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3 years ago

How was Bohr's atomic model different from those of previous scientists?

1 answer:

4 0

Answer:Bohr placed the electrons in distinct energy levels. Rutherford described the atom as consisting of a tiny positive mass surrounded by a cloud of negative electrons. Bohr thought that electrons orbited the nucleus in quantised orbits.

Explanation: also rutherfords was just a hypothesis while Bhor took the time to make his an experiment

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What increase in temperature is needed to increase the length of an aluminum meterstick by 1.0 mm?

Fudgin [204]

Answer:

42k

Explanation:

4 0

2 years ago

You're driving your new sports car at 85 mph over the top of a hill that has a radius of curvature of 525 m.

Bumek [7]

Explanation:

It is given that,

Speed of the sports car, v = 85 mph = 37.99 m/s

The radius of curvature, r = 525 m

Let $W_N$ is the normal weight and $W_A$ is the apparent weight of the person. Its apparent weight is given by :

$W_A=mg-\dfrac{mv^2}{r}$

So, $\dfrac{W_A}{W_N}=\dfrac{mg-\dfrac{mv^2}{r}}{mg}$

$\dfrac{W_A}{W_N}=\dfrac{g-\dfrac{v^2}{r}}{g}$

$\dfrac{W_A}{W_N}=\dfrac{9.8-\dfrac{(37.99)^2}{525}}{9.8}$

$\dfrac{W_A}{W_N}=0.719$

$\dfrac{W_A}{W_N}=71.9\%$

Hence, this is the required solution.

5 0

3 years ago

Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missil

tatyana61 [14]

Answer:

The time taken by missile's clock is $4.6\times 10^{6} s$

Solution:

As per the question:

Speed of the missile, $v_{m = 6.5\times 10^{3}} m/s$

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

$T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}$ (1)

Using binomial theorem in the above eqn:

We know that:

$(1 + x)^{a} = 1 + ax$

Thus eqn (1) becomes:

$1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}$

$T = \frac{2c^{2}}{v_{m}^{2}}$

Now, putting appropriate values in the above eqn:

$T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}$

$T = 4.6\times 10^{6} s$

4 0

3 years ago

A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e

mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

∑ M = 0

mass of the log will be concentrated at the center

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

$m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)$

$200 \times 0.1 L = 53.5 \times (x - 0.6 L)$

$0.374 L =x - 0.6 L$

x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0

3 years ago

Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their

Vera_Pavlovna [14]

Answer:

A = 2.36m/s

B = 3.71m/s²

C = 29.61m/s2

Explanation:

First, we convert the diameter of the ride from ft to m

10ft = 3m

Speed of the rider is the

v = circumference of the circle divided by time of rotation

v = [2π(D/2)]/T

v = [2π(3/2)]/4

v = 3π/4

v = 2.36m/s

Radial acceleration can also be found as a = v²/r

Where v = speed of the rider

r = radius of the ride

a = 2.36²/1.5

a = 3.71m/s²

If the time of revolution is halved, then radial acceleration is

A = 4π²R/T²

A = (4 * π² * 3)/2²

A = 118.44/4

A = 29.61m/s²

7 0

3 years ago