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Nostrana [21]
3 years ago
12

Can someone help because I am very confused

Physics
1 answer:
Rzqust [24]3 years ago
3 0
I can't see ot carefuly
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Which of the following waves travel at the greatest speed
NeTakaya
HEY mate here your answer

p waves travel at the greatest speed and these waves can travel at solid and liquid also. p waves means primary waves.
4 0
3 years ago
On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he
Liula [17]

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

      v = v₀ - a t

     a = (v₀- v) / T

On planet X

    v = v₀ - a' t’

    a ’= (v₀-v) / 2T

Let's substitute the land values ​​in plot X

     a’= a / 2

Now let's use Newton's second law

       W = ma

      m g = m a

      a = g

We substitute

      a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

4 0
2 years ago
Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen
Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

\lambda = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}

sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m

For first dark fringe

dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

3 0
2 years ago
At what speed do a bicycle and its rider, with a combined mass of 100 kg, have the same momentum as a 1400 kg car traveling at 6
Ilia_Sergeevich [38]
Momentum of car

Given: Mass m= 1,400 Kg;    V = 6.0 m/s

Formula:  P = mv    

                P = (1,400 Kg)(6.0 m/s)

                P = 8,400 Kg.m/s

Velocity of the rider to have the same momentum as a car.

Mass of rider and bicycle  m = 100 Kg

P = mv

V = P/m

V = 8,400 Kg.m/s/100 Kg

V = 84 m/s









7 0
2 years ago
Hii please help i’ll give brainliest!!
den301095 [7]

Answer:

C

Explanation:

6 0
2 years ago
Read 2 more answers
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