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vampirchik [111]
3 years ago
14

You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While

there, you notice that the elf Sneezy, when hanging from a rope, produces a tension of 445 N in the rope.
If Sneezy hangs from a similar rope while delivering presents at the earth's equator, what will the tension in it be?
Physics
1 answer:
Fynjy0 [20]3 years ago
3 0

Answer:

Explanation:

The question here is that if sneezy hands from a similar rope while delivering presents at the earth's equator, what will be the tension in the rope be. Here is the solution:The tension on the rope when it is at pole, T= 455 NTo find, the tension, t= mgTo solve for mass, m= t/g. Substituting this we have, m=455/9.8. m=46.43 kgAssume that the downwards acceleration is, a= -46.43 m/s^2.T = mg + maT = (46.43 kg) ( 9.8 m/s^2) - (46.43 kg) (-46.43 m/s^2)T = 455.01 kg-m/s^2 - -2155.74 kg-m/s^2T = 2610.75 kg-m/s^2 = 2610.75 N

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joja [24]

ANSWER:100dB

f<em>rom the  sound intensity level,the sound intensity is calculated as:</em>

<em />\beta<em>₁=</em>\beta<em>=(10dB)㏒₁₀(l₁/l₀)</em>

<em>inserting numbers:</em>

<em>120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²</em>

<em>Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which </em>

<em>can be used to solve for l₁ and get</em>

<em>          l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²</em>

<em>since the sound  intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in</em>

<em>ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:</em>

<em>l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²</em>

<em>since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:</em>

<em>     </em>\beta<em>₂=</em>\beta<em>=(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)</em>

<em>    </em>\beta<em>₂=(10dB)㏒₁₀(2.04×10¹⁰)</em>

<em />\beta<em> =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB</em>

<em />

3 0
4 years ago
How many significant figures does 0.09164500561 have?
kolbaska11 [484]

Answer:

10 Sig Figs

Explanation:

Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.

7 0
4 years ago
4. Compare the change in initial and final gravitational force in
mylen [45]
Gravitational force equals GMm/r^2, where G is constant, M and m are the masses, and r is distance.

For I, if both masses double, the formula becomes G2M2m/r^2, or 4GMm/r^2. Therefore, the gravitational force will quadruple or 4x.

For II, if the distance between the object doubles, the formula becomes GMm/(2r)^2 or GMm/4r^2. In this case, the gravitational force is 1/4x the initial force.
4 0
3 years ago
A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di
Lyrx [107]

Answer:

D

Explanation:

From the information given:

The angular speed for the block \omega = 50 \ rad/s

Disk radius (r) = 0.2 m

The block Initial velocity is:

v = r \omega \\ \\  v = (0.2  \times 50) \\ \\  v= 10 \ m/s

Change in the block's angular speed is:

\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s

However, on the disk, moment of inertIa is:

I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2

The time t = 10s

∴

Frictional torques by the wall on the disk is:

T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10})  \\ \\ =0.6 \ N.m

Finally, the frictional force is calculated as:

F = \dfrac{T}r{}

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8 0
3 years ago
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Iteru [2.4K]

Answer:

15 nC and 10 nC

Explanation:

qA + qB = 25 nC = 25 x 10^-9 C

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5.4 \times 10^{-4}=\frac{9 \times 10^{9}q_{A}q_{B}}{0.05^{2}}

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So, q_{A}\left ( 25 \times 10^{-9}-q_{A} \right )=1.5 \times 10^{-16}

q_{A}^{2}-25 \times 10^{-9}q_{A}+1.5 \times 10^{-16}=0

q_{A}=\frac{25\times 10^{-9} \pm \sqrt{6.25\times 10^{-16}-6 \times10^{-16}}}{2}

q_{A}=\frac {25\times 10^{-9} \pm 5 \times 10^{-9}}}{2}

qA = 15nC or 10nC

So, qB = 10 nC or 15 nC

8 0
4 years ago
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