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NeX [460]
2 years ago
11

What is weight?

Physics
2 answers:
OverLord2011 [107]2 years ago
8 0

Answer:

C

Explanation:

The answer is C) the mass of an object

suter [353]2 years ago
4 0

Answer:

C

Explanation:

it would be the mass of an object

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Can someone help me with this?
oksian1 [2.3K]

Answer:

Yes

Explanation:

5 0
2 years ago
As shown in the diagram, threee equal charges are spaced evenly in a row. The magnitude of each charge is +2e, and the distance
Nataly [62]

Answer: 4.27 x 10^-10 N to the left

Explanation: I just took this quiz

4 0
3 years ago
Read 2 more answers
A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0
3 years ago
Madison was driving at 40 mph and went 80 miles. How long did it take madison?
monitta
v=40\ mph\\\\s=80\ miles\\\\t=?\\--------------\\v=\frac{s}{t}\to vt=s\to t=\frac{s}{v}\\--------------\\t=\frac{80\ miles}{40\ mph}=2\ h\leftarrow Answer
3 0
3 years ago
Read 2 more answers
Would a measured force of (46.5 0.8 N  ) be in agreement with a theoretically calculated force of (48.4 0.6 N  ) ? Show your w
OverLord2011 [107]

Answer:

A measured force of (46.5 0.8 N  ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N  )

Explanation:

From the question we are told that

  Measured force is  F_M  =  [46.5 \pm 0.8 \  N ]

   Calculated force is  F_c =  [48.4 \pm 0.6 \  N ]

Generally the measured force in interval form is

     46.5 - 0.8  < F_M  <  46.5 + 0.8

=>  45.7   < F_M  < 47.3

Generally the calculated  force in interval form is

     48.4 - 0.6  < F_c  <  48.4 + 0.6

=>  47.8   < F_M

Generally looking both interval we see that they do not intersect at any point Hence  

A measured force of (46.5 0.8 N  ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N  )      

8 0
2 years ago
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