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nika2105 [10]
3 years ago
6

Near the surface of the Earth, objects in free fall (but not terminal velocity) experience

Physics
2 answers:
Aleks [24]3 years ago
5 0
B (9.81 m/s^2)
Speed no, because acceleration isn't 0
Velocity, pretty much same as speed
Distance no, because it's getting closer

amid [387]3 years ago
4 0

Answer: constant acceleration

Explanation: constant acceleration is maintain when a body is falling neglecting air resistance.

The distance gets closer so it's impossible to say constant distance.

It can't experience constant speed because the distance is not constant.

It can't experience constant velocity because the displacement is not constant

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An unstrained horizontal spring has a length of 0.26 m and a spring constant of 180 N/m. Two small charged objects are attached
MakcuM [25]

Answer:

a)

two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)

b)

both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

Explanation:

Given that;

L = 0.26 m

k = 180 N/m

x = 0.039 m

a)

we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.

b)

Spring force F = kx

F = 180 × 0.039

F = 7.02 N

Now, Electrostatic force F = Keq²/r²

where r = L + x = ( 0.26 + 0.039 )

we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2

so from the equation; F = Keq²/r²

Fr² = Keq²

q = √ ( Fr² / Ke )

we substitute

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q =  √ (0.627595 / 9×10⁹)

q = √(6.97 × 10⁻¹¹)

q = 8.35 × 10⁻⁶ C

Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

5 0
3 years ago
I WILL MARK YOU THE BRAINLIEST NO LINKS<br>What goes were put them in the correct place
PIT_PIT [208]

Answer:

Fluid fricton goes to Static friction and sliding friction goes to rolling friction

Explanation:

6 0
3 years ago
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A certain type of laser emits light that has a frequency of 4.2 × 1014 Hz. The light, however, occurs as a series of short pulse
bogdanovich [222]

Explanation:

It is given that,

Frequency of the laser light, f=4.2\times 10^{14}\ Hz

Time, t=3.2\times 10^{-11}\ s

(a) Let \lambda is the wavelength of this light. It can be calculated as :

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8}{4.2\times 10^{14}}

\lambda=7.14\times 10^{-7}\ m

or

\lambda=714\ nm

(b) Let n is the number of the wavelengths in one pulse. It can be calculated as :

n=f\times t

n=4.2\times 10^{14}\times 3.2\times 10^{-11}

n = 13440

Hence, this is the required solution.

8 0
3 years ago
If riding a lawnmower engine exerts 19 hp in one minute to move the lawnmower how much work is done
ioda

Answer:

the work done by the lawnmower is 236.14 J.

Explanation:

Given;

power exerted by the lawnmower engine, P = 19 hp

time in which the power was exerted, t = 1 minute = 60 s.

1 hp = 745.7 watts

The work done by the lawnmower is calculated as follows;

Work = Energy = \frac{Power}{time} \\\\Work = \frac{(19 \times 745.7)}{60} = 236.14 \ J

Therefore, the work done by the lawnmower is 236.14 J.

6 0
3 years ago
Suppose you are on an airplane moving at high speed. If you flip a coin straight up it will land in your lap rather than a great
maxonik [38]
It will land in your lap because there's different frames of motion relative to yourself. For example, if you're running at a speed of 6 mph, it doesn't mean you'll run as fast as the Earth spins. Also, since you're on the interior of the plane, any kind of wind or weather on the outside will not affect the coin. A law to back up this claim is Einsteins Special Law of Relativity.
8 0
3 years ago
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