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2 years ago

12

A ball is released from a hot air balloon moving downward with a velocity of

1 answer:

padilas [110]2 years ago

4 0

The answer is b. 12.1 seconds

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Which statement best describes why an island's food web could be considered a closed system?

garik1379 [7]

I would say Option B) because Option C) is wrong since matter cannot be created. A closed system does not exchange matter so it's not Option D). Since an island is an isolated area, Option A) is wrong.

3 0

3 years ago

Read 2 more answers

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is: $w_{c} =m_{c} g$ and maximum tension is $T=2.50 m_{c} g=1.41*10^4N$

total mass and weight is :

$M =m_{c}+ m_{b} =740kg+550kg=1290 kg$

$w_{M} =1.2650*10^4N$

∑ $F_{y} =ma_{y}$

$T-M_{g} =Ma_{y}$

$a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg$

$=0.645m/s^2$

B)

maximum acceleration

$a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0$

using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

to solve for t

$t=\sqrt{2(y-y_{0} )/a_{y} }$

$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

6 0

3 years ago

11. A 3.8 kg object is lifted 12 meters. Approximately how much work is performed during the lifting?

Marianna [84]

I found the answer for you if u need any help ask anytime!

3 0

2 years ago

3 hours into minutes

otez555 [7]

Answer:

180

60 minutes in a hour

60x3 is 180minutes

6 0

2 years ago

A particle undergoes two displacements. The first has a magnitude of 11 m and makes an angle of 82 ◦ with the positive x axis. T

Luden [163]

Answer:

$\theta = 211.7 degree$

Explanation:

First displacement of the particle is given as

$r_1$ = 11 m at 82 degree with positive X axis

so we can say

$\vec r_1 = 11 cos82 \hat i + 11 sin82 \hat j$

$\vec r_1 = 1.53\hat i + 10.9 \hat j$

resultant displacement of the particle after second displacement is given as

r = 8.7 m at 135 degree with positive X axis

so we can say

$r = 8.7 cos135\hat i + 8.7 sin135\hat j$

$r = -6.15 \hat i + 6.15 \hat j$

now we know that

$r = r_1 + r_2$

now we have

$r_2 = r - r_1$

so we will have

$r_2 = (-6.15 \hat i + 6.15 \hat j) - (1.53\hat i + 10.9 \hat j)$

$r_2 = -7.68 \hat i - 4.75 \hat j$

so angle of the second displacement is given as

$tan\theta = \frac{r_y}{r_x}$

$tan\theta = \frac{-4.75}{-7.68}$

$\theta = 211.7 degree$

8 0

2 years ago