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steposvetlana [31]
3 years ago
14

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 catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.6 m/s at ground level.
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Answer:

The rocket is motion above the ground for 44.7 s.

Explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engine fails:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 =  initial height

v0 = initial speed

t=  time

a = acceleration due to the engines

g = acceleration due to gravity

First, let´s calculate the time the rocket is being accelerated by the engines:

(The center of the framer of reference is located at the ground, y0 = 0).

y = y0 + v0 · t + 1/2 · a · t²

1190 m = 0 m + 79.6 m/s · t + 1/2 · 4.10 m/s² · t²

0 = 2.05  m/s² · t² + 79.6 m/s · t - 1190 m

Solving the quadratic equation:

t = 11.5 s  (the other value of t is discarded because it is negative).

At that time, the engines fail and the rocket starts to fall. The equation of the height will be:

y = y0 + v0 · t + 1/2 · g · t²

The initial velocity (v0) will be the velocity acquired during 11.5 s of acceleration plus the initial velocity of launch:

v = v0 + a · t

v = 79.6 m/s + 4.10 m/s² · 11.5 s

v = 126.8 m/s

Now, we can calculate the time it takes the rocket to reach the ground (y = 0) from a height of 1190 m:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 1190 m + 126.8 m/s · t - 1/2 · 9.80 m/s² · t²

Solving the quadratic equation:

t = 33.2 s

Then, the total time the rocket is in motion is (33.2 s + 11.5 s) 44.7 s

 

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A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 25.0 kg of water at 20.0°C. What is the final te
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Answer:

Te =  23.4 °C

Explanation:

Given:-

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- The initial temperature of horseshoe, Ti_h = 550°C

- The specific heat capacity of iron, ci = 448 J/kgC

- The mass of water, M = 25 kg

- The initial temperature of water, Ti_w = 20°C

- The specific heat capacity of water, cw = 4186 J/kgC

Find:-

What is the final temperature of the water–horseshoe system?

Solution:-

- The interaction of horseshoe and water at their respective initial temperatures will obey the Zeroth and First Law of thermodynamics. The horseshoe at higher temperature comes in thermal equilibrium with the water at lower temperature. We denote the equilibrium temperature as (Te) and apply the First Law of thermodynamics on the system:

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                             Te = [ m*ci*( Ti_h ) + M*cw*( Ti_w ) ] / [ m*ci + M*cw ]

- Plug in the values and evaluate (Te):

                             Te = [1.5*448*550 + 25*4186*20 ] / [ 1.5*448 + 25*4186 ]

                             Te = 2462600 / 105322

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