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Paladinen [302]
2 years ago
8

9. A Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table. How long does it take to hit the floor if no one sneezes? Wh

at is its velocity when it hits the floor?
Physics
1 answer:
Paha777 [63]2 years ago
8 0

By Considering the vertical distance and both vertical and horizontal final velocity, the time t = 0.45 s and Velocity V = 6.7 m/s

Given that a Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table.

Height h = 1.0 m

As the ball rolls off the table, it will be fallen under gravity. Where

g = 9.8 m/s^{2}

Initial vertical velocity u_{y} = 0

Initial horizontal velocity u_{x} = 5 m/s

Considering the vertical distance, the formula to use to calculate the time will be;

h = ut + 1/2gt^{2}

1 = 0 + 1/2 x 9.8t^{2}

1 = 4.9t^{2}

t^{2} = 1/4.9

t = \sqrt{0.204}

t = 0.45 seconds

It takes 0.45 seconds to hit the floor if no one sneezes.

To calculate its velocity when it hits the floor, we will need to calculate for both vertical and horizontal final velocity and find the resultant velocity of the two.

Vertical component

V_{y} = U_{y} + gt

V_{y} = 0 + 9.8(0.45)

V_{y} = 4.41 m/s

Horizontal component

V_{x} = u_{x} + at

but a = 0

V_{x} = 5 m/s

Final velocity V = \sqrt{5^{2} + 4.41^{2}  }

V = 6.67 m/s

Therefore, it will hit the floor at a velocity of 6.7 m/s

Learn more here: brainly.com/question/5063616

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If a ball that is 10 meters above the ground is thrown horizontally at 5.51 meters per second. a. how long will it take for the
GalinKa [24]

Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height = -10 m

y_{0}: is the initial height = 0

v_{0_{y}}: is the initial speed in the vertical direction = 0

g: is the acceleration due to gravity = 9.81 m/s²

By solving the above equation for "t" we have:

t = \sqrt{\frac{2y_{f}}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s

Hence, the ball will hit the ground in 1.43 s.

b. The distance in the horizontal direction can be found as follows:

x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2}

Where:

x₀: is the initial position in the horizontal direction = 0

a: is the acceleration in the horizontal direction = 0 (it is moving at constant speed)

x_{f} = 5.51 m/s*1.43 s = 7.88 m

Therefore, the ball will travel 7.88 m before it hits the ground.

I hope it helps you!

4 0
2 years ago
The line graph below shows the number of downloads of two songs after
daser333 [38]

Answer: D

1200

Explanation:

Song 1 is spotted with a cube sign.

At 3 minute, trace the spot to the vertical axis. And you will notice that it a little bit above 10.

Since it is above 10, let assume it is equal to 12.

The number of song downloaded are in hundreds. Therefore, multiply the 12 by 100

12 × 100 = 1200 downloads

Approximately, song 1 has 1200 downloads at minute 3

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PLEASE HELP ME WITH THIS ONE QUESTION
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Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>

[A] = Our incognite

k is constant decay:

k = ln 2 / Half-life

k = ln 2 / 4.96 x 10^3 s

k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]

ln[A] = 37.538

[A] = 2.01x10¹⁶ nuclei remain ≈

<h3>c) 2.02 x 10^16 nuclei</h3>
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12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it
vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

v_{f} = v_{o}+a\cdot (t-t_{o}) (1)

Where:

v_{o} - Initial velocity, measured in meters per second.

v_{f} - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t_{o} - Initial time, measured in seconds.

t - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust (v_{o} = 0\,\frac{m}{s}, a = 30\,\frac{m}{s^{2}}, t_{o} = 0\,s, 0\,s\le t< 30\,s)

v = 30\cdot t (2)

Free fall (v_{o} = 900\,\frac{m}{s}, a = -9.807\,\frac{m}{s}, t_{o} = 30\,s, 30\,s \le t \le 120\,s)

v = 900-9.81\cdot (t-30) (3)

Now we created the graph speed-time, which can be seen below.

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