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3 years ago

What is the basis for rutherford's planetary model?

1 answer:

3 0

The basis for Rutherford's Planetary model, was the results he got from experiments.

He observed that most of the alpha particles he fired at a gold foil, passed through the foil, but only few were deflected back. So he concluded that most of the Atom would be empty space, with a positive entity at the center.

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When falling through the air, which of the following objects will hit the ground first?

Pavel [41]

Answer:

a penny

Explanation:

5 0

4 years ago

Read 2 more answers

Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th

8090 [49]

Answer:

$r_{cm} = 0.074 m$ from the position of the center of the Sun

Explanation:

As we know that mass of Sun and Jupiter is given as

$M_s = 1.98 \times 10^{30} kg$

$M_j = 1.89 \times 10^{27} kg$

distance between Sun and Jupiter is given as

$r = 7.78 \times 10^{11} m$

now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them

so we will have

$r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}$

$r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}$

$r_{cm} = 0.074 m$ from the position of the center of the Sun

3 0

4 years ago

When making fitness purchases what should you not do?

astraxan [27]

Of these options the one which best answers the question is 'Use whatever professionals claim to use.'
It is not that professional recommendations should not be consired or valued. The reason you should not just use whatever professionals claim to use is that their needs, as professionals, may be very different from your own. In order to know whether your own needs will be met you need to know what those needs are and whether the product meets those needs. All of the other options given would help you to figure this out, but the testimony of professionals would not.

7 0

3 years ago

Read 2 more answers

Energy transferred as heat always moves from an object

nata0808 [166]

Well, if we are being technical, yes and the only reason I say yes is because the sun is a hot gas rock. And it gives off UV rays which is heat on earth. So, yes. It does

3 0

3 years ago

A 35 kg trunk is pushed 4.8 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic fric

netineya [11]

Answer:

We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

$F_x = F\cos(30^\circ)\\F_y = F\sin(30^\circ)$

Only the x-component of the applied horizontal force does work on the trunk.

But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

$0 = F\cos(30^\circ) - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0.86F = 35(9.8)(0.5) + 35(9.8)(0.86)(0.19)\\F =264.5N$

$F_x = 264.5\cos(30^\circ) = 227.5N\\F_y = 264.5\sin(30^\circ) = 132.25N$

$W_{F_x} = F_xd = 227.5\times 4.8 = 1092J$

The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

$W_g = -mg\sin(30^\circ)d$

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

$W_g = -823J$

The energy dissipated by the frictional force can be found as follows:

$W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J$

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.

In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.

$0 = F - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0 = F - 35(9.8)(0.5) - 35(9.8)(0.86)(0.19)\\0 = F - 171.5 - 56\\F = 227.5N$

$W_F = Fd = 227.5\times 4.8 = 1092J$

$W_g = -mg\sin(30^\circ)d = -35(9.8)(0.5)(4.8) = -823J$

$W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J$

Explanation:

As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.

8 0

4 years ago