I need help question 1 of science.

A ball is traveling uphill with an initial velocity of 5.0 m/s and an acceleration of -2.0 m/s^2. A) How fast is the ball travel

Rzqust [24]

Answer:

A) The ball is traveling at 5.0 m/s (magnitude) when the ball returns to its release point.

B) The maximum uphill position is at 6.25 m from the release point.

C) On the way up, the velocity of the ball at x = 6.0 m is 1 m/s and on the way down it is - 1m/s.

Explanation:

Hi there!

The position and velocity of the ball can be calculated using the following equations:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of th ball at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

v = velocity at time t.

A) Let´s place the origin of the frame of reference at the point at which the ball has a velocity of 5.0 m/s. Then, x0 = 0.

When the ball returns to the initial point, its position will be 0. Then using the equation of position we can calculate at which time the ball is at x = 0:

x = x0 + v0 · t + 1/2 · a · t²

0 m = 5.0 m/s · t - 1/2 · 2.0 m/s² · t²

0 m = 5.0 m/s · t - 1.0 m/s² · t²

0 m = t (5.0 m/s - 1.0 m/s² · t)

t = 0 (this is logic becuase the ball starts at x = 0)

and

5.0 m/s - 1.0 m/s² · t = 0

t = -5.0 m/s / -1.0 m/s²

t = 5.0 s

With this time, we can calculate the velocity of the ball:

v = v0 + a · t

v = 5.0 m/s - 2.0 m/s² · 5.0 s

v = -5.0 m/s

The ball is traveling at 5.0 m/s when the ball returns to its release point.

B) Let´s use the equation of velocity to obtain the time at which the ball is at its maximum uphill position:

v = v0 + a · t

0 = 5.0 m/s - 2.0 m/s² · t

-5.0 m/s/ -2.0 m/s² = t

t = 2.5 s

Now, using the equation of position, let´s find the position of the ball at t = 2.5 s. This position will be the maximum uphill position because at that time the velocity is 0:

x = x0 + v0 · t + 1/2 · a · t²

x = 5.0 m/s · 2.5 s - 1/2 · 2.0 m/s² · (2.5 s)²

x = 6.25 m

The maximum uphill position is at 6.25 m from the release point.

C) First, let´s find the time at which the ball is 6.0 meters uphill from the releasing point:

x = x0 + v0 · t + 1/2 · a · t²

6.0 m = 5.0 m/s · t - 1/2 · 2 m/s² · t²

0 = -1 m/s² · t² + 5.0 m/s · t - 6.0 m

Solving the quadratic equation using the quadratic formula:

a = -1

b = 5

c = -6

t = [-b ± √(b² - 4ac)]/2a

t₁ = 2 s (on its way up)

t₂ = 3 s (on its way down)

Now, let´s calculate the velocity of the ball at those times:

v = v0 + a · t

v = 5.0 m/s - 2 m/s² · 2 s = 1 m/s

v = 5.0 m/s - 2 m/s² · 3 s = -1 m/s

On the way up, the velocity of the ball at x = 6.0 m is 1 m/s and on the way down it is - 1m/s.

7 0

3 years ago

Technician A says test lights are great for quick tests on non-computerized circuits. Technician B says you can use a test light

Tpy6a [65]

Answer:

that technician A is right

Explanation:

The test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits, these two values are high and can light the bulb. In digital circuits the current is very small in the order of milliamps, so there is not enough power to turn on these lights.

From the above it is seen that technician A is right

4 0

4 years ago

The blade of a windshield wiper moves through an angle of 180.0 degrees in 0.500 s. The tip of the blade moves on the arc of a c

sweet-ann [11.9K]

Answer:

Centripetal acceleration will be equal to $10.66m/sec^2$

Explanation:

We have given time taken to cover 180° is 0.5 sec

So time taken by 360° is equal to = 2×0.5 = 1 sec

Radius of the circle r = 0.520 m

So distance $d=2\pi r=2\times3.14\times 0.520=3.2656m$

So velocity $v=\frac{distance}{time}=\frac{.2656}{1}=3.2656m/sec$

We have to find the centripetal acceleration

Centripetal acceleration will be equal to $a_c=\frac{v^2}{r}=\frac{3.2656^2}{1}=10.66m/sec^2$

So centripetal acceleration will be equal to $10.66m/sec^2$

7 0

4 years ago

Lora (of mass 43.6 kg) is an expert skier. She starts at 3.6 m/s at the top of the lynx run, which is 67 m above the bottom. Wha

Likurg_2 [28]

Explanation:

As per the law of conservation of energy, the final mechanical energy of Lora is equal to its initial mechanical energy. So, when Lora is at the bottom of ski run then her potential energy will change into kinetic energy.

Hence, $E_{initial} = \frac{mv_{i}^{2}}{2} + mgh
$

$E_{final} = \frac{mv_{f}^{2}}{2}$

Now, final kinetic energy that will be at the bottom of the ski run is as follows.

Let, $E_{k} = E_{final}$

$E_{intial} = E_{final}$

$E_{k} = \frac{mv_{i}^{2}}{2} + mgh
$

= $\frac{(43.6 \times (3.6)^{2}}{2} + (43.6 \times 9.81 \times 67)$

= 282.53 + 28656.97

= 28939.502 J

Thus, we can conclude that her final kinetic energy at the bottom of the ski run is 28939.502 J.

3 0

3 years ago

Components grouped together for a particular function form a _____ system.

Alina [70]

A natural system is formed when components are grouped together so as to perform a more specific function or use. In addition, this system is bound to exist without any human interference. Biological and non-biological components work hand-in-hand as they do nature processes.

4 0

3 years ago