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KIM [24]
3 years ago
8

suppose you use three different scales to weigh bag of oranges one scale says the bag weighs 2.1 lbs a second says it weighs 2.2

lbs and a third says it weighs 2.1 lbs the actual weight of the bag of oranges is 2.153 lbs which of the following best describes the results?
Physics
1 answer:
Ber [7]3 years ago
4 0

Answer:

It would be 2.2 if you have to round the five to the one but if your not rounding the number, it'd be 2.1.

Explanation:

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Why can’t solids flow
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Answer:

basically they have too much mass in them

Explanation:

They are held tightly together by strong forces of attraction. They are held in fixed positions but they do vibrate. Because the particles don't move, solids have a definite shape and volume, and can't flow. Because the particles are already packed closely together, solids can't easily be compressed.

6 0
3 years ago
A student walks 3 blocks east, 4 blocks north, and 3 blocks west. What is the displacement of the student?
12345 [234]
4 blocks north because he is it not asking for north east 
5 0
3 years ago
A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

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3 years ago
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