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kvasek [131]
3 years ago
9

What should i do if i'm stuck at home? Any thing except watching movies and going to sleep

Physics
1 answer:
Makovka662 [10]3 years ago
3 0

card games

board games

bird watch

write a song

make a game

count stuff

throw a ball with someone

play outside

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X-rays are used instead of visible light because X-rays have shorter ______________ than visible light, allowing them to produce
Novay_Z [31]
X-rays have shorter wavelength than visible light. But that's hardly the reason that they're used for medical imaging. xrays have much higher frequencies then visible light which means they have much greater penetrating ability. with xrays you can see inside the body. you can't do that with a visible flashlight no matter how bright and powerful it is.
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3 years ago
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When light shines through atomic hydrogen gas, it is seen that the gas absorbs light readily at a wavelength of 91.63 nm. What i
Artist 52 [7]

Answer:

D) 21

Explanation:

When gas absorbs light , electron at lower level jumps to higher level .

and the difference of energy of orbital is equal to energy of radiation absorbed.

Here energy absorbed is equivalent to wavelength of 91.63 nm

In terms of its energy in eV , its energy content is eual to

1243.5 / 91.63 = 13.57 eV. This represents the difference the energy of orbit .

Electron is lying in lowest or first level ie n = 1.

Energy of first level

= - 13.6 / 1² = - 13.6 eV.

Energy of n th level = - 13.6 / n². Let in this level electron has been excited

Difference of energy

= 13.6 - 13.6 / n² = 13.57 ( energy of absorbed radiation)

13.6 / n² = 13.6 - 13.57 = .03

n² = 13.6 / .03 = 453

n = 21 ( approx )

4 0
3 years ago
A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff
postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

F_c=\frac{mv^2}{R}

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

N=mg

Therefore, frictional force, f=\mu mg

Now, frictional force = centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0
3 years ago
Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa
tester [92]

Answer:

Part a)

t = 2.83 s

Part b)

Ball thrown downwards =v_f = 23.8 m/s

Ball thrown upwards =v_f = 23.8 m/s

Part c)

d = 22.24 m

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

\Delta y = 0 = v_y t + \frac{1}{2}at^2

0 = 13.9 t - \frac{1}{2}(9.81) t^2

t = 2.83 s

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

v_f^2 - v_i^2 = 2 a \Delta y

v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)

v_f^2 = 567.9

v_f = 23.8 m/s

Part c)

Relative speed of two balls is given as

v_{12} = v_1 - v_2

v_{12} = (13.9) - (-13.9) = 27.8 m/s

now the distance between two balls in 0.8 s is given as

d = v_{12} t

d = 27.8 \times 0.8

d = 22.24 m

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3 years ago
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