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4 years ago

13

The destruction of the ozone layer primarily occurs in the mesosphere. thermosphere. stratosphere. troposphere.

2 answers:

otez555 [7]4 years ago

7 0

Hello,

The correct answer is D) troposphere

Hope this helps!!!! :)

**(Vanessa)**

Fofino [41]4 years ago

4 0

I believe the answer would be Troposphere.

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For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

4 years ago

the goat weighs 900 N and is 1 meter from the fulcrum. The strongman pulls down on the lever 3 meters from the fulcrum. What is

garik1379 [7]

Answer: The smallest effort = 300N

Explanation:

Using one of the condition for the attainment of equilibrium:

Clockwise moment = anticlockwise moments

900 × 1 = 3 × M

Where M = the weight of the strong man

3M = 900

M = 900/3 = 300N

Therefore, 300N is the smallest effort that the strongman can use to lift the goat

6 0

3 years ago

Explain the conditions through which friction can be increased.

Rufina [12.5K]

Answer:

Friction can be increased the longer two objects are rubbed together, since the longer they are rubbed is the more heat produced.

Explanation:

8 0

3 years ago

Read 2 more answers

hovering mosquito is hit by a raindrop that is 45 times as massive and falling at 8.1 m/s , a typical raindrop speed. How fast i

Y_Kistochka [10]

Answer:

7.92 m/s

Explanation:

$m_1$ = Mass of raindrop = $45m_2$

$m_2$ = Mass of mosquito

$u_1$ = Initial Velocity of raindrop = 8.1 m/s

$u_2$ = Initial Velocity of mosquito = 0 m/s

$v$ = Velocity of center of mass

For elastic collision

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{45m_2\times 8.1 + m_2\times 0}{45m_2 + m_2}\\\Rightarrow v=\frac{364.5m_2}{46m_2}\\\Rightarrow v=7.92\ m/s$

Hence, the velocity of the attached mosquito, falling immediately afterward is 7.92 m/s

4 0

3 years ago

An 8 g bullet leaves the muzzle of a rifle with

Elena-2011 [213]

Answer: 1872 N

Explanation:

This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:

$V^{2}=V_{o}^{2} + 2ad$ (1)

$F=ma$ (2)

Where:

$V=611.9 m/s$ is the bullet's final speed (when it leaves the muzzle)

$V_{o}=0$ is the bullet's initial speed (at rest)

$a$ is the bullet's acceleration

$d=0.8 m$ is the distance traveled by the bullet before leaving the muzzle

$F$ is the force

$m=8 g \frac{1 kg}{1000 g}=0.008 kg$ is the mass of the bullet

Knowing this, let's begin by isolating $a$ from (1):

$a=\frac{V^{2}}{2d}$ (3)

$a=\frac{(611.9 m/s)^{2}}{2(0.8 m)}$ (4)

$a=234013.5063 m/s^{2} \approx 2.34(10)^{5} m/s^{2}$ (5)

Substituting (5) in (2):

$F=(0.008 kg)(2.34(10)^{5} m/s^{2})$ (6)

Finally:

$F=1872 N$

4 0

3 years ago