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frosja888 [35]
2 years ago
11

Identify the law, write the equation and calculate the answer to the problem below.

Physics
1 answer:
lyudmila [28]2 years ago
5 0

Find refractive index first

\\ \rm\Rrightarrow \mu=\dfrac{c}{v}

\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}

\\ \rm\Rrightarrow \mu =0.75

Now

\\ \rm\Rrightarrow \dfrac{sini}{sinr}=\mu

\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75

\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr

\\ \rm\Rrightarrow sinr=0.94

\\ \rm\Rrightarrow r=sin^{-1}(0.94)

\\ \rm\Rrightarrow r=70^{\circ}

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If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per​ second, then its height h aft
Marina CMI [18]

Explanation:

The equation of motion of an object is given by :

h(t)=-16t^2+112t+128

Where

t is the time in seconds

We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

-16t^2+112t+128=0

-t^2+7t+8=0

On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

8 0
3 years ago
If the atoms and molecules of a substance are moving very fast, the substance is _________.
lana66690 [7]
A. hot is the correct answer.
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5 0
3 years ago
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Identify the areas on the image where the force of repulsion is the least.​
aalyn [17]

Answer:

The central blue square in between the lower pair of magnet has the least force of repulsion.

Explanation:

We can explain this using the dual nature of magnets.

Each magnet must have two poles namely:

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We assume that the magnetic lines of forces enters from south pole and leaves from the north pole.

When brought together, like poles repel each other while opposite poles attract each other.

In the picture, the lower two magnets have opposite poles facing each other, hence the force of repulsion is minimum there and the force of attraction is maximum.

4 0
3 years ago
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Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1
kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}[/tex]

and v(0)=\hat{i}

r(0)=\hat{j}

we know a=\frac{\mathrm{d} v}{\mathrm{d} t}

\int dv=\int adt

v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt

v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c

at t=0

v(0)=0-1\cdot \hat{j}+0+c

c=\hat{i}+\hat{j}

v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}

and \frac{\mathrm{d} r}{\mathrm{d} t}=v(t)

\int dr=\int vdt

r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt

r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2

at t=0

r(0)=\hat{j}

r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}

       

4 0
3 years ago
A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what
irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

        \Delta L=\frac{PL}{AE}

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

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Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

       0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg  

Mass of the climber = 69.38 kg

6 0
3 years ago
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