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2 years ago

Can someone help me with the exercise 3 only c,d,e please!!!

Physics

2 answers:

saul85 [17]2 years ago

4 0

I'm not sure...
I feel ya.

fgiga [73]2 years ago

3 0

Im not sure, but i think you should try C

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A pendulum is set in motion and time. The time measured for 20 complete swings is 308.

laiz [17]

Answer:

solution

308=20 swings

?::: =1

( 308×1)÷20

308÷20

154÷10

=15.4

=15.4

6 0

3 years ago

Si un balón tiene una masa aproximada de 0.5 kg ¿con que fuerza hay que patearlo para imprimirle una aceleración de 1.5m/s2?

ExtremeBDS [4]

Answer:

translation

Explanation:

If a ball has an approximate mass of 0.5 kg, with what force must be kicked to give it an acceleration of 1.5m / s2?

8 0

2 years ago

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T

motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

$KE_r = \dfrac{1}{2}I\omega^2$

I of the moment of inertia of the sphere

$I = \dfrac{2}{5}MR^2$

v = R ω

using conservation of energy

$KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2$

$KE_r = \dfrac{1}{5}MV^2$

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

$\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h$

$0.7 V^2 = 0.7 V'^2 + gh$

$0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53$

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

6 0

3 years ago

You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen

Nikolay [14]

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂ = d₁ / d₂

But

λ₁ < λ₂

d₁ < d₂

Therefore no of lines per unit length of grating A will be more because

no of lines per unit length ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

λ₁ D/ d₂

As λ₁ < λ₂

λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

5 0

2 years ago

This table shows four examples of experiments.

yarga [219]

Answer

32000

Explanation:

120394

5 0

2 years ago

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