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2 years ago

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Write a hypothesis about the effect of the angle of the track on the acceleration of the cart. Use the "if . . . then . . . beca

use . . .” format and be sure to answer the lesson question: "How does an object's position and velocity change as the object accelerates?”

1 answer:

seropon [69]2 years ago

5 0

Answer:

Sample Response: If the angle of the track increases, then the acceleration of the cart increases because a greater track angle increases the cart's velocity.

Explanation:

because.

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A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from

Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat ( $U_{g}$ ), in joules, is equal to the sum of the final translational kinetic energy ( $K$ ), in joules, and work losses due to air resistance ( $W_{l}$ ), in joules:

$U_{g} = K +W_{l}$ (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

$m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l}$ (2)

Where:

$m$ - Mass of the cat, in kilograms.

$g$ - Gravitational acceleration, in meters per square second.

$h$ - Initial height of the cat, in meters.

$v$ - Final speed of the cat, in meters per second.

If we know that $m = 2.70\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h = 5.20\,m$ and $W_{l} = 120\,J$ , then the final speed of the cat is:

$v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }$

$v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }$

$v \approx 7.586\,\frac{m}{s}$

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

4 0

2 years ago

Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi

AleksandrR [38]

The horizontal component is calculated as:
Vhorizontal = V · cos(angle)

In your case Vhoriontal = 16 · cos(40) = 12.3 m/s

Answer: 12.3 m/s

7 0

2 years ago

Read 2 more answers

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed?

sweet [91]

Answer:

Speed of the electron will be $v=5.896\times 10^7m/sec$

Explanation:

We have given that charge on electron $e=1.6\times 10^{-19}C$

Mass of electron $m=9.11\times 10^{-31}kg$

Potential difference = $V=9.9KV=9.9\times 10^3volt$

Now according to energy conservation $eV=\frac{1}{2}mv^2$

$1.6\times 10^{-19}\times 9.9\times 10^3=\frac{1}{2}\times 9.11\times 10^{-31}v^2$

$v=5.896\times 10^7m/sec$

4 0

1 year ago

The suspension cable of a 1,000 kg elevator snaps, sending the elevator moving downward through its shaft. The emergency brakes

tester [92]

Answer:

option (E) 1,000,000 J

Explanation:

Given:

Mass of the suspension cable, m = 1,000 kg

Distance, h = 100 m

Now,

from the work energy theorem

Work done by the gravity = Work done by brake

or

mgh = Work done by brake

where, g is the acceleration due to the gravity = 10 m/s²

or

Work done by brake = 1000 × 10 × 100

or

Work done by brake = 1,000,000 J

this work done is the release of heat in the brakes

Hence, the correct answer is option (E) 1,000,000 J

4 0

2 years ago

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init

Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

$h=\frac{gt^2}{2}$

For second stone

$h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$ ---2

Equating 1 &2 we get

$\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$

$\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$26.46t-35.721-52.92t+142.884=0$

$t=4.05 s$

4 0

2 years ago