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liubo4ka [24]
4 years ago
5

Li is riding her bicycle at 8.0 m/s. She slows down to 4.0 m/s. Her change in velocity is m/s. If Li takes 2 seconds to make thi

s change, her acceleration is m/s2.
Physics
2 answers:
malfutka [58]4 years ago
8 0
Her change in speed is (4 m/s - 8 m/s) = -4 m/s.

Her acceleration is

(change in speed) / (time for the change)

= ( -4 m/s ) / (2 sec)

= -2 m/s² .
forsale [732]4 years ago
4 0
You will have to use this formula:
v = vo + a \times t

Final Velocity (V) = 4m/s
Initial Velocity (Vo) = 8m/s
Acceleration (a) = ? m/s^2
Time (t) = 2 secs

Then:

-> 4 = 8 + a x 2
-> 4 - 8 = 2a
-> -4 = 2a
-> a = -4/2
-> a = -2 m/s^2

Ps: It's value is negative because the she was in retrograde motion.

Answer: Her acceleration is -2 m/s^2.
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A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal
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(a) The ball has a final velocity vector

\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j

with horizontal and vertical components, respectively,

v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}

v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}

The horizontal component of the ball's velocity is constant throughout its trajectory, so v_{x,i}=v_{x,f}, and the horizontal distance <em>x</em> that it covers after time <em>t</em> is

x=v_{x,i}t=v_{x,f}t

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s

The vertical component of the ball's velocity at time <em>t</em> is

v_{y,f}=v_{y,i}-gt

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}

So, the initial velocity vector is

\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j

which carries an initial speed of

\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}

and direction <em>θ</em> such that

\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

y=v_{y,i}t-\dfrac12gt^2

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}

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3 years ago
What is acceleration, and what units do we use to measure it? *
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Answer:

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).

Explanation:

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You are part of the designing team for a new solar thermal power plant. The plant will use thermal energy storage to produce ele
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Explanation:

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3 years ago
Which best describes the differences between a resume and a CV?
son4ous [18]

Answer:

The resume will be tailored to each position whereas the CV will stay put and any changes will be in the cover letter. ... I would say the main difference between a resume and a CV is that a CV is intended to be a full record of your career history and a resume is a brief, targeted list of skills and achievements

Explanation:

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3 years ago
A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal. (a) Determi
sattari [20]

Answer:

(a) t = 3.74 s

(b) H = 136.86 m

(c) Vₓ = 41.83 m/s,  Vy = 0 m/s

(d) ax = 0 m/s²,  ay = 9.8 m/s²

Explanation:

(a)

Time to reach maximum height by the projectile is given as:

t = V₀ Sinθ/g

where,

V₀ = Launching Speed = 55.6 m/s

Angle with Horizontal = θ = 41.2°

g = 9.8 m/s²

Therefore,

t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)

<u>t = 3.74 s</u>

<u></u>

(b)

Maximum height reached by projectile is:

H = V₀² Sin²θ/g

H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)

<u>H = 136.86 m</u>

<u></u>

(c)

Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:

Vₓ = V₀ₓ = V₀ Cos θ

Vₓ = (55.6 m/s)(Cos 41.2°)

<u>Vₓ = 41.83 m/s</u>

Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.

<u>Vy = 0 m/s</u>

<u></u>

(d)

Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.

<u>ax = 0 m/s²</u>

The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:

<u>ay = 9.8 m/s²</u>

3 0
3 years ago
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