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3 years ago

7

What is the critical angle for a ray trying to pass from glass (n = 1.52) into water (n = 1.33)? (Water n = 1.33, Air n = 1.00)

(Unit = deg)

2 answers:

ivann1987 [24]3 years ago

7 0

Answer:

61.05

Explanation:

correct on acellus

horrorfan [7]3 years ago

3 0

Answer:

59.52

Explanation:

59.5 in sigfigs

got it right on Acellus

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A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, veloci

Damm [24]

It will take a shorter amount of time for the cylinder to go down the plane down off the plane Because more pressure is applied one going up then going down there’s no pressure at all it’s the gravity is helping

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3 years ago

The aorta is the main artery from the heart. a typical aorta has an inside diameter of 1.8 cm and carries blood at speeds of up

defon

Answer:

Explanation:

Volume per unit time flowing will be conserved

a₁v₁ = a₂ v₂

π r₁² x v₁ = π r₂² x v₂

(0.9 x 10⁻²)² x .35 = ( .45 x 10⁻² )² x v₂

v₂ = 1.4 m / s

3 0

3 years ago

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was

tigry1 [53]

Answer:

$\Delta V=316167V$

Explanation:

The difference of electric potential between two points is given by the formula $\Delta V=Ed$ , where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.

The electric field formula is $E=\frac{F}{q}$ , where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.

Putting this together we have $\Delta V=\frac{Fd}{q}$ , so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force $N=W-F$ . On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.

Using the equation for accelerated motion (departing from rest) $d=\frac{at^2}{2}$ , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

$a_1=\frac{2d}{t_1^2}$

$a_2=\frac{2d}{t_2^2}$

These relate with the forces by Newton's 2nd Law:

$W=ma_1$

$N=ma_2$

Putting all together:

$N=W-F=ma_1-F=ma_2$

Which means:

$F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

And finally we substitute:

$\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

Which for our values means:

$\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V$

7 0

3 years ago

Hii please help i’ll give brainliest!!

antiseptic1488 [7]

The answer is C Because it is just C

8 0

3 years ago

Read 2 more answers

In a maneuver, the jet comes in for a landing on solid ground with a speed of 115 m/s, and its acceleration can have a maximum m

V125BC [204]

Answer:

17.1130952381 s

No

Explanation:

t = Time taken

u = Initial velocity = 115 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -6.72 m/s² (negative as it is decelerating)

From the equations of motion

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-115}{-6.72}\\\Rightarrow t=17.1130952381\ s$

The minimum time required to stop is 17.1130952381 s

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-115^2}{2\times -6.72}\\\Rightarrow s=984.00297619\ m$

The distance that is required for the jet to stop is 0.98400297619 km which is greater than 0.8 km. So, the jet cannot land on a small tropical island airport.

4 0

3 years ago