A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration

Simple machines are widely used in our daily life. why?<br>

HACTEHA [7]

Answer:

Simple machines are useful because they reduce effort or extend the ability of people to perform tasks beyond their normal capabilities. Simple machines that are widely used include the wheel and axle, pulley, inclined plane, screw, wedge and lever.

7 0

3 years ago

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A red ball is thrown down with an initial speed of 1.6 m/s from a height of 25 meters above the ground. Then, 0.4 seconds after

Radda [10]

Answer:

1. v = 22.2 m/s

2. t = 2.25 seconds

3. h = 27.05 m

4. t = 1.16 seconds

Explanation:

The questions involve motion under the influence of gravity

1. Using the formula v² = u² + 2gh

where u = 1.6 m/s; g= 9.81 m/s²; h = 25 m; v = ?

v² = (1.6)² + 2 * 9.81 * 25

√v² = √493.06

v = 22.2 m/s

2. Using h = ut + 1/2 gt²

where h = 25 m; u = 0 (since velocity on reaching the ground is zero); g = 9.81 m/s²; t = ?

therefore, h = 1/2 gt²

making t subject of the formula, t = √ (2*h /g)

t = √ (2 * 25 / 9.81)

t = 2.25 seconds

3. Time of travel for the blue ball, t = 2 - 0.4 = 1.6s

using h = ut - gt²

u = 24 m/s; t = 1.6 s; g = 9.81 m/s²

note: since the ball is travelling against gravity, g is negative

h = 24 * 1.6 - 11/2 * 9.81 * 1.6²

h = 38.4 - 12.55 = 25.85 m

since height above the ground is 1.2 m,

total height h = 25.85 m + 1.2 m

h = 27.05 m

4. Let the time of travel of the red ball be t seconds.

So the time of travel of the blue ball = (t - 0.4) seconds.

Both the balls are at the same height :

25 - s = 1.2 + h where s & h are the displacements of the red & the blue ball respectively.

25 - (ut + 1/2 gt2) = 1.2 + (ut - 1/2 gt2)

25 - (1.6 t + 0.5 * 9.8 t²) = 1.2 + (24(t-0.4) - 0.5*9.8*(t-0.4)²)

solving the equation above for the time after which both the balls are at the same height.

25 - 1.6t - 4.9t² = 1.2 + 24t - 9.6 - 4.9t² + 3.92t - 0.784

collecting like terms

(25 - 1.2 + 9.6 + 0.784) = (24 + 3.92 + 1.6) * t

t = 34.184 / 33.44

t = 1.16 seconds

6 0

3 years ago

After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.

GarryVolchara [31]

Answer:

$Imp = 11.666\,\frac{kg\cdot m}{s}$

Explanation:

Speed experimented by the ball before and after collision are determined by using Principle of Energy Conservation:

Before collision:

$(0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (19\,m) = \frac{1}{2}\cdot (0.320\,kg)\cdot v_{A}^{2}$

$v_{A} \approx 19.304\,\frac{m}{s}$

After collision:

$\frac{1}{2}\cdot (0.320\,kg)\cdot v_{B}^{2} = (0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (15\,m)$

$v_{B} \approx 17.153\,\frac{m}{s}$

The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

$Imp = (0.32\,kg)\cdot [(17.153\,\frac{m}{s} )-(-19.304\,\frac{m}{s} )]$

$Imp = 11.666\,\frac{kg\cdot m}{s}$

5 0

3 years ago

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What is the difference between a tropical storm and a hurricane

vazorg [7]

Answer and Explanation:

Tropical storm:

The point at which the tropical depression intensifies and can sustain a maximum wind speed in the range of 39-73 mph, it is termed as a tropical storm.

Hurricane:

A hurricane is that type of tropical cyclone which includes with it high speed winds and thunderstorms and the intensity of the sustained wind speed is 74 mph or more than this.

Hurricanes are the most intense tropical cyclones.

The only difference between the tropical storm and hurricane is in the intensity.

7 0

4 years ago

An astronaut and his space suit have a combined mass of 157 kg. The

alexgriva [62]

Answer:

v₃ = 9.62[m/s]

Explanation:

To solve this type of problem we must use the principle of conservation of linear momentum, which tells us that the momentum is equal to the product of mass by velocity.

We must analyze the moment when the astronaut launches the toolkit, the before and after. In order to return to the ship, the astronaut must launch the toolkit in the opposite direction to the movement.

Let's take the leftward movement as negative, which is when the astronaut moves away from the ship, and rightward as positive, which is when he approaches the ship.

In this way, we can construct the following equation.

$-(m_{1}+m_{2})*v_{1}=(m_{1}*v_{2})-(m_{2}*v_{3})$

where:

m₁ = mass of the astronaut = 157 [kg]

m₂ = mass of the toolkit = 5 [kg]

v₁ = velocity combined of the astronaut and the toolkit before throwing the toolkit = 0.2 [m/s]

v₂ = velocity for returning back to the ship after throwing the toolkit [m/s]

v₃ = velocity at which the toolkit should be thrown [m/s]

Now replacing:

$-(157+5)*0.2=(157*0.1)-(5*v_{3})\\(5*v_{3})= 15.7+32.4\\v_{3}=9.62[m/s]$

6 0

3 years ago