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Shalnov [3]
3 years ago
5

Several objects were dropped from a balcony. Use the data below to draw conclusions about the results of these drops. Be sure to

:
Describe the rate of speed for each object and tell when or in what order they will hit the ground with and without factoring in air resistance.
Describe the force each object will exert on the ground.
Use the data to support your conclusion statements.




Object

Mass

Diameter/Surface area

Height of drop

Tennis ball

0.06 kg

6.6 cm

5 m

Baseball

0.15 kg

7.4 cm

5 m

Bowling ball

5 kg

21 cm

5 m

Golf ball

0.04 kg

4.3 cm

5 m



i mainly need to know the rate of speed, acceleration, and force of mass
Physics
1 answer:
levacccp [35]3 years ago
3 0

The answer is wait can u explain iagaing Again*

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A rifle fires a 2.10 x 10^-2 kg pellet straight upward, because the pellet rests on a compressed spring when the trigger is is p
yKpoI14uk [10]

m = mass of pellet = 2.10 x 10⁻² kg

x = compression of spring at the time launch = 9.10 x 10⁻² m

h = height gained by the pellet above the initial position = 6.10 m

k = spring constant

using conservation of energy

spring potential energy = gravitational potential energy of pellet

(0.5) k x² = m g h

inserting the values

(0.5) k (9.10 x 10⁻²)² = (2.10 x 10⁻²) (9.8) (6.10)

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3 years ago
Use the data provided to calculate the gravitational
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Answer:

3 kg:   14.7 kJ

6 kg:   29.4 kJ

9 kg:  44.1 kJ

Explanation:

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8 0
4 years ago
A student rings a brass bell with a frequency of 200 Hz. The sound wave
djverab [1.8K]

Answer:

Option D. 23.5 m

Explanation:

From the question given above, the following data were obtained:

Frequency = 200 Hz

Speed of sound in brass = 4700 m/s

Wavelength of sound in brass =?

We can obtain the wavelength of the sound in the brass by using the following formula as illustrated below:

Wave speed = wavelength × frequency

4700 = wavelength × 200

Divide both side by 200

Wavelength = 4700 / 200

Wavelength = 23.5 m

Thus, the wavelength of the sound in the brass is 23.5 m

7 0
3 years ago
A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular
bagirrra123 [75]

To solve this problem we will use the concepts related to energy conservation. Both potential energy, such as rotational and linear kinetic energy, must be conserved, and the gain in kinetic energy must be proportional to the loss in potential energy and vice versa. This is mathematically

PE = KE_{lineal} + KE_{rotational}

mgh = \frac{1}{2}mv^2 +\frac{1}{2} I\omega^2

Where,

m = mass

v = Tangential Velocity

\omega = Angular velocity

I = Moment of Inertia

g = Gravity

Replacing the value of Inertia in a Disk and rearranging to find h, we have

mgh = \frac{1}{2}mv^2 +\frac{1}{2} I\omega^2

mgh = \frac{1}{2}mr^2\omega^2 + \frac{1}{2}(\frac{1}{2}mr^2)\omega^2 )

h = \frac{3}{4} \frac{r^2\omega^2}{g}

Replacing,

h = \frac{3}{4} \frac{(1.6)^2 (5.35)^2}{9.8}

h = 5.607m

Therefore the height of the inclined plane is 5.6m

3 0
3 years ago
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