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kolbaska11 [484]
4 years ago
10

When light waves pass through the lenses of a pair of glasses, the light waves ______________ .

Physics
1 answer:
e-lub [12.9K]4 years ago
4 0

refracts. Hope this helped.

When light enters a pair of glasses, it ______________ when it hits the glass.

When light waves pass through the lenses of a pair of glasses, the light waves ______________ .


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A van moves with a constant speed of 20 miles per hour. How far can it travel in 3 1/2 hours
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A measurement must include both a number and a(an)
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A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
4 years ago
What is the acceleration of an object with a mass of 15 kg and a coefficient of friction of 0.18
11111nata11111 [884]

Answer:

a = 1.764m/s^2

Explanation:

By Newton's second law, the net force is F = ma.

The equation for friction is F(k) = F(n) * μ.

In this case, the normal force is simply F(n) = mg due to no other external forces being specified

F(n) = mg = 15kg * 9.8 m/s^2 =  147N.

F(k) = F(n) * μ = 147N * 0.18 = 26.46N.

Assuming the object is on a horizontal surface, the force due to gravity and the normal force will cancel each other out, leaving our net force as only the frictional one.

Thus, F(net) = F(k) = ma

26.46N = 15kg * a

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Answer: its 50

Explanation:

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