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2 years ago

Which of the following is an application of of the technology developed by Charles kuen kao

Physics

2 answers:

amid [387]2 years ago

7 0

The answer would be D

Arada [10]2 years ago

6 0

D because it’s most wanted and I see the people through it and I know because I love it

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A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw

meriva

Answer:

As 28m/s = 28m/s

Explanation:

r = the radius of the curve

m = the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

$a = \frac{v^{2}}{r}$

therefore

$\mu mg = m \frac{v^{2}}{r} \\\\ \mu = \frac{v^{2}}{rg}$

$v = \sqrt{\mu rg}$

$\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s$

As 28m/s = 28m/s

8 0

2 years ago

. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it

kap26 [50]

Answer:

29.4 N/m

0.1

Explanation:

a) From the restoring Force we know that :

F_r = —k*x

the gravitational force :

F_g=mg

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force

xi , x2 are the displacement made by the two masses.

Givens:

m1 = 1.29 kg

m2 = 0.3 kg

x1 = -0.75 m

x2 = -0.2 m

g = 9.8 m/s^2

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m

Plugging known infromation to get :

l= x1 — (F_1/k)

= 0.1

3 0

2 years ago

If you were trying to cross a river with the shortest possible time, would you aim your boat slightly upstream, directly across

Katyanochek1 [597]

I would say downstream since the stream can push your boat, then you would have momentum and would just have to row towards the land.

7 0

2 years ago

The critical angle for a certain air-liquid surface is 47.7°. What is the index of refraction of the liquid? Round to the neares

KengaRu [80]

Answer:

The index of refraction of the liquid is 1.35.

Explanation:

It is given that,

Critical angle for a certain air-liquid surface, $\theta_1=47.7^{\circ}$

Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1

Using Snell's law for air liquid interface as :

$n_1\ sin\theta_1=n_2\ sin(90)$

$n_1\ sin(47.7)=1$

$n_1=\dfrac{1}{sin(47.7)}$

$n_1=1.35$

So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.

5 0

2 years ago

Science is the process of asking scientific questions about which of the following?

Eva8 [605]

the answer to your question would be D.

4 0

2 years ago

Read 2 more answers