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What electromagnetic radiations can the human bodies senses detect??

BlackZzzverrR [31]

Answer:

Human senses include the ability to detect electromagnetic waves in the 3800-7600 angstrom range, air waves of 15 to 20,000 beats per second, air-borne and liquid-borne molecules of proper size, quantity and configuration, and to generate nerve impulses triggered by objects impinging on body surfaces with enough force

4 0

3 years ago

A 60 g ball is dropped from rest from a height of 2.4 m. It bounces off the floor and rebounds to a maximum height of 1.9 m. If

kap26 [50]

Answer:

The force is 1.34 newtons and its direction is upward.

Explanation:

Choosing positive direction pointing towards the floor in this collision we're going to use the momentum-impulse theorem that states:

$J=\Delta p$ (1)

with $\Delta p=p_f-p_i$ the change in the momentum and J the impulse, with pi the initial momentum that is the momentum just before the collision and pf the final momentum th is the momentum just after the collision. The impulse J is also defined as:

$J=F_{avg}\Delta t$ (2)

with $F_{avg}$ the average force and $\Delta t$ the time the collision lasts

We can equate expressions (2) and (1):

$\Delta p=p_f-p_i=F_{avg}\Delta t$

Using the definition of linear momentum as mass (m) time velocity (v):

$mv_f-mv_i=F_{avg}\Delta t$

We can solve for Favg:

$F_{avg}=\frac{m(v_f-v_i)}{\Delta t}$ (3)

Now we should find the velocities vf and vi, we should do this using conservation of energy:

For the velocity the ball has just before reaches the floor:

$U_i=K_f$

With Ui the initial potential energy (there is not initial kinetic energy) and Kf the final kinetic energy (there is not final potential energy), then:

$mgh=\frac{mv_i^2}{2}$

solving for vi:

$v_i=\sqrt{2gh}=\sqrt{2*9.81*2.4}=6.86\frac{m}{s}$

For the velocity the ball has just after bounces the floor:

$K_i=U_f$

There is not initial potential energy because it's a floor level at this instant, and the there is not final kinetic energy because the ball has instantly zero velocity at its maximum height (hm), then:

$\frac{mf_i^2}{2}=mgh_m$

solvig for vf:

$v_f=\sqrt{2gh_m}=\sqrt{2*9.81*1.9}=6.10\frac{m}{s}$

Using vf and vi on (3):

$F_{avg}=\frac{(0.06)(6.10-6.86)}{0.034}=-1.34 N$

The negative sign indicates the direction of the force is pointing away the floor

4 0

3 years ago

a stone is vertically thrown upward with the velocity of 72km/hr find the maximum height reached the height

erik [133]

Answer:

y = 20.38 [m]

Explanation:

In order to solve these problems, we must use the following kinematics equation.

$v_{f} ^{2} =v_{i} ^{2}-(2*g*y)$

where:

Vf = final velocity = 0

Vi = initial velocity = 72 [km/h]

g = gravity acceleration = 9.81 [m/s^2]

y = vertical elevation [m]

We need to convert [km/h] to [m/s]

$72[\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km} ] = 20 [m/s]$

Note: the negative sign of the equation means that the acceleration acts in the opposite direction to the movement of the body. And the final speed is zero, because when the body reaches the maximum height, the Stone does not move its speed has been reduced to its entirety.

0 = (20)^2 - (2*9.81*y)

20^2 = 2*9.81*y

y = 20.38 [m]

3 0

3 years ago

A particle with charge −5 µC is located on

Nataly [62]

Answer:

36.25 N

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the magnitude of the two charges

r is the separation between the two charges

Moreover:

- The force is repulsive if the two charges have same sign

- The force is attractive if the two charges have opposite sign

In this problem, we have 3 charges:

$q_1=-5\mu C = -5\cdot 10^{-6}C$ is the charge located at $x=+10 cm = +0.10 m$

$q_2=+6\mu C=+6\cdot 10^{-6}C$ is the charge located at $x=-8 cm =-0.08 m$

$q_3=+2\mu C=+2\cdot 10^{-6}C$ is the charge located at $x=-2 cm=-0.02 m$

The force between charge 1 and charge 3 is:

$F_{13}=\frac{kq_1 q_3}{(x_1-x_3)^2}=\frac{(9\cdot 10^9)(5\cdot 10^{-6})(2\cdot 10^{-6})}{(0.10-(-0.02))^2}=6.25 N$

And since the two charges have opposite sign, the force is attractive, so the force on charge 3 is to the right (towards charge 1).

The force between charge 2 and charge 3 is:

$F_{23}=\frac{kq_2 q_3}{(x_2-x_3)^2}=\frac{(9\cdot 10^9)(6\cdot 10^{-6})(2\cdot 10^{-6})}{(-0.08-(-0.02))^2}=30.0 N$

And since the two charges have same sign, the force is repulsive, so the force on charge 3 is to the right (away from charge 2).

So the two forces on charge 3 have same direction (to the right), so the net force is the sum of the two forces:

$F=F_{13}+F_{23}=6.25+30.0=36.25 N$

8 0

4 years ago

The hydrogen atom, changing from its first excited state to its lowest energy state, emits light with a wavelength of 122 nm. Th

tiny-mole [99]

Answer:

c. There would be a series of spectral lines in hydrogen with the longest wavelength one at 122 nm.

d. The hydrogen atom binds its electron more tightly than the sodium atom does, and would require more energy to remove its electron completely.

Explanation:

The hydrogen atom which changes from the excited state to the lower ground state, it emits light having a wavelength of 122 m. And the sodium atom also gets excited and emits light at 589 nm when it moves from the 1st excited state to the lowest excited state.

Therefore, when the electrons jumps from the 1st excited state to the ground state, only one wavelength is observed as there is only one transition.

The hydrogen atom will bind the electron tightly but the sodium atom does not and would require more energy to remove the electron the electron completely as the binding energy is higher when the electron is closer to the nucleus.

5 0

3 years ago