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lord [1]
3 years ago
7

Help me!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
Gemiola [76]3 years ago
8 0

Answer:a computer , machine forcery,0,push

Explanation:

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A horizontal beam of light of intensity 25 W/m2 is sent through two polarizing sheets. The polarizing direction of the first mak
Zina [86]

Answer:

option (B)

Explanation:

Intensity of unpolarised light, I = 25 W/m^2

When it passes from first polarisr, the intensity of light becomes

I'=\frac{I_{0}}{2}=\frac{25}{2}=12.5 W/m^{2}

Let the intensity of light as it passes from second polariser is I''.

According to the law of Malus

I'' = I' Cos^{2}\theta

Where, θ be the angle between the axis first polariser and the second polariser.

I'' = 12.5\times Cos^{2}15

I'' = 11.66 W/m^2

I'' = 11.7 W/m^2

7 0
3 years ago
Calculate the kinetic energy of an 80,000 kg airplane that is flying with a velocity of 167 m/s.
liq [111]

Answer:

1115560000 J

Explanation:

1/2 * 80,000 * 167^2 m/s = 1115560000 J

7 0
3 years ago
SOMEONE PLEASE HELP!!
Arisa [49]

Hello!

\large\boxed{\text{C. 7,350,000 J}}

Use the equation:

PE = mgh

Where:

m = mass of the object (kg)

g = acceleration due to gravity (≈9.8 m/s)

h = height above ground (m)

Plug the given values into the equation:

PE = 7500 · 9.8 · 100

PE = 7,350,000 Joules.

7 0
3 years ago
Please help This very important for me you will get 40 points if you help me and explain in deatil please.
posledela
Well, I think mostly it is oceanic crust well because <span>The scientist noticed a symmetrical pattern of positive and negative magnetic lines as they moved along the ocean floor, and the line of symmetry was at the mid-ocean ridge. that why it is seen as the youngest on our earth</span>
4 0
3 years ago
Read 2 more answers
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
3 years ago
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