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Anna11 [10]
3 years ago
14

The position of an electron is given by , with t in seconds and in meters. At t = 3.99 s, what are (a) the x-component, (b) the

y-component, (c) the magnitude, and (d) the angle relative to the positive direction of the x axis, of the electron's velocity (give the angle in the range (-180°, 180°])?
Physics
1 answer:
egoroff_w [7]3 years ago
7 0

Answer:

A. Vx = 3.63 m/s

B. Vy = -45.73 m/s

C. |V| = 45.87 m/s

D. θ = -85.46°

Explanation:

Given that position, r, is given as:

r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk

Velocity is the derivative of position, r:

V = dr/dt = 3.63 - 11.46t^j

A. x component of velocity, Vx = 3.63 m/s

B. y component of velocity, Vy = -11.46t

t = 3.99 secs,

Vy = - 11.46 * 3.99 = -45.73 m/s

C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]

|V| = √(2091.2329 + 13.1769)

|V| = √(2104.4098)

|V| = 45.87 m/s

D. Angle of the velocity relative to the x axis, θ is given as:

tanθ = Vy/Vx

tanθ = -45.73/3.63

tanθ = -12.6

θ = -85.46°

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Aloiza [94]

Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

F=ma

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

8 0
2 years ago
Read 2 more answers
The specific heat of fat is roughly 1,700 J/(kgoC) whereas that for water is around 4200 J/(kgoC). A 0.63 kg solution of lipids
BartSMP [9]

Answer:

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

Explanation:

Total heat content of the fat = heat content of water +heat content of the lipids

Let it be Q

the Q= (mcΔT)_lipids + (mcΔT)_water

total mass of fat  M= 0.63 Kg

Q= heat supplied = 100 W in 5 minutes

ΔT= 20°C

c_lipid= 1700J/(kgoC)

c_water= 4200J/(kgoC)

then,

100\times5\times60= m(1700)20+(0.63-m)(4200)20

solving the above equation we get

m= 0.46 kg

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

3 0
3 years ago
In a closed system, as kinetic energy increases, what happens to potential energy?
kupik [55]
<h2>Because kinetic energy is proportional to the velocity squared, increases in velocity will have an exponentially greater effect on translational kinetic energy. Doubling the mass of an object will only double its kinetic energy, but doubling the velocity of the object will quadruple its velocity.</h2>
8 0
2 years ago
Heat is added to a 2kg piece of ice at a rate of 793kW. How long will it take for ice to melt if it was initially 0?
Ede4ka [16]

Answer:

0.84 s

Explanation:

Step 1

Given information:

Mass of the ice (m) = 2.0 kg

Heat transfer rate (Q/T) = 793.0 kW

Latent heat of fusion of ice (Lf) = 334 kJ/kg

\frac{Q}{T}  =  \frac{mLf}{T}

Substituting the corresponding values we have:

793.0 kW=  \frac{2.0kg(334 kJ/kg)}{T} \\  T  =  \frac{2.0kg(334 kJ/kg)}{793.0kW}  =  \frac{668kJ}{793kW}   \\  = 0.84s

8 0
2 years ago
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notsponge [240]

Answer:

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Explanation:

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Standard form  

Ey=Eo cos[k x-ωt]  N/C

By comparing the given equation with the standard wave equation

Eo = 375 N/C

ω  = 2.20 × 10¹⁴ rad/s

We know that ω = 2 π f

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f=\dfrac{2.2\times 10^{14}}{2\pi }\ Hz

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V = f λ

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λ =8.57 μ m

8 0
3 years ago
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