A liquid thermometer can be used to test for fevers. When body temperature increases,

Electric charge is distributed over the disk x2 + y2 ≤ 4 so that the charge density at (x, y) is rho(x, y) = 4x + 4y + 4x2 + 4y2

maw [93]

Answer:

Q=185.84C

Explanation:

We have to take into account the integral

$Q=\int \rho dV$

In this case we have a superficial density in coordinate system.

Hence, we have for R: x2 + y2 ≤ 4

$Q=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\rho dydx$

but, for symmetry:

$Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C$

HOPE THIS HELPS!!

8 0

2 years ago

A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.

Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mb*g*H = 0.5*mb*vb^2

vb = √2*g*H

vb = ( 2*9.81*15 ) ^0.5

vb = 17.15517 m/s

- The angular momentum of system before collision is:

M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

M1 = M2

vo = 247.03448 / (5.10*3)

vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mo*g*h = 0.5*mo*vo^2

h = vo^2 / 2*g

h = 16.14604^2 / 2*(9.81)

h = 13.3 m

3 0

3 years ago

A certain moving electron has a kinetic energy of 0.991 × 10−19 J. Calculate the speed necessary for the electron to have this e

alisha [4.7K]

Answer: The speed necessary for the electron to have this energy is 466462 m/s

Explanation:

Kinetic energy is the energy posessed by an object by virtue of its motion.

$K.E=\frac{1mv^2}{2}$

K.E= kinetic energy = $0.991\times 10^{-19}J$

m= mass of an electron = $9.109\times 10^{-31}kg$

v= velocity of object = ?

Putting in the values in the equation:

$0.991\times 10^{-19}J=\frac{1\times 9.109\times 10^{-31}kg\times v^2}{2}$

$v=466462m/s$

The speed necessary for the electron to have this energy is 466462 m/s

4 0

2 years ago

Read 2 more answers

Could dark matter be a higher dimensional object? Why?

REY [17]

Well dark matter is , in a layman terms just a concept used to identify missing mass in the universe. (extremely simplified and probably wrong but look that up). It could be but then if it is then we would still be able to percive the matter in some form. All we know from instruments is that they have mass and absorb light. We would need more info before any conclusions but it is 100% a possibility. But remember that when someone says "3-d" for example they are talking about a certain space.. technically matter exists on all planes (at least all the planes we know of in our local universe.) The laws of the cosmos are not yet proven to be constant and even if they are what of outside it?

When you take high school physics ask your teacher.

3 0

2 years ago

If an object is accelerating to the right , then net force on the object must be directed towards the right ! True Or False ?

vichka [17]

Not so fast.

I think you're using 'accelerating' to mean 'speeding up', but you really need
to be more careful with it. "Acceleration" means ANY change in speed OR
direction.

If an object's speed to the left is decreasing, or its speed to the right is
increasing, then the net force on the object must be directed towards
the right.

If an object is moving with constant speed in a circular path, then it's
constantly accelerating, because its direction is constantly changing.
The force on it is always directed towards the center of the circle, so
there's one point on the path where the force is directed straight to the right.

8 0

2 years ago

Read 2 more answers