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3 years ago

What is the energy conversion in a hair dryer?

Physics

1 answer:

wel3 years ago

4 0

Answer:

C.Electrical energy to heat and kinetic energy

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Describe how Bohr’s model used the work of Maxwell.

pochemuha

Answer:

The bohr's model is the primitive model for the hydrogen atom, comparatively to the atom of valence shell. And it is derived from the hydrogen atom of the first approximation by using the quantum mechanics.

Basically, the model state that the electron revolved around in circular orbit in atom around the central nucleus. And it can be fixed in the circular orbit at the set of discrete distance at the nucleus.

3 0

2 years ago

A 1250 kg car is stopped at a traffic light. A 3550 kg truck moving at 8.33 m/s hits the car from behind. If bumpers lock, how f

Radda [10]

Answer:

the two vehicles will be moving at a speed of 6.16 m/s

Explanation:

This is a case of completely inelastic collision, therefore, the conservation of momentum can be written as:

$m_1\,v_1+m_2\,v_2=(m_1+m_2)\,v_f$

which given the information provided results into:

$m_1\,v_1+m_2\,v_2=(m_1+m_2)\,v_f\\(1250)\,(0)+(3550)\,(8.33)=(1250+3550)\,v_f\\29571.5=4800\,v_f\\v_f=6.16\,\,m/s$

7 0

3 years ago

A person is standing on an elevator initially at rest at the first floor of a high building. The elevator then begins to ascend

GREYUIT [131]

Answer:

The found acceleration in terms of h and t is:

$a=\frac{h}{5(t_1)^2}$

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

Constant acceleration, starts from rest.

Distance = $y = \frac{1}{2}a(t_1)^2$

Velocity = $v_1=at_1$

<h3>Stage 2</h3>

Constant velocity where

Velocity = $v_o=v_1=at_1$

Distance =

<h3> $y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\$ </h3><h3 /><h3>Stage 3</h3>

Constant deceleration where

Velocity = $v_0=v_1=at_1$

Distance =

$y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2$