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3 years ago

14

A ball is dropped and falls with an acceleration of 9.8m/s2 downward. it hit the ground with a velocity of 49 m/s downward. how

long did it take the ball to fall to the ground?

1 answer:

Lelechka [254]3 years ago

4 0

V=u+at
49=0+9.8t
t=49/9.8
t = 5 (seconds)

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A train moving with a velocity of 42.9 km/hour North, increases its speed with a uniform acceleration of 0.250 m/s^2 North until

ankoles [38]

Answer:

3658.24m

Explanation:

Hello!

the first thing that we must be clear about is that the train moves with constant acceleration

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

$\frac {Vf^{2}-Vo^2}{2.a} =X$

Vf = final speed =160km/h=44.4m/s

Vo = Initial speed =42.9km/h=11.92m/s

A = acceleration =0.25m/s^2

X = displacement

solving

$\frac {44.4^{2}-11.92^2}{2.(0.25)} =X\\X=3658.24m$

the distance traveled by the train is 3658.24m

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3 years ago

is a climate cycle where the temperature of the Ocean changes because of movements of air and ocean currents. During this period

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<h2>Answer: <u>El Niño</u> phenomenon </h2>

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To understand it better, it is necessary to know the following:

Normally in the referred area of the Pacific, trade winds blow from east to west, which move the warm water from the area to the west, allowing deeper and colder water to emerge. But, during El Niño the trade winds weaken or even flow in reverse, from west to east. It is then, when the warm water accumulates in front of the pacific coasts of South America provoking a change in the habitual patterns of precipitation and temperature.

As a consequence, hot water evaporates and condenses, causing an increase in rainfall. Due to this change in the trade winds, the climate pattern is altered, and these rains affect more the eastern part of the Pacific, while at the other end of the ocean (in Australia and Southeast Asia) the climate becomes colder and dry.

It should be noted that this is a global phenomenon that affects and extends directly or indirectly in most regions of the planet. However, its greatest impact occurs in the American countries of the Pacific coast; and in Southeast Asia and Australia.

This phenomenom begins during the final months of the year, from October and generally may last until January of the following year, although its effects are often maintained even until March.

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2 years ago

A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises

atroni [7]

Answer:

a) The velocity of the ball before it hits the floor is -6.3 m/s

b) The velocity of the ball after it hits the floor is 3.1 m/s

c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.

Explanation:

First, let´s calcualte how much time it takes the ball to hit the floor:

The equation for the position of the ball is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration due to gravity

We take the ground as the origin of the reference system.

a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s² * t²

-2.0 m = -4.9 m/s² * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The equation for the velocity of a falling object is:

v = v0 + g * t where "v" is the velocity

since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

now we replace t in the equation for position, since we know that the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t² y = 1.5 m y0 = 0 m t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration will be:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

the direction of the acceleration is upward perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

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2 years ago

The graph below show how natural processes and human activities affect climate

Kaylis [27]

Is it just me or can anyone else not see a graph? (Because I can’t)

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3 years ago

Read 2 more answers

A 5.31 kg object is swung in a vertical circular path on a string 2.99 m long. The acceleration of gravity is 9.8 m/s 2 . If the

11111nata11111 [884]

Answer:

T = 120.3 N

Explanation:

Since, the tension in the rope is acting against both the centripetal force and the weight of the stone. As both act downward towards center of the circle and tension acts towards point of support that is upward. So, tension will be equal to the sum of centripetal force and weight of the stone:

Tension = Centripetal Force + Weight of Stone

T = mv²/r + mg

where,

m = mass of stone = 5.31 kg

r = radius of circle = length of string = 2.99 m

g = 9.8 m/s²

Therefore,

T = (5.31 kg)(6.2 m/s)²/(2.99 m) + (5.31 kg)(9.8 m/s²)

T = 68.27 N + 52.03 N

<u>T = 120.3 N</u>

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2 years ago