The Number above the elements symobls is the elements atomic number .

Complete this equation that represents the process of nuclear fission.

tekilochka [14]

Answer:

A: 146

B: 56

Explanation:

5 0

3 years ago

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Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration

stepan [7]

Answer:

The magnitude of the acceleration ae of the earth due to the gravitational pull of the moon is $\mathbf{3.3187\times10^{-5}}\frac{m}{s^{2}}$

Explanation:

By Newton's gravitational law, the magnitude of the gravitational force between two objects is:

$F=G\frac{Mm}{r^{2}}$ (1)

With G the gravitational constant, M the mass of earth, m the mass of the moon and r the distance between the moon and the earth, a quick search on physics books or internet websites give us the values:

$M=5.972\times10^{24}\,kg$

$m=7.34767309\times10^{22}\,kg$

$r=384400\,km$

$G=6.674\times10^{-11}\,\frac{N\,m^{3}}{kg^{2}}$

Using those values on (1)

$F=(6.674\times10^{-11})*\frac{(5.972\times10^{24})(7.34767309\times10^{22})}{(384400\times10^{3})^{2}}$

$F\approx1.98193\times10^{20}N$

Now, by Newton's second Law we can find the acceleration of earth ae due moon's pull:

$F=M*ae\Longrightarrow ae=\frac{F}{M}=\frac{1.98193\times10^{20}}{5.972\times10^{24}}\approx\mathbf{3.3187\times10^{-5}}$

6 0

4 years ago

The picture shows a loop of wire rotating in a magnetic field in a generator. Determine the direction of the current through the

BaLLatris [955]

I believe the answer is b

3 0

3 years ago

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May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

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

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$