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Oksi-84 [34.3K]

2 years ago

14

3. A girl is running the 200 m dash. She starts by acceleration at 8m/s^2 for 7s. Then continues at this speed until the end of

the race. How long did it take for her to complete the race?

1 answer:

antoniya [11.8K]2 years ago

7 0

Answer:

10.6 s

Explanation:

Given that a girl is running the 200 m dash. She starts by acceleration at 8m/s^2 for 7s. Then continues at this speed until the end of the race. How long did it take for her to complete the race?

Solution.

If she accelerated for 7s, the velocity at which she accelerated will be:

Acceleration = velocity/time

8 = V/7

Make V the subject of the formula by cross multiplying.

V = 8 × 7

V = 56 m/s

She maintains the speed through out the journey.

Speed = distance/time

Make time the subject of formula

Time = distance/speed

Time = 200 / 56

Time = 3.57s

Therefore, she will complete the race by 7 + 3.6 = 10.6 s

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A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a

SOVA2 [1]

Answer:

a) x = v₀² sin 2θ / g

b) t_total = 2 v₀ sin θ / g

c) x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

sin θ = $v_{oy}$ / vo

cos θ = v₀ₓ / vo

v_{oy} = v_{o} sin θ

v₀ₓ = v₀ cos θ

v_{oy} = 13.5 sin 32 = 7.15 m / s

v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

v₀ₓ = x / t

x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

v_{y} = v_{oy} - gt

0 = v₀ sin θ - gt

t = v_{o} sin θ / g

we substitute

x = v₀ cos θ (2 v_{o} sin θ / g)

x = v₀² /g 2 cos θ sin θ

x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

v_{y} = v_{oy} - gt

v_{y} = 0

t = v_{oy} / g

t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

t_total = 2 t

t_total = 2 v₀ sin θ / g

c) we calculate

x = 13.5 2 sin (2 32) / 9.8

x = 16.7 m

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3 years ago

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Lera25 [3.4K]

Answer:

Before sled starts to move it has a potential energy due to the elevation...and then that potential energy converted to kinetic energy due to presence of a velocity...the sled will continue to move if their is no resesive force...but however friction force is presence that cause the sled to stop....

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3 years ago

Two traveling sinusoidal waves are described by the wave functions y1 = 4.85 sin [(4.35x − 1270t)] y2 = 4.85 sin [(4.35x − 1270t

Tamiku [17]

Answer:

Approximately $9.62$ .

Explanation:

$y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0]$ .

$y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)]$ .

Notice that sine waves $y_1$ and $y_2$ share the same frequency and wavelength. The only distinction between these two waves is the $(-0.250)$ in $y_2\!$ .

Therefore, the sum $(y_1 + y_2)$ would still be a sine wave. The amplitude of $(y_1 + y_2)\!$ could be found without using calculus.

Consider the sum-of-angle identity for sine:

$\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b)$ .

Compare the expression $\sin(a + b)$ to $y_2$ . Let $a = (4.35\, x - 1270)$ and $b = (-0.250)$ . Apply the sum-of-angle identity of sine to rewrite $y_2\!$ .

$\begin{aligned}y_2 &= 4.85\, \sin[(\underbrace{4.35\, x - 1270\, t}_{a}) + (\underbrace{-0.250}_{b})]\\ &= 4.85 \, [\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Therefore, the sum $(y_1 + y_2)$ would become:

$\begin{aligned}& y_1 + y_2\\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t) \\ &\quad \quad \quad\;+\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Consider: would it be possible to find $m$ and $c$ that satisfy the following hypothetical equation?

$\begin{aligned}& (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)\\&= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Simplify this hypothetical equation:

$\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\&=\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)\end{aligned}$ .

Apply the sum-of-angle identity of sine to rewrite the left-hand side:

$\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\[0.5em]&=m\, \sin(4.35\, x - 1270\, t)\cdot \cos(c) \\ &\quad\quad + m\, \cos(4.35\, x - 1270\, t)\cdot \sin(c) \\[0.5em] &=\sin(4.35\, x - 1270\, t)\cdot (m\, \cos(c)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot (m\, \sin(c)) \end{aligned}$ .

Compare this expression with the right-hand side. For this hypothetical equation to hold for all real $x$ and $t$ , the following should be satisfied:

$\displaystyle 1 + \cos(-0.250) = m\, \cos(c)$ , and

$\displaystyle \sin(-0.250) = m\, \sin(c)$ .

Consider the Pythagorean identity. For any real number $a$ :

${\left(\sin(a)\right)}^{2} + {\left(\cos(a)\right)}^{2} = 1^2$ .

Make use of the Pythagorean identity to solve this system of equations for $m$ . Square both sides of both equations:

$\displaystyle 1 + 2\, \cos(-0.250) + {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2$ .

$\displaystyle {\left(\sin(-0.250)\right)}^{2} = m^2\, {\left(\sin(c)\right)}^2$ .

Take the sum of these two equations.

Left-hand side:

$\begin{aligned}& 1 + 2\, \cos(-0.250) + \underbrace{{\left(\cos(-0.250)\right)}^2 + {\left(\sin(-0.250)\right)}^2}_{1}\\ &= 1 + 2\, \cos(-0.250) + 1 \\ &= 2 + 2\, \cos(-0.250) \end{aligned}$ .

Right-hand side:

$\begin{aligned} &m^2\, {\left(\cos(c)\right)}^2 + m^2\, {\left(\sin(c)\right)}^2 \\ &= m^2\, \left( {\left(\sin(c)\right)}^2 + {\left(\cos(c)\right)}^2\right)\\ &= m^2\end{aligned}$ .

Therefore:

$m^2 = 2 + 2\, \cos(-0.250)$ .

$m = \sqrt{2 + 2\, \cos(-0.250)} \approx 1.98$ .

Substitute $m = \sqrt{2 + 2\, \cos(-0.250)}$ back to the system to find $c$ . However, notice that the exact value of $c\!$ isn't required for finding the amplitude of $(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)$ .

(Side note: one possible value of $c$ is $\displaystyle \arccos\left(\frac{1 + \cos(0.250)}{\sqrt{2 \times (1 + \cos(0.250))}}\right) \approx 0.125$ radians.)

As long as $\! c$ is a real number, the amplitude of $(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)$ would be equal to the absolute value of $(4.85\, m)$ .

Therefore, the amplitude of $(y_1 + y_2)$ would be:

$\begin{aligned}|4.85\, m| &= 4.85 \times \sqrt{2 + 2\, \cos(-0.250)} \\&\approx 9.62 \end{aligned}$ .

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3 years ago

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bagirrra123 [75]

Answer:

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Explanation:

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Answer:

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Explanation:

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