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gregori [183]
2 years ago
11

find the work done be the force field f in moving an object along the curve pictured in the graph. do this by computing the work

on each piece-wise smooth portion of c.
Physics
1 answer:
Anton [14]2 years ago
7 0

Find the amount of labor necessary to move an object along the specified orientated curve given the force field F. y = 3x2 on the parabola from (0, 0) to F = (y, x) (4, 48).

<h3>How can you determine the work a force field performs along a curve?</h3>

A variable force's infinitesimal work can be defined in terms of the force's components and the displacement along the path,

<h3>How do you assess the force field F's work?</h3>

(c) Next, determine the work that the force field F performed on the object to move it through the field by moving it along the vector d. W = F d is the formula for work.

To know more about force field visit:-

brainly.com/question/13488023

#SPJ4

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All but one of the following is an important component of soil.
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Air is the answer i do believe
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4 years ago
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A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of
Sunny_sXe [5.5K]

Answer:

5.8\; {\rm kg\cdot m \cdot s^{-1}}.

Explanation:

If the mass of an object is m and the velocity of that object is v, the linear momentum of that object would be m\, v.

Assume that the initial velocity of the mass is positive (6.0\; {\rm m\cdot s^{-1}}.) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be (-4.0\; {\rm m\cdot s^{-1}}).

Thus, the change in the velocity of the mass would be:

\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}.

The change in the linear momentum of the mass would be:

\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\  =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}.

Thus, the magnitude of the change of the linear momentum would be 5.8\; {\rm kg \cdot m \cdot s^{-1}}.

7 0
3 years ago
The product of an object's mass and velocity is it's
Leviafan [203]

Answer:

C) Momentum

Explanation:

Refers to an objects mass in motion.

5 0
2 years ago
what is the energy of a single photon of ultraviolet light with a wavelength of 30.0 nm? 1 nm=10^-9 m
Arada [10]
From the planks equation
E=hv
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V= 3×10^8/30×10^-9
=1×10^16
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4 0
3 years ago
A large crate is at rest on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.500. A f
Nookie1986 [14]

Answer:

93.4 kg

Explanation:

Draw a free body diagram.  There are three four forces:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

Applied force F pulling up and to the right, 30.0° above the horizontal.

Sum of forces in the y direction:

∑F = ma

N + F sin 30.0° − mg = 0

N = mg − ½ F

Sum of forces in the x direction:

∑F = ma

F cos 30.0° − Nμ = 0

½√3 F = Nμ

Substitute:

½√3 F = (mg − ½ F) μ

½√3 F / μ = mg − ½ F

½√3 F / μ + ½ F = mg

½F (√3 / μ + 1) = mg

m = F (√3 / μ + 1) / (2g)

Plug in values:

m = 410 N (√3 / 0.500 + 1) / (2 × 9.8 m/s²)

m = 93.4 kg

4 0
3 years ago
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