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Tasya [4]
3 years ago
13

A 0.203-kg plastic ball moves with a velocity of 0.30 m/s. it collides with a second plastic ball of mass 0.092 kg, which is mov

ing along the same line at a speed of 0.10 m/s. after the collision, both balls continue moving in the same, original direction. the speed of the 0.092-kg ball is 0.26 m/s. what is the new velocity of the 0.203-kg ball?
Physics
1 answer:
dolphi86 [110]3 years ago
7 0
<span><span>A 0.200 kg plastic ball moves with a velocity of0.30 m</span>s<span>A 0.205-kg plastic ball moves with a velocity of0.30 m</span><span>A 0.199 kg plastic ball moves with a velocity of0.30 m</span><span>A 0.204-kg plastic ball moves with a velocity of0.30 m</span><span>A 100 g ball moving to the right at 4.0 m</span>s collides<span>have less momentum if the velocities</span><span>the same</span><span>A ball with a momentum of 4.0 kg•<span>m</span></span></span>
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What is the voltage at resistor #3? (must include unit - V)
Romashka-Z-Leto [24]

Explanation:

I = 120/30=2A

Voltage across R3= current × resistance

= 2A × 15 ohm = 30 volt

5 0
3 years ago
What cannot be used to dry utensils?​
Mashcka [7]

Answer:

Wiping cloth

Explanation:

When the wiping cloth is used to “clean” another surface, it will contaminate it with the bacteria. To avoid bacterial growth on the wiping cloth, a wet wiping cloth must be stored in a proper sanitizing solution between uses. Provide Proper Concentration of. Sanitizing Solution for Wiping Cloths.

wiping cloth cannot be used on a surface where raw meat was cleaned, utensils, plates and other kitchen eating accessories because of it's ability to contact germs and bacteria.

3 0
3 years ago
Drew observed in an experiment that algae in a nearby lake could be found at six meters below the water's surface on a clear day
babymother [125]

In this question the options are missing; here are the options:

What was the dependent variable in Drew's experiment?

A. The depth at which the algae were found

B. The sky conditions on a particular day

C. The amount of algae measured

D. The lake that was being observed

The answer to this question A. The depth at which the algae were found

Explanation:

In an experiment, it is common there are at least two factors or variables. Additionally, the variable that is modified by others or that depend on others is always the dependent variable.

In the case of the experiment presented, there are two main factors: sky conditions and depth at which algae can be found. From these, the dependent factor is the depth because this depth changes with the sky condition or depends on the sky conditions. Also, the dependent variable is always the factor being studied, for example, in this case, Drew's focus is to study how the location of algae in terms of depth changes.

8 0
3 years ago
It is known that the gravitational force of attraction between two alpha particles is much weaker than the electrical repulsion.
natali 33 [55]

Answer:

<em>The ratio of gravitational force to electrical force is 3.19 x 10^-36 </em>

<em></em>

Explanation:

mass of an alpha particle = 6.64 x 10^{-27} kg

charge on an alpha particle = +2e = +2(1.6 x 10^{-19} C) = 3.2 x 10^{-19} C

distance between particles = d

For gravitational attraction:

The force of gravitational attraction F = \frac{Gm^{2} }{r^{2} }

where G = gravitational constant = 6.67 x 10^{-11} m^3 kg^-1 s^-2

r = the distance between the particles = d

m = the mass of each particle

therefore, gravitational force = \frac{6.67*10^{-11}*(6.64*10^{-27} )^{2}  }{d^{2} } = \frac{2.94*10^{-63} }{d^{2} }  Newton

For electrical repulsion:

Electrical force between the particles = \frac{-kQ^{2} }{r^{2} }

where k is the Coulomb's constant = 9.0 x 10^{9} N•m^2/C^2

r = distance between the particles = d

Q = charge on each particle

therefore, electrical force = \frac{-9*10^{9}*(3.2*10^{-19} )^{2}  }{d^{2} } = \frac{-9.216*10^{-28} }{d^{2} } Newton

the negative sign implies that there is a repulsion on the particles due to their like charges.

Ratio of the magnitude of gravitation to electrical force = \frac{2.94*10^{-63} }{9.216*10^{-28} }

==> <em>3.19 x 10^-36 </em>

8 0
3 years ago
Which of these properties is the best measure of a star's brightness? (1 point)
storchak [24]

Answer:

ABSOLUTE MAGNITUDE, C

Explanation:

Absolute magnitude is how bright the star is from a distance of 10 parsecs. Apparent magnitude is how bright the star appears from Earth, which can vary, because some stars may be farther or closer away. Age and size are not very accurate either. For example, at the end of their lifetimes, very big start may explode in a violent and bright supernova, and outshine the stars near them. As for size, as many stars several times smaller than the sun shine much brighter than it, making it inaccurate.

6 0
2 years ago
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