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3 years ago

13

You throw a baseball straight up in the air so that it rises to a maximum height much greater than your height. Is the magnitude

of the ball’s acceleration greater while it is being thrown or after it leaves your hand? Explain.

1 answer:

Montano1993 [528]3 years ago

5 0

The ball's acceleration is constant in magnitude and direction, from the instant it leaves your hand, until the instant it hits the ground, no matter what direction or speed you throw it.

It's the acceleration of gravity, on whatever planet you happen to be standing when you throw the ball.

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Facts about conolizing into mars?

Stolb23 [73]

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3 years ago

A 10.3 kg block of ice slides without friction down a long track. The start of the track is 4.2 m higher than the end of the tra

Agata [3.3K]

Hello

1) Since there is no friction between the ice and the track, there is no loss of energy in the motion, so we can apply the law of conservation of energy.
The total energy E (sum of potential energy P and kinetic energy K) must be conserved:
$E=P+K$

2) At the beginning of the motion, the total energy of the object is just potential energy:
$E_1=P=mgh$
where m is the mass, $g=9.81~m/s^2$ is the gravitational acceleration, and $h=4.2~m$ is the initial height of the body.

3) At the end of the motion, this potential energy has converted into kinetic energy, and so the total energy at this point is
$E_2= \frac{1}{2}mv^2$
where m is the mass and v is the final velocity of the object.

4) We said that the total energy must be conserved, therefore we can write
$E_1 = E_2$
and so:
$mgh= \frac{1}{2}mv^2$
from which we can find v, the velocity:
$v= \sqrt{2gh}= \sqrt{2\cdot9.81~m/s^2 \cdot 4.2~m}=9.08~m/s$

8 0

3 years ago

Which statement about how light travels is true? A Light waves need a medium to travel, and they travel at the same speed even i

creativ13 [48]

The answer is a B Light wave can travel in a vacuum and travel at a contest speed even if the light source is moving

8 0

3 years ago

Read 2 more answers

A 1-kg book is at rest on a desk. Determine the force the desk exerts on the book<br>

Katena32 [7]

<h3>Answer:</h3>

$\displaystyle F_n = 9.8 \ N$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Physics</u>

<u>Forces</u>

SI Unit: Newtons N

Free Body Diagrams

Gravitational Force: $\displaystyle F_g = mg$

m is mass (in kg)
g is Earth's gravity (<em>9.8 m/s²</em>)

Normal Force: $\displaystyle F_n$

Newton's Law of Motions

Newton's 1st Law of Motion: An object at rest remains at rest and an object in motion stays in motion
Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration)
Newton's 3rd Law of Motion: For every action, there is an equal and opposite reaction

<h3>Explanation:</h3>

<u>Step 1: Define</u>

1 kg book at <em>rest</em>

<u>Step 2: FBD</u>

<em>See Attachment</em>

<em>Draw a free body diagram to label the forces acting upon the book. We see that we would have gravitational force from Earth pointing downwards and normal force from the surface of the desk pointing upwards.</em>

<em>Since the book is not moving, we know that ∑F = 0 (sum of forces equal to 0).</em>

<u>Step 4: Find Normal Force</u>

Define Forces [Newton's Law of Motions]: $\displaystyle \sum F = 0$
[Newton's Law of Motions] Substitute in forces: $\displaystyle F_g - F_n = 0$
[Newton's Law of Motions] [Addition Property of Equality] Isolate $\displaystyle F_n$ : $\displaystyle F_g = F_n$
[Newton's Law of Motions] Substitute in $\displaystyle F_g$ : $\displaystyle mg = F_n$
[Newton's Law of Motions] Rewrite: $\displaystyle F_n = mg$
[Newton's Law of Motions] Substitute in variables: $\displaystyle F_n = (1 \ kg)(9.8 \ \frac{m}{s^2})$
[Newton's Law of Motions] Multiply: $\displaystyle F_n = 9.8 \ N$

7 0

3 years ago

The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.56 m and 1.00 m, respectively. The respective

valentina_108 [34]

Answer:

Explanation:

Given

Diameter of main rotor $d_1=7.56 m$

Tail rotor $d_2=1 m$

$N_1=453 rev/min$

$N_2=4137 rev/min$

$\omega =\frac{2\pi N}{60}$

$\omega _1=\frac{2\pi 453}{60}=47.44 rad/s$

$\omega _2=\frac{2\pi 4137}{60}=433.28 rad/s$

Speed of the tip of main rotor $=\omega _1\times r_1=47.44\times \frac{7.56}{2}=179.32 m/s$

Speed of tail rotor $=\omega _2\times r_2=433.28\times \frac{1}{2}=216.64 m/s$

7 0

3 years ago