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2 years ago

The block shown above has a mass of 105 g. What is the density of the block? SC.8.P.8.3

Physics

1 answer:

zaharov [31]2 years ago

4 0

The density of the block is 1.25 cm³

The correct answer to the question is Option B. 1.25 cm³

To solve this question, we'll begin by calculating the volume of the block. This can be obtained as follow:

Length = 7 cm

Height = 4 cm

Width = 3 cm

<h3>Volume =? </h3>

Volume = Length × Width × Height

Volume = 7 × 3 × 4

<h3>Volume = 84 cm³</h3>

Thus, the volume of the block is 84 cm³

Finally, we shall determine the density of the block. This can be obtained as follow:

Density is defined as mass per unit volume i.e

$Density = \frac{mass}{volume} \\\\$

Mass of block = 105 g

Volume of block = 84 cm³

<h3>Density of block =? </h3>

$Density = \frac{mass}{volume}\\\\Density = \frac{105}{84}\\\\$

<h3>Density of block = 1.25 cm³</h3>

Therefore, the density of the block is 1.25 cm³.

Hence, Option B. 1.25 cm³ gives the correct answer to the question.

Learn more: brainly.com/question/2040396?referrer=searchResults

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Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .

Let the electric field intensity due to $Q_2$ be + $E_2$ and that due to $Q_3$ be - $E_3$ since the charge is negative. Hence at the origin;

$+E_2-E_3=0..................(1)$

From equation (1) above, we obtain the following;

$E_2=E_3.................(2)$

From Coulomb's law the following relationship holds;

$+E_2=\frac{kQ_2}{r_2^2}\\$

$-E_3=\frac{kQ_3}{r_3^2}$

where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

$\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)$

k can cancel out from both side of equation (3), so that we finally obtain the following;

$\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)$

Given;

$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

$Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C$

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3 years ago

At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counsel

sleet_krkn [62]

Answer:

47 m

Explanation:

Data obtained from the question include the following:

Length of dry leg 1 (L1) = 40 m

Length of dry leg 2 (L2) = 25 m

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The length of the swimming course can be obtained by using pythagoras theory as shown below:

L² = L1² + L2²

L² = 40² + 25²

L² = 1600 + 625

L² = 2225

Take the square root of both side.

L = √2225

L = 47.1 ≈ 47 m

Therefore, the length of the swimming course is approximately 47 m.

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3 years ago

a ball rolls along the floor with a constant velocity of 3 m/s. How far will it have gone after 234 seconds?

sergij07 [2.7K]

Answer:

The ball travel for 702m

Explanation:

distance = speed × time

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time = 234s

distance = 3 × 234

= 702m

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3 years ago

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Answer:

1) The greatest height attained by the ball equals 20.387 meters.

2) The time it takes for the ball to reach 15 meters approximately equals 1 second.

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thus using third equation of kinematics we obtain the height attained as

$v^2=u^2+2as$

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4 years ago

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uranmaximum [27]

Answer:

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3 years ago

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