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3 years ago

What is centripetal force that allows a car to move around a sharp curve in a roadway?

1 answer:

7 0

<span> The centripetal force that allows a car to move around a sharp curve in a roadway is the frictional force between the tires of its wheels and the surface of the road. </span>

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RANGKAIAN RLC SERI PADA TEGANGAN BOLAK BALIK 220V R=80 OHM XL= 90 OHM XC = 30 OHM MAKA TEGANGAN PADDA INDUKTOR ADALAH???

Whitepunk [10]

Okay, 90% of this is nonsense besides the numbers maybe.

8 0

3 years ago

A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle

Yuri [45]

Answer:

(a). Index of refraction are $n_{red}$ = 1.344 & $n_{violet}$ = 1.406

(b). The velocity of red light in the glass $v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

The velocity of violet light in the glass $v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

Explanation:

We know that

Law of reflection is

$n_1 \sin\theta_{1} = n_2 \sin\theta_{2}$

Here

$\theta_1$ = angle of incidence

$\theta_2$ = angle of refraction

(a). For red light

1 × $\sin 56.6$ = $n_{red}$ × $\sin 38.4$

$n_{red}$ = 1.344

For violet light

1 × $\sin 56.6$ = $n_{violet}$ × $\sin 36.4$

$n_{violet}$ = 1.406

(b). Index of refraction is given by

$n = \frac{c}{v}$

$n_{red}$ = 1.344

$v_{red} = \frac{c}{n_{red} }$

$v_{red} = \frac{3(10^{8} )}{1.344}$

$v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

$v_{violet} = \frac{3(10^{8} )}{1.406}$

$v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

This is the velocity of violet light in the glass.

8 0

3 years ago

What is called gravity

Harrizon [31]

Gravity is also called gravitation

3 0

4 years ago

Optimal cardiorespiratory fitness requires a BMI of __________.

PtichkaEL [24]

Answer:

A. 18.5 to 24.9

Explanation:

5 0

3 years ago

Read 2 more answers

A student measures the speed of yellow light in water to be 2.00x10^8

max2010maxim [7]

NOTE: The given question is incomplete.

<u>The complete question is given below.</u>

A student measures the speed of yellow light in water to be 2.00 x 10⁸ m/s. Calculate the speed of light in air.

Solution:

Speed of yellow light in water (v) = 2.00 x 10⁸ m/s

Refractive Index of water with respect to air (μ) = 4/3

Refractive Index = Speed of yellow light in air / Speed of yellow light in water

Or, The speed of yellow light in air = Refractive Index × Speed of yellow light in water

or, = (4/3) × 2.00 x 10⁸ m/s

or, = 2.67 × 10⁸ m/s ≈ 3.0 × 10⁸ m/s

Hence, the required speed of yellow light in the air will be 3.0 × 10⁸ m/s.

7 0

3 years ago