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konstantin123 [22]
3 years ago
15

Factors that affect the force of gravity between objects include their distance apart and their

Physics
1 answer:
mr_godi [17]3 years ago
6 0

Answer:

factors that affect the force of gravity between objects include their distance apart and their mass.

Explanation:

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A stone was dropped off a cliff and hit the ground with a speed of 96 ft/s. What is the height of the cliff? (Use 32 ft/s2 for t
nadezda [96]
The initial velocity of the stone is 0 ft/s. Given the initial velocity (Vi), final velocity (Vf), and acceleration due to gravity (g), the distance may be calculated through the equation,
                                     d = ((Vf)² - (Vi)²) / 2g
Substituting the known values,
                                     d = ((96 ft/s)² - 0))/ (2x32.2)
The value of d is 143.10 ft. 
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3 years ago
What is mesothermal climates
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Mesothermal regions have moderate climates. They are not cold enough to sustain a layer of winter snow but are also not remain warm enough to support flowering plants (and, thus, evapotranspiration) all year.

Explanation:

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1 year ago
Definition of waves ​
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A wave can be described as a disturbance that travels through a medium from one location to another location.

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2 years ago
A scientist shines light from a source onto a piece of metal, and no electrons are released by the metal. Increasing the intensi
Nezavi [6.7K]

This is the photoelectric effect, and it is best explained by the particle model of light.

<h3>What is the photoelectric effect?</h3>

The photoelectric effect refers to the emission of negatively charged particles and electromagnetic radiation that hits an object.

The photoelectric effect shows how electrons can be released from a given object when this material is absorbing electromagnetic radiation.

The photoelectric effect is a fundamental piece of evidence for understanding the nature of light particles.

Learn more about the photoelectric effect here:

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7 0
2 years ago
A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito
Nezavi [6.7K]

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

c =  \frac{c1 \times c2}{c1 + c2}

=  \frac{2 \times 4}{2 + 4}

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

= 1.33 \times 10 {}^{ - 6}  \times 8

= 10.64 \times 10 {}^{ - 6}

charge on 2.0μf capacitor is

\frac{Qeq}{2 \times 10 {}^{ - 6} }

=  \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} }

=5.32v

learn more about series capacitance from here: brainly.com/question/28166078

#SPJ4

3 0
2 years ago
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