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rusak2 [61]
3 years ago
10

A car starts from rest and accelerates at a constant rate in a straight line. In the first second the car moves a distance of 2.

0 meters. How fast will the car be moving at the end of the second second
Physics
1 answer:
goblinko [34]3 years ago
5 0

Answer:

The speed of the car at the end of the 2nd second = 8.0 m/s

Explanation:

The equations of motion will be used to solve this problem.

A car starts from rest,

u = initial velocity of the car = 0 m/s

Accelerates at a constant rate in a straight line,

a = constant acceleration of the car = ?

In the first second the car moves a distance of 2.0 meters,

t = 1.0 s

x = distance covered = 2.0 m

x = ut + (1/2)at²

2 = 0 + (1/2)(a)(1²)

a = 4.0 m/s²

How fast will the car be moving at the end of the second second

Now,

a = 4.0 m/s²

u = initial velocity of the car at 0 seconds = 0 m/s

v = final velocity of the car at the end of the 2nd second = ?

t = 2.0 s

v = u + at

v = 0 + (4×2)

v = 8.0 m/s

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An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th
jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

      R = Vo² sin (2θ) / g

      sin 2θ = g R / Vo²

      sin 2θ = 9.8 75/35²

      2θ = sin⁻¹ (0.6)

      θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

       X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

       X = Vox t

       t2 = X2 / Vox = X2 / (Vo cosθ)

        t2 = 37.5 / (35 cos 18.4)

        t2 = 1.13 s

With this time we calculate the height at this point

        Y = Voy t - ½ g t²

        Y = 35 sin 18.4   1.13 - ½ 9.8 1,13²

        Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

8 0
3 years ago
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49 = 1/2gt²
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Answer:

e  son

Explanation:

se lo on

7 0
3 years ago
a boy takes 15min to reach his school on bicycle.if the bicycle have speed of 2m/s the what is the distance between the house an
11111nata11111 [884]
When looking for distance you multiply speed by time

So 15 x 2 = 30

30 is the distance between his house and school
6 0
3 years ago
When are car velocity is positive and acceleration is negative, what is happening to the cars motion
DanielleElmas [232]
The motion of the car is negative
7 0
3 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
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