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Mashutka [201]
3 years ago
9

A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon

is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v = 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
Physics
1 answer:
Tpy6a [65]3 years ago
3 0

Answer:

The answer is 3.48 seconds

Explanation:

The kinematic equation

y= y0+V0*t+1/2*a*(t*t)

-50=0+(0)t+1/2(-9.8)*(t*t)

t=3.194 seconds

During ribbons ball,

x=x0+ Vt+1/2*a*(t*t)

x= 0+(15)*(3.194)+1/2*(0)* (3.194*3.194)

x= 47.9157m

So, distance (D) = 100-47.9157= 52.084m

52.084m=0+15(t)+1/2*(0)(t*t)

t=52.084/15=3.472286= 3.48seconds

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Scorpion4ik [409]

Answer:

(a) Current is 2831.93 A

(b) 8.40A/m^2

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Explanation:

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Resistance R=11.9mohm=11.9\times 10^{-3}ohm

Potential difference V = 33.7 volt

(A) current is equal to

i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A

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J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2

(c) Resistance is equal to

R=\frac{\rho l}{A}

11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}

\rho =15.52\times 10^{-9}ohm-m

4 0
3 years ago
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musickatia [10]
The frictional force is given by F = μmg 

<span>where μ is the coeficient of friction. </span>

<span>Work done by frictional force = Fd = μmgd </span>

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<span>Fd = μmgd = 1/2 mv² </span>

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<span>d = v² / 2μg </span>




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</span>
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3 years ago
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You are driving due north on i-81 to come to jmu with a speed of 10 m/s, suddenly you realize you forgot your book. You make a u
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8 0
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klio [65]

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Explanation:

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