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2 years ago

A car is traveling at a velocity of 22 m/s when the driver puts on the brakes

Physics

1 answer:

Brums [2.3K]2 years ago

8 0

The car’s velocity at the end of this distance is 18.17 m/s.

Given the following data:

Initial velocity, U = 22 m/s

Deceleration, d = 1.4 $m/s^2$

Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

$V^2 = U^2 + 2dS$

Substituting the values into the formula, we have;

$V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}$

Final velocity, V = 18.17 m/s

Therefore, the car’s velocity at the end of this distance is 18.17 m/s.

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2 years ago

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2 years ago

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marysya [2.9K]

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2 years ago

In a Young's double-slit experiment, 610-nm-wavelength light is sent through the slits. The intensity at an angle of 2.95° from

Sliva [168]

Answer:

spacing between the slits is 405.32043 × $10^{-9}$ m

Explanation:

Given data

wavelength = 610 nm

angle = 2.95°

central bright fringe = 85%

to find out

spacing between the slits

solution

we know that spacing between slit is

I = 4 $I_{0}$ × cos²∅/2

I/4 $I_{0}$ = cos²∅/2

here I/4 $I_{0}$ is 85 % = 0.85

0.85 = cos²∅/2

cos∅/2 = √0.85

∅ = 2 × $cos^{-1}$ 0.921954

∅ = 45.56°

∅ = 45.56° ×π/180 = 0.7949 rad

and we know that here

∅ = 2π d sinθ / wavelength

d = ∅× wavelength / ( 2π sinθ )

put all value

d = 0.795 × 610× $10^{-9}$ / ( 2π sin2.95 )

d = 405.32043 × $10^{-9}$ m

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7 0

2 years ago

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

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2 years ago