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zimovet [89]
3 years ago
7

Sketch the solid whose volume is given by the integral and evaluate the integral. $ \int_{0}^{{\color{red}6}}\int_{0}^{2\pi }\in

t_{{\color{red}4} r}^{{\color{red}24}}\,r\,dz\,d\theta \,dr

Physics
1 answer:
Serggg [28]3 years ago
8 0

Answer:

Explanation:

Attached is the evaluation

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One factor of x^3-3x^2-4x+12 is x+2. find the remaining factors
telo118 [61]
Dividing  x^3 - 3x^2 - 4x + 12 by x + 2 gives

x^2 -5x + 6

= (x - 3)(x - 2)

so remaining factors are    x - 3 and x - 2 answer
6 0
2 years ago
Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and
Anna11 [10]

Kepler’s third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496 × 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/( 86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The closest answer is 1.99 × 10^30

(it may vary a little with rounding – the difference is less than 1%)

5 0
3 years ago
Read 2 more answers
A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
Bas_tet [7]

Answer:

Explanation:

Total momentum of the system before the collision

.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left

If v be the velocity of the stuck pucks

momentum after the collision = 2 v

Applying conservation of momentum

2 v = -  .75

v =  - .375 m /s

Let after the collision v be the velocity of .5 kg puck

total momentum after the collision

.5 v + 1.5 x .231 = .5v +.3465

Applying conservation of momentum law

.5 v +.3465 = - .75

v = - 2.193 m/s

2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.

Kinetic energy before the collision

= 2.25 + 1.6875

=3.9375 J

kinetic energy after the collision

= .04 + 1.2 =1.24 J

So kinetic energy is not conserved . Hence collision is not elastic.

3 ) Change in the momentum of .5 kg

1.5 - (-1.0965 )

= 2.5965

Average force applied = change in momentum / time

= 2.5965 / 25 x 10⁻³

= 103.86 N

5 0
3 years ago
How can you degauss a magnet temporarily?
EleoNora [17]

Answer:

Demagnetization processes include heating past the Curie point, applying a strong magnetic field, applying alternating current, or hammering the metal.

Explanation:

7 0
2 years ago
A 20-liter container contains 2.0 moles of oxygen at a pressure of 92 kpa. the average kinetic energy of translation of oxygen m
Vikentia [17]

From ideal gas law, PV=nRT

where P is the pressure, V is the volume of the container, n is number of moles, R is the gas constant and T is the temperature.

Hence, T=\frac{PV}{nR}=\frac{92*10^{3}*2.0*10^{-3}}{2*3.314}

T= 110.65 k

Kinetic Energy = \frac{3}{2}KT=\frac{3}{2}  (1.38*10^{-23})(110.65)

K.E=  2.2*10^{-21}J

<h3>What is a kinetic energy? </h3>

The energy an object has as a result of motion is known as kinetic energy.

A force must be applied to an object in order to accelerate it. We must put in effort in order to apply a force. After the work is finished, energy is transferred to the item, which then moves at a new, constant speed. Kinetic energy is the type of energy that is transferred and is dependent on the mass and speed attained.

Kinetic energy can be converted into other types of energy and transported between objects. A flying squirrel may run into a chipmunk that is standing still, for instance. Some of the squirrel's initial kinetic energy may have been transferred to the chipmunk or changed into another kind of energy after the collision.

To know more about kinetic energy, visit:

brainly.com/question/22174271

#SPJ4

4 0
1 year ago
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