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2 years ago

An object moving in the xy-plane is subjected to the force f⃗ =(2xyı^ 3yȷ^)n, where x and y are in m

Physics

1 answer:

Svetllana [295]2 years ago

5 0

The magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

<h3>Work done by the force experienced by the object</h3>

The magnitude of the work done by force experience by the object is calculated as follows;

W = f.d

where;

F is the applied force (2xyi + 3yj), where x and y are in meters
d is the displacement of the object = (a, b)

The work done by the force is determined from the dot product of the force and the displacement of the object.

F = (2xyi + 3yj).(a + b)

W = (2abi + 3bj).(ai + bj)

W = (2a²b + 3b²)J

Thus, the magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

The complete question is below:

The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis.

How much work does the force do?

Learn more about work done here: brainly.com/question/8119756

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A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$

$\displaystyle Y_r=15+14\times3.69$

$\displaystyle \boxed{Y_r=66.62\ m}$

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2 years ago

A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.6

Dafna1 [17]

Answer:

45.89m/s²

Explanation:

Given

Distance S = 305m

Time t = 3.64s

To get the acceleration during this run, we will apply the equation of motion:

S = ut+1/2at²

Substitute the given parameters into the formula and calculate the value of a

305 = 0+1/2 a(3.64)²

304 = 1/2(13.2496)a

304 = 6.6248a

a = 304/6.6248

a = 45.89m/s²

Hence the average acceleration during this run is 45.89m/s²

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What roles do the musk ox play in the tundra ecosystem? (Select all that apply.)

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Answer:

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A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before

nadya68 [22]

Answer:

Explanation:

Given

mass of steel ball $m=0.2\ kg$

initial speed of ball $u=10\ m/s$

Final speed of ball $v=-10\ m/s$ (in upward direction)

Impulse imparted is given by change in the momentum of object

therefore impulse J is given by

$J=\Delta P$

$\Delta P=m(v-u)$

$\Delta P=0.2(-10-10)$

$\Delta =-4\ N-s$

so magnitude of Impulse =4 N-s

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2 years ago

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What happens to the gravitational attraction between 2 objects

amm1812

Answer:

It decreases.

Explanation:

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