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lana [24]
2 years ago
11

If a apple has a a weight of 1 N, what is the mass of the apple?

Physics
1 answer:
mixas84 [53]2 years ago
7 0

Answer:

The answer is approximately 0.1 kg

\frac{1 \: newton}{9.8}   \: is \: approx. \: 0.10204081

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Radio Station W as the slower the frequency the longer the wave length
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2 years ago
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A particle executes simple harmonic motion with an amplitude of 2.18 cm.
Bogdan [553]

Answer:

The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.

Explanation:

Maximum speed is  :

                          v (max) = Aω

Speed v at any displacement y is given by  

v^{2} = w^{2} (A^{2} - y^{2})   ........................................................  i

And,

               v = \frac{1}{2} v (max)  

          or,  2 × v = Aω     ....................................................   ii

Eliminating  ω from equations i and ii,

                       \frac{1}{4} A^{2}  w^{2}  =  w^{2}  ( A^{2}  - y^{2})

                     or, y^{2} =  (\frac{3}{4}) A^{2}  =(\frac{3}{4}) 2.18^{2}

                    or,  y =  3.56 cm.

3 0
2 years ago
A slab of glass has a 0.600 cm thick layer of water on top of it. A light ray strikes the water at an incident angle of 59.0°. A
Dimas [21]

Answer:

49.63 degree

Explanation:

thickness of glass slab, t = 0.6 cm

angle of incidence = 59 degree

Let r be the angle of refraction

The refractive index of glass, ng = 3/2

refractive index of water, nw = 4/3

refarctive index of glass with respect to water = ng / nw = 3 /2 ÷ 4 /3 = 9 / 8

So, by use of Snell's law

Refractive index of glass with respect to water = Sin i / Sin r

9 / 8 = Sin 59 / Sin r

9 / 8 = 0.857 / Sin r

Sin r = 0.7619

r = 49.63 degree

4 0
3 years ago
A sharp raises a note by
vivado [14]

The answer should be B. A half step

8 0
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If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?
Anon25 [30]
<h2>Answer:50ms^{-1}</h2>

Explanation:

Let v_{a} be the airspeed.

Let v_{w} be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If v_{1} and v_{2} are two perpendicular vectors,the resultant vector has the magnitude \sqrt{|v|_{1}^{2}+|v|_{2}^{2}}

Given,

v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}

So,the ground speed is \sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}

6 0
3 years ago
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