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3 years ago

How much farther will a car traveling at 100 km/s skid than the same car traveling at 50 km/s?

Physics

1 answer:

Roman55 [17]3 years ago

6 0

Answer:

option A

Explanation:

given,

speed of car 1 = 100 Km/s

speed of car 2 = 50 Km/s

final speed = 0 Km/s

now,

skid mark when the car is moving at 100 Km/s

using equation of motion

v² = u² + 2 a s₁

0² = 100² - 2 a s₁

100² = 2 a s₁

$s_1 = \dfrac{100^2}{2a}$ ..........(1)

now,

skid mark when the car is moving at 50 Km/s

using equation of motion

v² = u² + 2 a s₂

0² = 50² - 2 a s₂

50² = 2 a s₂

$s_2 = \dfrac{50^2}{2a}$ ..........(2)

dividing equation (1)/ (2)

$\dfrac{s_1}{s_2} = \dfrac{100^2}{50^2}$

$\dfrac{s_1}{s_2} =4$

s₁ = 4 s₂

Hence, the skid mark for 100 Km/s is four time as far as 50 Km/s

The correct answer is option A

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3 years ago

If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder

kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

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3 years ago

Light is shone on a diffraction grating

Pani-rosa [81]

Answer:

λ = 482.05 nm

Explanation:

The diffraction phenomenon and the diffraction grating is described by the expression

d sin θ = m λ

where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction

in this case they indicate the distance between slits, the angle and the order of diffraction

λ = $\frac{d sin \theta }{m}$ d sin θ / m

let's calculate

λ = 1.00 10⁻⁶ sin 74.6 / 2

λ = 4.82048 10⁻⁷ m

Let's reduce to nm

λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)

λ = 482.05 nm

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3 years ago

How much force is needed to keep the 750000 Newton Space Shuttle moving at a constant speed of 28000 km/h, in a straight line?

Alika [10]

The force needed to keep the space shuttle moving at constant speed is 0.

The given parameters;

<em>weight of the space shuttle, F = 750,000 N</em>
<em>constant speed of the space shuttle, v = 28,000 km/h</em>

The mass of the space shuttle is calculated as follows;

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The force needed to keep the space shuttle moving at constant speed is calculated as follows;

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where;

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F = 76,530.61 x 0

F = 0

Thus, the force needed to keep the space shuttle moving at constant speed is 0.

Learn more here:brainly.com/question/16374764

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2 years ago