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3 years ago

A falling object accelerates from -10.0 m/s to -30.0 m/s. how much time does it take?

Physics

2 answers:

Zepler [3.9K]3 years ago

8 0

Answer:

2.04 s

Explanation:

v = at + v₀

(-30.0 m/s) = (-9.8 m/s²) t + (-10.0 m/s)

t = 2.04 s

KIM [24]3 years ago

8 0

<u>Answer</u>:

The time taken by the falling object is 2.04 sec

<u>Solution:</u>

The initial velocity $V = -10 m/s$

The final velocity $V_1 = -30m/s$

Let us assume that the time taken is t sec

We know that the acceleration due to gravity $g = 9.8m/sec^2$

As per the free falling object motion,

$V_1 = V -gt$

$-30 = -10 -9.8\times t$

$-30+10 = -9.8\times t$

$-20 = -9.8\times t$

$t =\frac{20}{9.8} = 2.04$

So, it takes total 2.04 sec

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A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10 m/s. The instan

gizmo_the_mogwai [7]

Answer:

The velocity just before hitting the ground is $v_f = 30 m/s$

Explanation:

From the question we are told that

The initial speed is $u = 10 m/s$

The final speed is $v = 30 \ m/s$

From the equations of motion we have that

$v^2 =u^2 + 2as$

Where s is the distance travelled which is the height of the cliff

So making it the subject of the the formula we have that

$s = \frac{v^2 - u^2 }{2a}$

Where a is the acceleration due to gravity with a value $a = 9.8m/s^2$

$s = \frac{30^2 - 10^2 }{2 * 9.8 }$

$s = 40.8 \ m$

Now we are told that was through horizontally with a speed of

$v_x =10 m/s$

Which implies that this would be its velocity horizontally through out the motion

Now it final velocity vertically can be mathematically evaluated as

$v_y = \sqrt{2as}$

Substituting values

$v_y = \sqrt{(2 * 9.8 * 40.8)}$

$v_y = 28.3 \ m/s$

The resultant final velocity is mathematically evaluated as

$v_f = \sqrt{v_x^2 + v_y^2}$

Substituting values

$v_f = \sqrt{10^2 + 28.3^2}$

$v_f = 30 m/s$

5 0

2 years ago

A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp

Anna71 [15]

$\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}$

Explanation:

Natural length of a spring is $0.5\text{ }m$ . The spring is streched by $0.02\text{ }m$ . The resultant energy of the spring is $0.5\text{ }J$ .

The potential energy of an ideal spring with spring constant $k$ and elongation $x$ is given by $\dfrac{1}{2}kx^{2}$ .

So, in the current problem, the natural length of the spring is not required to find the spring constant $k$ .

$\text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}$