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LUCKY_DIMON [66]
3 years ago
13

A painter lifts a 2.8 kg bucket using a rope. if the acceleration of the bucket in 3.1 m/s2, what is the tension in the rope (in

n)? selected:
a. 27.4 n
b. 9.8 n
c. 8.68 n
d. 36.1 n
e. 2.94 n f. 0 n

Physics
2 answers:
ad-work [718]3 years ago
7 0
Refer to the diagram shown below

Given:
m = 2.8 kg, the mass of the bucket
a =3.1 m/s², the acceleration of the bucket
Therefore
W = 2.8*9.8 = 27.44 N, the weight of the bucket.

Let T =  the tension in the rope.

From the free body diagram, the net force accelerating the bucket is
T - W = m*a
That is,
T = W + m*a
   = 27.44 + 2.8*3.1 N
   = 36.12 N

Answer:  d. 36.1 N

olganol [36]3 years ago
4 0
The problem indicate that the mass is moving upward, calculate first the "Weight" of an object, W=mg W=(2.8Kg)(3.2m/s²)=8.68N. The tension on the wire will be proportional to the weight of an object because "Tension" is acting upward and weight is acting "downward".
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Answer:

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Explanation:

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a man exerts 700 newtons of force to move a piece of furniture 4 meters. if it takes him 2 seconds to move the furniture, how mu
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                               Work = (force) x (distance)

The work he did:    Work = (700 N) x (4m)  =  2,800 joules

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3 years ago
A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f
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Answer:

8.8 m and 52.5 m

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The vertical component and horizontal component of water velocity leaving the hose are

v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s

v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

s = v_vt + gt^2/2

10 = 19.97t - 9.8t^2/2

4.9t^2 - 19.97t + 10 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}

t= \frac{19.9658877511823\pm14.24}{9.8}

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

s_1 = v_ht_1 = 15*0.58 = 8.8 m

s_2 = v_ht_2 = 15*3.49 = 52.5m

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