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1 month ago

Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+βF(x)=αx3+β.

Physics

1 answer:

Sav [3.1K]1 month ago

8 0

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To calculate the kinetic energy variation, we can utilize the work-energy theorem.

W = ΔK

∫ F .dx = K - K₀

If the object starts from rest, then K₀ = 0.

So, ∫ F dx cos θ = K.

As the force and displacement directions align, the angle is zero, and hence the cosine is 1.

Now we can substitute and perform integration:

α ∫ x³ dx + β ∫ dx = K.

Thus, α x⁴ / 4 + β x = K.

Next, we evaluate from the limits F = 0 to F:

α (x⁴ / 4 - 0) + β (x - 0) = K.

Consequently, K = αX⁴ / 4 + β x.

This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.

To finalize the computation, we need to ascertain the displacement.

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Wire A has the same length and twice the radius of wire B. Both wires are made of the same material and carry the same current.

serg [3582]

Answer:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

Consequently, we find that:

$V_A = \frac{1}{4} V_B$

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

$L_A = L_B =L$ both wires share the same length

both wires carry an identical current $I_A = I_B =I$

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

$n_A = n_B =n$

We also know that $r_A = 2 r_B$ where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

$A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B$

$A_B = \pi r^2_B$

Thus, we conclude that

$A_A = 4 A_B$

Now we are aware that the drift velocity of an electron in a wire can be described by:

$v_d = \frac{I}{neA}$

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

So we observe that:

$V_A = \frac{1}{4} V_B$

Thus, the most fitting answer is:

a. vA = vB/4

6 0

22 days ago

A circuit contains a 6.0-v battery, a 4.0-w resistor, a 0.60-µf capacitor, an ammeter, and a switch all in series. what will be

Softa [3030]

<span>You are presented with a circuit that includes a 6.0-v battery, a 4.0-ohm resistor, a 0.60 microfarad capacitor, an ammeter, and a switch all connected in series. Your task is to determine the current reading once the switch is closed. Ohm's law should be used, which states V = IR where V signifies voltage, I indicates current, and R represents resistance.</span>

V = IR
I = V/R
I = 6 volts / 4 ohms
I = 1.5A

Upon closing the switch, the cathode side plate starts accumulating electrons if it was previously empty. As this process continues, the current diminishes. Eventually, when the capacitor reaches its maximum electron retention, the current will cease. An increased capacitance means a greater capacity for electron storage.

4 0

27 days ago

Read 2 more answers

If the diameter of a hole is 4.500 mm, what should be the diameter of a rivet at 23.0 ∘C, if its diameter is to equal that of th

kicyunya [3294]

Answer:

Explanation:

at 23 degrees Celsius, the diameter measures 4.511 mm

GIVEN DATA:

diameter of hole = 4.500 mm

T_1 = 23.0 degrees Celsius

T_2 = - 78.0 degrees Celsius

the expansion coefficient of aluminum is 2.4*10^{-5} (degrees Celsius)^{-1}

the diameter at 23 degrees Celsius is stated as

$d = d_o (1+ \alpha \delta T)$

$= 4.5 (1+2.4*10^{-5} *(23-(-78)))$

= 4.511 mm

the diameter of the rivet after temperature change is given as

$d= d_o + \Delta d$

$= d_o(1+ \alpha \delta T)$

$= 0.4500 *(1+2.4*10^{-5} *(23-(-78)))$

= 0.4511 cm

8 0

22 days ago

When 999mm is added to 100m ______ is the result

Keith_Richards [3271]

Answer:

The outcome of adding 999mm to 100m is 101m.

Explanation:

That's my belief.

6 0

1 month ago

Read 2 more answers

A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p

Keith_Richards [3271]

Response:

$Q = 8,345 * v$

Clarification:

We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:

$Q = 8,345 * v$

Where v signifies the number of gallons emptied from the tub.

Have a great day! Let me know if there's anything else I can assist with.

8 0

1 month ago