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AVprozaik [17]
3 years ago
8

Changing which factor would NOT have an influence on the kinetic energy of a moving van loaded with 100 kg bags mulch of with a

total mass of 1,500 kg. The vehicle is traveling across an open parking lot at a speed of 5 m/s.
Question 8 options:


A: direction the moving van is going across the parking lot.


B: increasing the rate of speed without altering the mass of the vehicle or its contents.


C: emptying the moving van one bag at a time at a constant rate.


D: adding more contents (increasing the overall mass) to the van while the van is in motion traveling 5 m
Physics
2 answers:
Lemur [1.5K]3 years ago
8 0

Answer:

A

Explanation:

i took the test

xenn [34]3 years ago
3 0

Answer:

A. The direction

Explanation:

I did the test lol

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It is less than the acceleration of the backpack because abs has a greater mass.

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A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

F_g = F_c

\frac{GmM}{r^2} = \frac{mv^2}{r}

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0
2 years ago
High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of th
murzikaleks [220]

Answer:

Lifetime = 4.928 x 10^-32 s  

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s  

4 0
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