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lyudmila [28]
2 years ago
10

A disgruntled physics student, frustrated with

Physics
1 answer:
Nata [24]2 years ago
4 0

Answer:

Explanation:

The balloon would require a time of

t = d/v = 13.5/ (23.6cos38) = 0.7259...s

to travel the horizontal distance.

the vertical position relative to the throw point at that time is

h = 0 + (23.6sin38)(0.7259) + ½(-9.8)(0.7259²)

h = 7.9652...

so as long as the adjacent building is at least 8.0 m higher than the student position, the balloon is in the air for 0.726 s.

If the building is shorter than 8.0 m above the student, the balloon will land on the building roof and will be in the air for a longer period of time

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umka21 [38]

Answehmmm

Explanation:

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A charge q1 of -5.00X10^-9 C and a charge q2 of -2.00X10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium posit
Alex73 [517]

Answer:

The equilibrium position for the third charge is 69.28 cm

Explanation:

Given;

q₁ = -5.00 x 10⁻⁹ C

q₂ = -2.00 x 10⁻⁹ C

q₃ = 15.00 x 10⁻⁹ C

distance between q₁  and q₂ = 40.0 cm = 0.4 m                                    

(-q₁)--------------------------------------(-q₂)---------------------------------(+q₃)

At equilibrium the repulsive force between q₁ and q₂ must be equal to attractive force between q₂ and q₃

According to Coulomb's law, repulsive or attractive force between charges is calculated as;

F = \frac{Kq_1q_2}{r_1^2} =  \frac{Kq_2q_3}{r_2^2}

where;

F is repulsive or attractive force between charges

K is Coulomb's constant = 8.99 x 10⁹ Nm²/c²

r₁ is the distance between q₁ and q₂

q₁, q₂ and q₃ are the charge

distance between q₂ and q₃, r₂ is calculated as;

\frac{Kq_1q_2}{r_1^2} = \frac{Kq_2q_3}{r_2^2}\\\\\frac{q_1q_2}{r_1^2} = \frac{q_2q_3}{r_2^2}\\\\r_2^2= \frac{r_1^2q_2q_3}{q_1q_2}\\\\r_2^2= \frac{r_1^2q_3}{q_1} = \frac{0.4^2*15*10^{-9}}{5*10^{-9}} = 0.48\\\\r_2 = \sqrt{0.48} = 0.6928 \ m

Therefore, the equilibrium position for the third charge is 69.28 cm

3 0
3 years ago
. In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscilla
cluponka [151]

Answer:

Spring constant, k = 1.16 N/m

Explanation:

It is given that,

Mass of the air track, m = 0.2 kg

Time, t = 2.6 s

Let T is the time period of the spring. The expression for the time and spring constant is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.2}{(2.6)^2}

k = 1.16 N/m

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What is <br> 0.02204 in scientific notation
Studentka2010 [4]

Answer:

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