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Allushta [10]
2 years ago
12

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:
Elan Coil [88]2 years ago
8 0

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To find the variation of kinetic energy, let's use the work energy theorem

            W = ΔK

           ∫ F .dx = K -K₀

If the body starts from rest K₀ = 0

           ∫ F dx cos θ = K

Since the force and displacement are in the same direction, the angle is zero, so the cosine is 1

we substitute  and integrate

          α ∫ x³ dx + β ∫ dx = K

          α x⁴ / 4 + β x / 1 = K

we evaluate from the lower limit F = 0 to the upper limit F

         α (x⁴ / 4 -0) + β (x -0) = K

        K = αX⁴ / 4 + β x

        K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

in order to finish the calculation we must know the displacement

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zhannawk [14.2K]

The velocity of Ned as measured by Pam is the interpretation of v.

Answer: Option D

<u>Explanation:</u>

According to question, we know that this is an issue depending on the logical and translation of the factors. From the measured information taken what is gathered by the two people is communicated and we have given as:

The Ned reference framework : (x, t)  

The Pam reference framework :  \left(x^{\prime}, t^{\prime}\right)

From the reference framework, we realize that ν is the speed of Pam (the other reference framework) as estimation by Ned.  

At that point, v^{\prime} is the speed of Ned (from the other arrangement of the reference) as estimation by Pam.

3 0
2 years ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
2 years ago
. Imagine that you are standing at the center of a giant bowl of gelatin. What type of wave will you make across the top of the
vichka [17]
Transverse wave as the wave is going up and down no compressions
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2 years ago
A test car and its driver, with a combined mass of 600 kg, are moving along a straight,horizontal track when a malfunction cause
ANEK [815]

Answer:

The two of the following measurements, when taken together, would allow engineers to find the total mechanical energy dissipated during the skid

B. The contact area of each tire with the track.

C. The co-efficent of static friction between the tires and the track.

D. The co-efficent of static friction between the tires and the track.

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4 0
1 year ago
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light w
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Answer:

Explanation:

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= 63.63 mph .

( b )

yes Car A was crossing speed limit by a difference of

63.63 - 35 = 28.63 mph.

7 0
2 years ago
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