₂at²

s = 0.5 x 2.5 x (2.5)²

s = 7.8 m

Therefore, the distance moved by the puck after 2.5 s is 7.8 m

Answer:

Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.

Answer:

cobina

Explanation:

me 2

Answer:

c) charges exert forces on other charges.

Explanation:

When two different materials are rubbed together, there is a transfer of electrons from one material to the other material so this causes one object to become positively charged and the other object is negatively charged so they will attract each other not repel each other. Charges exert forces on other charges i.e. opposite charges attract each other whereas similar charges repel each other so in both cases force are exerted on one another.

Answer:

253.85°C

Explanation:

Here is the formula for converting K to °C

527K − 273.15 = 253.85°C

Answer:

2Al + 2H2O + 2NaOH ⟶ 3H2 + 2NaAlO2

Chất rắn màu xám bạc của nhôm (Al) tan dần trong dung dịch, sủi bọt khí là hidro (H2).

Explanation:

Answer:

Octabromine tetraphosphide

Explanation:

This compound has in its formula:

- Eight bromines

- Four phosphorous

8 → octa prefix

4 → tetra prefix

Right answer is Octabromine tetraphosphide

Answer:

speed is the gradient of the graph

Answer:

Speed is the slope of a distance time graph.

Explanation:

Speed= d/t

Slope is equal to rise/run

If the rise of the graph is the distance and the run is the time, calculating slope is the equivalent of calculating average speed.

Answer:

the  speed of the electron at the given position is 106.2 m/s

Explanation:

Given;

initial position of the electron, r = 9 cm = 0.09 m

final position of the electron, r₂ = 3 cm = 0.03 m

let the speed of the electron at the given position = v

The initial potential energy of the electron is calculated as;

[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]

When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;

[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]

Apply the principle of conservation of energy;

ΔK.E = ΔU

[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]

Therefore, the  speed of the electron at the given position is 106.2 m/s

Answer:

89.87m/s

Explanation:

Given the velocity function

v(t)=−t²+3t−2

In order to get the displacement function, we will integrate the velocity function as shown:

[tex]\int\limits^5_{-2} {v(t)} \, dt \\d(t)= \int\limits^5_{-2}{(-t^2+3t+2)} \, dt \\\\d(t)=[\frac{-t^3}{3}+\frac{3t^2}{2}+2t ]^5_{-2}\\[/tex]

at t = 5

[tex]d(5)=[\frac{-5^3}{3}+\frac{3(5)^2}{2}+2(5) ]\\d(5)=[\frac{-125}{3}+\frac{75}{2}+10 ]\\d(5)=-41.7+37.5+10\\d(5)=89.2m/s[/tex]

at t = -2

[tex]d(-2)=[\frac{-(-2)^3}{3}+\frac{3(-2)^2}{2}+2(-2) ]\\d(-2)=[\frac{-8}{3}+\frac{12}{2}+(-4) ]\\d(-2)=-2.67+6-4\\d(-2)=-0.67m/s[/tex]

Required displacement = d(5) - d(-2)

Required displacement = 89.2 - (-0.67)

Required displacement = 89.2 + 0.67

Required displacement = 89.87m/s

Answer:

In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.

Answer:

a)  [tex]P=0.80[/tex]

b)  [tex]1.25Hz[/tex]

c)  [tex]A=25cm[/tex]

Explanation:

From the question we are told that:

Travel Time [tex]T=0.40s[/tex]

Distance [tex]d=50cm[/tex]

a)

Period

Time taken to complete one oscillation

Therefore

[tex]P=2*T\\\\P=2*0.40[/tex]

[tex]P=0.80[/tex]

b)

Frequency is

[tex]F=\frac{1}{T}\\\\F=\frac{1}{0.80}[/tex]

[tex]1.25Hz[/tex]

c)

Amplitude:the distance between the mean and extreme position

[tex]A=\frac{50}{2}[/tex]

[tex]A=25cm[/tex]

Answer:

0.125

Explanation:

Density =mass/volume

Density =2.5/20

Density =0.125

Answer:

D=m/v

=2.5kg/(20/1000000)m^3

2.5kg÷0.000002m^3

1250000kgm^-3

Explanation:

20 is divided by 1000000 because m^3 is its si unit

Answer:

The specific latent heat of a substance is the amount of energy required to change the state of one kilo of the substance without change in it temperature.The latent heat of vaporization or evaporation is the heat given to some mass to convert if from the liquid to the vapor phase.

Explanation:

For air, n1 = 1.00003; for water, n2 = 1.3330

Given: θ2 = 30 degrees, then

θ1 = arcsin [(n2/n1) sin θ2]

= arcsin [(1.3330/1.0003) sin (40)]

= 58.93 degrees

Note that since, in this example, light is traveling from a medium of higher density (water; n2 = 1.3330) to a medium of lower density (air; n1 = 1.0003), then n2 > n1, and the angle of refraction (θ1) is larger than the angle of incidence (θ2), thus the light bends away from the normal (in this example, the vertical) as it leaves the water and enters the air.

Answer:

The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.

Explanation:

Concepts and reason

The concept required to solve this problem is Newton’s second law of motion.

Initially, write an expression for the force according to the Newton’s second law of motion. Later, rearrange the expression for the acceleration. Finally, substitute the value of the acceleration obtained to find the new force.

Fundamentals

According to the Newton’s second law of motion, the net force is equal to the product of the mass and the acceleration of an object. The expression for the Newton’s second law of motion is as follows:

F = maF=ma

Here, m is mass and a is the acceleration.

(a)

Rearrange the equation F = maF=ma for a.

a = \frac{F}{m}a=  

m

F

​

Substitute 18,000 N for F and \left( {2300{\rm{ kg + 2400 kg}}} \right)(2300kg+2400kg) for m in the equation a = \frac{F}{m}a=  

m

F

​

 .

\begin{array}{c}\\a = \frac{{18,000{\rm{ N}}}}{{\left( {2300{\rm{ kg + 2400 kg}}} \right)}}\\\\ = \frac{{18,000{\rm{ N}}}}{{\left( {4700{\rm{ kg}}} \right)}}\\\\ = 3.83{\rm{ m/}}{{\rm{s}}^2}\\\end{array}  

a=  

(2300kg+2400kg)

18,000N

​

=  

(4700kg)

18,000N

​

=3.83m/s  

2

​

(b)

Substitute 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s  

2

 for a and 2400 kg for m in the equation F = maF=ma .

\begin{array}{c}\\F = \left( {2400{\rm{ kg}}} \right)\left( {3.83{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 9120{\rm{ N}}\\\end{array}  

F=(2400kg)(3.83m/s  

2

)

=9120N

​

Ans: Part a

The maximum possible acceleration the truck can give the SUV is 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s  

2

 .

Part b

The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.

The maximum possible acceleration the truck can give the SUV is equal to 4 m/s².

The force of the SUV's bumper on the truck's bumper is 10000N

Acceleration of an object can be described as as the change in the velocity of an object w.r.t. time. The acceleration is a vector quantity, contains both magnitude and direction. Acceleration is the second derivative of position w.r.t. time and the first derivative of velocity w.r.t. time.

According to Newton's second law of motion, the force is equal to the product of acceleration and mass of an object.

F = ma

And, a =  F/m

Given, the mass of the ruck , m = 2000 Kg

The mass of the SUV, M = 2500 Kg

The total mass of the both = 2000 + 2500 = 4500 Kg

The maximum force on the trick , F = 18000 N

The maximum acceleration of the truck can give the SUV:

[tex]a_{max} = \frac{F_{max}}{m+M}[/tex]

a = 18000/4500

a = 4 m/s²

The force of the SUV's bumper on the truck's bumper will be:

[tex]F_{max} -f= ma_{max}[/tex]

[tex]f= 18000-2000\times 4[/tex]

[tex]f =10000N[/tex]

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Answer:

Explanation:

The formula for angular velocity is

[tex]\omega=\frac{\theta}{t}[/tex] where omega is the angular velocity, theta is the change in the angular rotation, and t is the time in seconds. First and foremost, we have the angular rotation in minutes and the time in seconds, so that's a problem we have to amend. Let's change the angular rotation to rotations per second:

[tex]281.133\frac{r}{min}*\frac{1min}{60s}=4.68555\frac{r}{s}[/tex]

Now we're ready to set up the problem:

[tex]4.68555=\frac{\theta}{3.686}[/tex] and we multiply both sides by 3.686 to get the rotations per seconds:

θ = 17.27 rotations

Answer:

a)  [tex]F_g=1.5*10^9Ibf[/tex]

b)  [tex]F_t=12490Ibf/ft^2[/tex]

     [tex]F_b=0[/tex]

Explanation:

From the question we are told that:

Height [tex]h=200ft[/tex]

Width [tex]w=1200ft[/tex]

a)

Generally the equation for Dam's Hydro static force is mathematically given by

[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]

Where

[tex]\rho=Density\ of\ water[/tex]

[tex]\rho=62.4Ibm/ft^3[/tex]

Therefore

[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]

[tex]F_g=1.5*10^9Ibf[/tex]

b)

Generally the equation for Dam's Force per unit area is mathematically given by

[tex]F=\rho*g*h[/tex]

For Top

[tex]F_t=\rho*g*h[/tex]

[tex]F_t=62.4*32.2*200[/tex]

[tex]F_t=12490Ibf/ft^2[/tex]

For bottom

[tex]Here \\H=0 zero[/tex]

Therefore

[tex]F_b=0[/tex]

The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].

The force per unit area near the top is 86.74 psi.

The force per unit area near the bottom is zero.

The hydrostatic force on the dam is the force exerted on the dam by the column of the water.

[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]

The force per unit area is the pressure exerted near the top of the dam.

[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]

where;

P is pressure in PSI

ρ is density of water in lb/gal

h is the vertical height in ft

[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]

The pressure near the bottom is zero, become the vertical height is zero.

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