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1 year 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]1 year 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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Answer:

Option D is the correct answer.

Explanation:

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Centripetal acceleration

$a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2$

We have radius = 7.112 cm = 0.07112 m

Frequency, f = 1975 rpm = 32.92 rps

Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s

Substituting in centripetal acceleration equation,

$a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2$

Option D is the correct answer.

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3 years ago

A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig

trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

$V=\frac{MU+mu}{M+m}$

Substituting values:

$V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s$

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

$V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134$

The coefficient of kinetic friction between the sled and the snow is 0.0134

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2 years ago

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Answer:

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work:

PE= 65kg * 9.8 kg *8,800 m

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PtichkaEL [24]

Answer:

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