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2 years ago

What are the rules for setting up an integral of rotation?

1 answer:

3 0

Setting up an integral of rotation is used as a method of of calculating the volume of a 3D object formed by a rotated area of a 2D space. Finding the volume is similar to finding the area, but there is one additional component of rotating the area around a line of symmetry.

<span>First the solid of revolution should be defined. The general function is y=f(x), on an interval [a,b].</span>

Then the curve is rotated about a given axis to get the surface of the solid of revolution. That is the integral of the function.

<span>It all depends of the function f(x), which must be known in order to calculate the integral.</span>

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What foxes need to eat to<br> stay alive

11Alexandr11 [23.1K]

Answer:

<h3>They eat meat and vegetation. </h3>

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2 years ago

Read 2 more answers

a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

AysviL [449]

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, $M=75\ kg$

Mass of the ball is, $m=4\ kg$

Initial velocity of the man is, $u_m=0\ m/s(rest)$

Initial velocity of the ball is, $u_b=0\ m/s(rest)$

Final velocity of the ball is, $v_b=3.50\ m/s$

Final velocity of the man is, $v_m=?\ m/s$

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of man and ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Now, initial momentum = final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

3 0

2 years ago

which has a greater kinetic energy, a bowling ball that has a mass of 5kg travelling at 6m/s, or a ship that has a mass of 12000

AVprozaik [17]

Answer:bowling ball has greater kinetic energy

Explanation:

Kinetic energy of bowling ball:

mass=m=5kg

Velocity=v=6m/s

Kinetic energy =ke

Ke=0.5 x m x v x v

Ke=0.5 x 5 x 6 x 6

Ke=90J

Kinetic energy of ship:

mass=m=120000kg

velocity=v=0.02m/s

Ke=0.5 x m x v x v

Ke=0.5 x 120000 x 0.02 x 0.02

Ke=24J

6 0

2 years ago

Compare these two collisions of a PE student with a wall.

Stolb23 [73]

1) The variable that is different in the two cases is $\Delta t$ , the duration of the collision

2) The change in momentum is the same in the two cases

3) The impulse is the same in the two cases

4) Case B will experience a greater force

Explanation:

The variable that is different in the two cases is $\Delta t$ , the duration of the collision.

In fact, in the first case the wall is padded: this means that the collision will be "softer" and therefore will last longer, so the duration of the collision, $\Delta t$ , will be larger.

In the second case instead, the wall is unpadded: this means that the collision is "harder" and so it will last less time, therefore the duration of the collision $\Delta t$ will be smaller.

The change in momentum in the two cases is the same.

In fact, the change in momentum is given by:

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

where:

m is the mass of the student

u is the initial velocity

v is the final velocity

In both cases, we have:

m = 75 kg

u = 8 m/s

v = 0 (they both comes to rest)

Therefore, the change in momentum is

$\Delta p = (75)(0-8)=-600 kg m/s$

The impulse in the two cases is the same.

In fact, impulse is defined as the product of force applied, F, and duration of the collision, $\Delta t$ :

$J=F \Delta t$

However, the force can be rewritten as product of mass (m) and acceleration (a), according to Newton's second law:

$F=ma$

So the impulse is

$J=ma\Delta t$

The acceleration can be rewritten as rate of change of velocity:

$a=\frac{\Delta v}{\Delta t}$

So the impulse becomes

$J=m\frac{\Delta v}{\Delta t}\Delta t = m\Delta v$

So, the impulse is equal to the change in momentum: and since in the two cases the change in momentum is the same, the impulse is the same as well.

The force in the collision is related to the impulse by

$J=F\Delta t$

where

J is the impulse

F is the force

$\Delta t$ is the duration of the collision

The equation can be rewritten as

$F=\frac{J}{\Delta t}$

In the two situations described in the problem (A and B), we already said that the impulse is the same (because the change in momentum is the same). However, in case A (padded wall) the time $\Delta t$ is longer, while in case B (unpadded wall) the time $\Delta t$ is shorter: since the force F is inversely proportional to the duration of the collision, this means that in case B the student will experience a greatest force compared to case A.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

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2 years ago

What three types of tempature marine climate have?

AnnyKZ [126]

The answer is marine west coast, humid subtropical and Mediterranean. <span />

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3 years ago