Dirt is classified as an: a)element b) compound c)heterogeneous mixture d) solution

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

ryzh [129]

The work done by the applied force on the object is (2ab²i + 3b²j) J.

<h3>Magnitude of the force on the object</h3>

The magnitude of the force on the object is calculated as follows;

f = (2xyi + 3yj)

when;

x = a, and y = b

f = (2abi + 3bj)

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

The work done the applied force is the product of force and displacement of the object.

W = fΔs

where;

Δs is displacement of the object

Δx = a - a = 0

Δy = 0 - b = -b

Δs = √(Δx² + Δy²)

Δs = √(-b)²

Δs = b

W = (2abi + 3bj) x b

W = (2ab²i + 3b²j) J

Thus, the work done by the applied force on the object is (2ab²i + 3b²j) J.

The complete question is below;

An object moving in the xy-plane is subjected to the force f = (2xyi + 3yj), where x and y are in m. 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

6 0

2 years ago

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

leva [86]

To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

$MV=mv$

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

$V = \frac{mv}{M}$

Also we have that $m/M = 1/7 times$

On the other hand we have from law of conservation of energy that

$W_f = KE$

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

$F_f*S = \frac{1}{2}MV^2$

$\mu M*g*S = \frac{1}{2}MV^2$

$\mu g*S = \frac{1}{2}( \frac{mv}{M})^2$

$\mu = \frac{1}{2} (\frac{1}{7}v)^2$

$\mu = \frac{1}{98}v^2$

$\mu = \frac{1}{g(98)(5.1)}v^2$

Here we can apply the law of conservation of energy for light mass, then

$\mu mgs = \frac{1}{2} mv^2$

Replacing the value of $\mu$

$\frac{1}{g(98)(5.1)}v^2 mgs = \frac{1}{2}mv^2$

Deleting constants,

$s= \frac{(98*5.1)}{2}$

$s = 249.9m$

7 0

4 years ago

What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radiu

alexira [117]

The expression for the flux through a sphere is
flux = charge/Enot (i.e. epsillon not: a constant)

this shows that flux doesn't vary with radius but with the charge of the points

7 0

3 years ago

Calculate the decrease in gravitational potential energy of the ball as it moves down through the 0.90m the mass of the ball is

icang [17]

Answer:

G.P.E = 5.292 Joules

Explanation:

Given the following data;

Mass = 0.6 kg

Height = 0.9 m

To find the gravitational potential energy;

Gravitational potential energy (GPE): it is an energy possessed by an object or body due to its position above the earth.

Mathematically, potential energy is given by the formula;

$G.P.E = mgh$

Where,

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

But we know that acceleration due to gravity is equal to 9.8m/s²

G.P.E = 0.6*9.8*0.9

G.P.E = 5.292 Joules

8 0

3 years ago

Why might you want to use a single fixed pulley?

irakobra [83]

C to raise or lower lightweight objects

3 0

3 years ago