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Kazeer [188]
3 years ago
7

An automobile traveling on a straight, level road has an initial speed v when the brakes are applied. In coming to rest with a c

onstant acceleration, it travels a distance x. How far would the automobile travel in coming to rest if it had the same acceleration but an initial speed 2v ?
Physics
1 answer:
Molodets [167]3 years ago
3 0

Answer:

4x

Explanation:

Use v^{2} = u^{2} +2as to do the question.

For first instance,

0 = v^{2} +2ax -------------------( 1 )

for second instance,

0 = (2v)^{2} +2as-----------------( 2 )

So by (1) and (2),

s = 4x

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In a classroom demonstration, students are using a Slinky to observe and learn about wave properties. If the Slinky has a period
Andrej [43]

Answer:

Frequency = 3.0 Hertz

Explanation:

Given the following data;

Period = 0.333 seconds

To find the frequency;

Mathematically, frequency of a wave is given by the formula;

Frequency = 1/period

Substituting into the formula, we have;

Frequency = 1/0.333

Frequency = 3.0 Hertz

3 0
3 years ago
If a ball is launched horizontally at 40 m/s
vaieri [72.5K]

Answer:

40m/s

Explanation:

The horizontal component of velocity remains constant because there are no external forces in that direction

By applying motion equations, V= U+ at

where ,

  • v - final velocity
  • u - initial velocity
  • a-acceleration
  • t - time

v = u +at

As no force act on the ball ( we neglect air resistance here) no acceleration is seen,

So v = u = 40m/s

4 0
3 years ago
Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.
masya89 [10]
The velocity would be the same for all ramps.
5 0
3 years ago
Read 2 more answers
A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
3 years ago
Calculate the missing variable....
OlgaM077 [116]

Answer:

option b

Explanation:

from the given formula, s=d/t

make t the subject of the formula we have

t=d/s

5/100

0.5

6 0
2 years ago
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