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Finger [1]
3 years ago
14

A child does 350J of work while pulling a box from the ground up to his tree house with a rope. The tree house is 5.2 m above th

e ground. What is the mass of the box? A) 4.1 kg B) 6.9 kg C) 6.2 kg D) 5.2 kg
Physics
1 answer:
Advocard [28]3 years ago
3 0

I think the answer to that questions is B.

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If there is 8 g of a substance before a physical change , how much will there be afterwards?
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A substance undergoing a physical change will still weigh the same even after the change. This is in accordance to the law of conservation of mass which states that mass is neither created nor destroyed. so an 8 g substance remains of the same weight even after undergoing a physical change.
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How is a scientific explanation evaluated
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You test it over and over again. so basically you experiment on it more. 
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The mass of 1 dozen dimes is 27.22 g. the numerical value of the mass of 5 dozen dimes can be obtained by
Jobisdone [24]

Answer:

by obtaining the total mass of the dimes present:

d = 27.22 g / dozen       the density of dimes

M = n * d = 5 dozen * 27.22 g / dozen = 126.1 g

4 0
2 years ago
A golf club hits a 0.0459 kg golf ball at rest; the club is in contact with the ball for 0.00138 s. Afterwards the ball leaves a
Advocard [28]

Answer: 2120 N is correct

Explanation: correct answer for acellus

7 0
3 years ago
) Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) . What are (a)
natali 33 [55]

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

\tau = \vec{F} \times \vec{r}

\tau = (8i+6j)\times(-3i+4j)

\tau = (8*4)(i\times j)+(6*-3)(j\times i)

\tau = 32k +18k

\tau = 50 k

Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}

cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}

cos\theta = -0.24

\theta = cos^{-1} (-0.24)

\theta = 103.88\°

Therefore the angle between the ratio and the force is 103.88°

5 0
3 years ago
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