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Allushta [10]
3 years ago
10

When did ernest rutherford make his discovery

Physics
1 answer:
oee [108]3 years ago
5 0

Answer:

1911

Explanation:

"In 1911, he was the first to discover that atoms have a small charged nucleus surrounded by largely empty space, and are circled by tiny electrons, which became known as the Rutherford model (or planetary model) of the atom."

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When warm air rises, cold air will _____.
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sink and fill in space..........

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Explain what relative weight is and why it is important to the calculation of atomic mass.
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Relative weight- weight that is relevant, and it is important to calculate atomic mass because that's the answer
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a bus is moving with the velociity of 36 km/hr . after seeing a boy at 20 m ahead on the road, the driver applies the brake and
Klio2033 [76]

Answer:

Assumption: the acceleration of this bus is constant while the brake was applied.

Acceleration of this bus: approximately \left(-6.0\; \rm m \cdot s^{-2}\right).

It took the bus approximately 1.7\;\rm s to come to a stop.

Explanation:

Quantities:

  • Displacement of the bus: x = 10\; \rm m.
  • Initial velocity of the bus: \displaystyle u = 36\; \rm km \cdot hr^{-1} = 36\; \rm km \cdot hr^{-1}\times \frac{1\; \rm m \cdot s^{-1}}{3.6\; \rm km\cdot hr^{-1}} = 10\; \rm m \cdot s^{-1}.
  • Final velocity of the bus: v = 0\; \rm m\cdot s^{-1} because the bus has come to a stop.
  • Acceleration, a: unknown, but assumed to be a constant.
  • Time taken, t: unknown.

Consider the following SUVAT equation:

\displaystyle x = \frac{1}{2}\, \left(a\, t^2\right) + u\, t.

On the other hand, assume that the acceleration of this bus is indeed constant. Given the initial and final velocity, the time it took for the bus to stop would be inversely proportional to the acceleration of this bus. That is:

\displaystyle t = \frac{v - u}{a}.

Therefore, replace the quantity t with the expression \displaystyle \left(\frac{v - u}{a}\right) in that SUVAT equation:

\displaystyle x = \frac{1}{2}\, \left(a\, \left(\frac{v -u}{a}\right)^2\right) + u\, \left(\frac{v - u}{a}\right).

Simplify this equation:

\begin{aligned}x &= \frac{1}{2}\, \left(a\, {\left(\frac{v -u}{a}\right)}^2\right) + u\, \left(\frac{v - u}{a}\right) \\ &= \frac{1}{2}\left(\frac{{(v - u)}^2}{a}\right) + \frac{u\, (v - u)}{a} =\frac{1}{a}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)\end{aligned}.

Therefore, \displaystyle a= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right).

In this question, the value of x, u, and v are already known:

  • x = 10\; \rm m.
  • \displaystyle u =10\; \rm m \cdot s^{-1}.
  • v = 0\; \rm m\cdot s^{-1}.

Substitute these quantities into this equation to find the value of a:

\begin{aligned} a &= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right) \\ &= \frac{1}{10\; \rm m}\times \left(\frac{{\left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)}^2}{2} + \left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)\times 10\; \rm m \cdot s^{-1}\right)\\ &\approx -6.0\; \rm m \cdot s^{-2}\end{aligned}.

(The value of acceleration a is less than zero because the velocity of the bus was getting smaller.)

Substitute a \approx -6.0\; \rm m \cdot s^{-2} (alongside u = 10\; \rm m \cdot s^{-1} and v = 0\; \rm m \cdot s^{-1}) to estimate the time required for the bus to come to a stop:

\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}}{-6.0\; \rm m \cdot s^{-2}} \approx 1.7\; \rm s\end{aligned}.

8 0
3 years ago
Plz tell me the answer to this problem!!
elena-s [515]

Answer:

B

Explanation:

They are in same group

7 0
3 years ago
A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
Hoochie [10]

Answer:

The ball moves from lowest to highest point:

W = M g h = 3 * 9.8 * 4 = 118 J

This is work done "against" gravity so work done by gravity is -118 J

The tension of the string does no work because the tension does not

move thru any distance   W = T * x = 0 because the length of the string is fixed.

5 0
3 years ago
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