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Oliga [24]
3 years ago
12

What is the length of a string with a mass of 2.5 kg, with a

Physics
1 answer:
natulia [17]3 years ago
4 0
B 0.4 m should be correct if not it is A
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3.315 moles of MgCl2
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Bettina spoke into a microphone during the school play to increase the sound of her voice so the audience could hear her speak.
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A woman walked 115 m. As she did so, her speed increased from 4.20 m/s to 5.00 m/s. How long did it take her to walk this distan
ASHA 777 [7]

Answer:

25 seconds

Explanation:

Assuming the woman is accelerating at a constant rate of a \;\;m/s^2 from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.

Let she takes t seconds to cover the distance, s=115 m.

As acceleration, a=\frac{v-u}{t}=\frac{5-4.2}{t}

\Rightarrow at=0.8\cdots(i)

Now, from the equation of motion

s=ut+\frac 12 at^2

\Rightarrow s=ut+\frac 12 at(t)

\Rightarrow 115=4.2t+\frac 12 \times 0.8 t [ from equation (i)]

\Rightarrow 115=(4.2+0.4)t

\Rightarrow t= 115/4.6 = 25 seconds.

Hence, she takes 25 seconds to walk the distance.

5 0
3 years ago
When is the absorption spectrum of an electron generated
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Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe
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Answer:

1) v₀x = 13.76 m/s

2) v₀y = 19.66 m/s

3) ymax = 21.199 m

4) X = 55.1746 m

5) and 6) y = 18.4 m

Explanation:

1) v₀x = v₀*Cos α = 24 m/s* Cos 55° = 13.76 m/s

2) v₀y = v₀*Sin α = 24 m/s* Sin 55° = 19.66 m/s

3) ymax = y₀ + (v₀y²/(2g)) = 1.5 m + ((19.66 m/s)²/(2*9.81 m/s²)) = 21.199 m

4) We can use this equation

y = y₀ + (tan α)*x – (g / (2* v₀x²))*x²

where y = y₀ = 1.5 m

then

1.5 = 1.5 + tan (55°)*x - (9.81 / (2* (13.76)²))*x²

⇒   0.02588 x² - 1.42815 x = 0

Solving this equation we get

x₁ = 0     and    x₂ = 55.1746 m

The distance between the two girls is 55.1746 m

5) and 6) If   v₀x = 15 m/s = vx   and   ymax = 24 m

y = ?   when x = (xmax/2)

ymax = y₀ + (v₀y²/(2g)) ⇒ v₀y = √(2g*(ymax - y₀))

⇒ v₀y = √(2(9.81 m/s²)(24 m - 1.5 m)) = 21.01 m/s

then we get α' as follows

α' = tan⁻¹(v₀y/v₀x) = tan⁻¹(21.01 m/s/15 m/s) = 54.47°

v₀ = √(v₀x² + v₀y²) = √((15 m/s)² + (21.01 m/s)²) = 25.81 m/s

Now we can apply the equation of the path

y = ymax - ((gx²)/(2v₀²))

⇒  y = 24m - ((9.81)(55.1746/2)²/(2*25.81²))

⇒  y = 18.4 m

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3 years ago
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