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Alex787 [66]
3 years ago
8

Strzała o masie 20g tuz po wystrzale ma prędkość 50m/s. Oblicz pracę wykonana przez zawodniczkę. Ile wynosi energia potencjalna

sprężystości napiętej cięciwy?
Physics
1 answer:
garri49 [273]3 years ago
6 0

Answer:

Explanation:

masie m = 20g = 20/1000 = 0.02kg

prędkość v = 50m/s

P.E = K.E = ½mv²

P.E = ½ × 0.02 × 50²

P.E = 25 J

pracę wykonana = P.E = 25J

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Papessa [141]

Answer:

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2 years ago
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3 years ago
Researcher measures the thickness of a layer of benzene (n = 1.50) floating on water by shining monochromatic light onto the fil
earnstyle [38]

Answer:

Minimum thickness; t = 9.75 x 10^(-8) m

Explanation:

We are given;

Wavelength of light;λ = 585 nm = 585 x 10^(-9)m

Refractive index of benzene;n = 1.5

Now, let's calculate the wavelength of the film;

Wavelength of film;λ_film = Wavelength of light/Refractive index of benzene

Thus; λ_film = 585 x 10^(-9)/1.5

λ_film = 39 x 10^(-8) m

Now, to find the thickness, we'll use the formula;

2t = ½m(λ_film)

Where;

t is the thickness of the film

m is an integer which we will take as 1

Thus;

2t = ½ x 1 x 39 x 10^(-8)

2t = 19.5 x 10^(-8)

Divide both sides by 2 to give;

t = 9.75 x 10^(-8) m

8 0
3 years ago
A 0.5-kg ball moving at 5 m/s strikes a wall and rebounds in the opposite direction with a speed of 2 m/s. If the impulse occurs
mart [117]

<em>Given that:</em>

                       mass of the ball (m) = 0.5 Kg ,

                    ball strikes the wall (v₁) = 5 m/s ,

rebounds in opposite direction (v₂) = 2 m/s,

                                time duration (t) = 0.01 s,

        <em> Determine the force (F) = ?</em>

We know that from Newton's II law,

                                <em>F = m. a</em>  Newtons  

                                  (velocity acting in opposite direction, so <em>a = ( (v₁ + v₂)/t</em>

                                   = m × (v₁ + v₂)/t

                                   = 0.5 × (5 + 2)/0.01

                                  = 350 N

<em>The force acting up on the ball is 350 N</em>

                                     

6 0
3 years ago
Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

Answer:

E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

Then the electric field at the point of interest is estimated as:

E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}

6 0
3 years ago
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