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Alex777 [14]
3 years ago
8

A wooden block measuring 50 cm by 100 cm by 200 cm has a mass of 8000 g.

Physics
1 answer:
photoshop1234 [79]3 years ago
7 0

The density of the block is mathematically given as

\phi=8e-4

<h3>Density of block</h3>

Question Parameters:

50 cm by 100 cm by 200 cm has a mass of 8000 g.

Generally the equation for the density  is mathematically given as

phi=\frac{m}{v}

Where

v= 50 cm * 100 cm * 200 cm

v=1000000cm

v=10000m

Therefore

phi=\frac{m}{v}\\\\phi=\frac{8}{10000}

phi=8e-4

For more information on density

brainly.com/question/14010194

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how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at
DiKsa [7]

Answer:

|a|=2.83\ m/s^2

\theta=75^o

Explanation:

<u>Net Force And Acceleration </u>

The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

\vec F_n=m\vec a

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

\vec F_1=

\vec F_1=\ N

The components of the second force are

\vec F_2=

\vec F_2=\ N

The net force is

\vec F_n=

\vec F_n=\ N

The magnitude of the net force is

|F_n|=\sqrt{14.64^2+54.64^2}

|F_n|=\sqrt{3200}=56.57\ N

The acceleration has a magnitude of

\displaystyle |a|=\frac{|F_n|}{m}

\displaystyle |a|=\frac{56.57}{20}

|a|=2.83\ m/s^2

The direction of the acceleration is the same as the net force:

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5 0
3 years ago
A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh
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Answer:

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Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

mass = volume x density

mass = area x thickness x density

170.097 = 46451.5 x t x 2.70

t = 1.36 x 10^-3 cm

Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.

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An automobile with tires that have a radius of 0.260 m travels 76000 km before wearing them out. How many revolutions do the tir
Tomtit [17]

Answer:

Explanation:

circumference of the tyre = 2πr = 2 x 3.14 x 0.26 = 1.6328m

76000km = 76000000m

no of revolutions required

= 76000000/1.6328 = 46546 revolutions.

3 0
3 years ago
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