5 kg dog accelerates at a rate of 3 m/s2. What is the net force acting on the dog?

This can be explained as in Rutherford's model of atom the electrons orbits the nucleus which means that they will travel around the nucleus with some velocity and hence radiate electromagnetic waves which results in the loss of energy due to which the electron keeps coming closer and eventually falls into the nucleus.

But Bohr came up with a better explanation as according to the Bohr's atomic model, electrons stay fixed in orbit with certain energy in different shells around the nucleus and can only jump from an energy level to another if that specific amount of energy is supplied to it.

This model is based on the quantization of energy thus giving an explanation why electrons do not fall into the nucleus of an atom.

6 0

3 years ago

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Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra

vredina [299]

Answer:

At low pressure- $\mu_{k}=0.02315$

At high pressure- $\mu_{k}=0.00445$

Explanation:

Initial speed, $V_{i}=3.3 m/s$

Final speed, $V_{f}=3.3/2= 1.65 m/s$

Net horizontal force due to rolling friction $F_{net}=\mu_{k}$ mg where m is mass, g is acceleration due to gravity, $\mu_{k}$ is coefficient of rolling friction

From kinematic relation, $V_{f}^{2}- V_{i}^{2}=2ad$

For each tire,

$V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd$

Making $\mu_{k}$ the subject

$\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}$

Under low pressure of 40 Psi, d=18 m

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315$

Therefore, $\mu_{k}=0.02315$

At a pressure of 105 Psi, d=93.7

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445$

Therefore, $\mu_{k}=0.00445$

4 0

3 years ago

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 20 cm apart. The sound

Sophie [7]

Answer:

a. Wavelength = λ = 20 cm

b. Next distance of maximum intensity will be 40 cm

Explanation:

a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.

Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.

This we get the equation:

(nλ + λ/2) - nλ = 30-20

λ/2 = 10

λ = 20 cm

b. at what distance, sound intensity will be maximum again.

For next point calculation for maximum sound intensity, the path difference must be increased (n+1) λ. The distance must increase by λ/2 from the point of zero intensity.

= 30 + λ/2

= 30 + 20/2

=30+10

=40 cm

7 0

3 years ago