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aleksklad [387]
2 years ago
11

A conservative force on a particle moving along the x axis is given by F = (3x^2-2x)i. Which of the following is a potential tha

t is associated with this force?
A. (6x-2)i
B. (-6x+2)i
C. x^3-x^2+3
D. -x^3+x^2+3
E. No answer given above is correct.
Physics
1 answer:
ANTONII [103]2 years ago
5 0

pck u

report me if you can

tex]\purple{\rule{45pt}{7pt}}\blue{\rule{45pt}{999999pt}}

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A 0.75 kg rock is projected from the edge of the top of a building with an initial velocity of 11.9 m/s at an angle 59◦ above th
Vesna [10]

Answer:

12.76 m.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 11.9 m/s

Angle of projection (θ) = 59°

Acceleration due to gravity (g) = 9.8 m/s²

Horizontal distance (i.e Range) (R) =?

The horizontal distance other wise known as range from the base of the building to which the rock will strike the ground can be obtained as follow:

R = u²Sine2θ / g

R = 11.9² × Sine (2×59) / 9.8

R = 141.61 × Sine 118 / 9.8

R = 12.76 m

Therefore, the horizontal distance from the base of the building to which the rock will strike the ground is 12.76 m

4 0
3 years ago
What part of a plant cell traps sunlight?_____________________
elixir [45]
It is the <span>Chloroplast in the cells.</span>
4 0
3 years ago
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A car parked on level pavement exerts a force of 10,000 newtons on the ground. What force does the pavement exert back on the ca
natka813 [3]

Answer:

Normal force of 10,000N

Explanation:

From the question, the weight the car exerts on the pavement is 10,000N.

The pavement exerts upward and perpendicular contact force called normal force on the car to support its weight. Also, the normal force is equal and opposite to the weigh of the car.

Hence the pavement exerts normal force of 10,000N back on the car to prevent it from passing through it.

7 0
3 years ago
Particles q1, 92, and q3 are in a straight line.
dangina [55]

Answer: 0

Explanation: Acellus

6 0
3 years ago
In a Broadway performance, an 83.0-kg actor swings from a R = 3.90-m-long cable that is horizontal when he starts. At the bottom
Veseljchak [2.6K]

Answer:

h = 2.821\,m

Explanation:

The speed of the actor before the collision is found by means of the Principle of Energy Conservation:

(83\,kg)\cdot(9.807\,\frac{m}{s})\cdot (3.90\,m) = \frac{1}{2}\cdot (83\,kg)\cdot v^{2}

v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (3.90\,m)}

v \approx 8.746\,\frac{m}{s}

The speed after the inelastic collision is obtained by using the Principle of Momentum Conservation:

(83\,kg)\cdot (8.746\,\frac{m}{s} )+(55\,kg)\cdot (0\,\frac{m}{s} ) = (83\,kg + 55\,kg)\cdot v

v = 5.260\,\frac{m}{s}

Lastly, the maximum height is determined by using the Principle of Energy Conservation again:

\frac{1}{2}\cdot (138\,kg)\cdot (5.260\,\frac{m}{s} )^{2} = (138\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot h

h = \frac{(5.260\,\frac{m}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }

h = 2.821\,m

8 0
3 years ago
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