All categories

3 years ago

Cube A has a volume of 2cm x 2cm x 2cm. what is its density

1 answer:

8 0

The volume of the cube is (2cm x 2cm x 2cm) = 8 cm³ .

The density of the cube is

(the mass of the cube, in grams) / (8 cm³) grams per cm³ .

That's the best I can do, because you haven't told us the cube's mass.

You might be interested in

A 16.7 kg block is dragged over a rough, horizontal surface by a constant force of 132 N acting

Alisiya [41]

The magnitude of the work done by the force of friction is 1045.68 joules

Explanation:

Work done = force × distance

Work done by friction force (W) = friction force (F) × distance (d)
Friction force (F) = coefficient of kinetic friction (μ) × normal reaction force (R)

A 16.7 kg block is dragged over a rough, horizontal surface by a constant force of 132 N acting at an angle of 25.8. above the horizontal. The block is displaced 57,8 m, and the coefficient of kinetic friction is 0.229

To find the normal reaction force of the block distribute The constant force into two component, horizontal component of 132 cos(25.8°) and vertical component 132 sin(25.8°), Then equate ∑vertical forces by 0 to find R

The weight of the block = mg, where m is its mass and g is 9.8 m/s²

→ ∑Vertical forces = R + 132 sin(25.8°) - mg

→ ∑Vertical forces = R + 132 sin(25.8°) - (16.7)(9.8)

→ ∑Vertical forces = R - 106.2

Equate it by 0

→ R - 106.2 = 0 ⇒ add 106.2 to both sides

→ R = 106.2 N

Now let us find the friction force

→ F = μ R

→ μ = 0.229

→ F = 0.229 (106.2) = 24.3198 N

Let us calculate the work don by the force of friction

→ W = F × d

→ d = 57.8 meters

→ W = 24.3198 × 57.8 = 1045.68 joules

The magnitude of the work done by the force of friction is 1045.68 joules

Learn more:

You can learn more about force of friction in brainly.com/question/6217246

LearnwithBrainly

6 0

3 years ago

A pencil rolls horizontally of a 1 meter high desk and lands .25 meters from the base of the desk. How fast was the pencil rolli

vovikov84 [41]

Answer: 0.55 m/s

Explanation:

This situation is related to projectile motion (also called parabolic motion), where the main equations are as follows:

$x=V_{o} cos\theta t$ (1)

$y=y_{o}+Vo sin \theta t + \frac{g}{2}t^{2}$ (2)

Where:

$x=0.25 m$ is the horizontal displacement of the pencil

$V_{o}$ is the pencil's initial velocity

$\theta=0\°$ since we are told the pencil rolls <u>horizontally</u> before falling

$t$ is the time since the pencil falls until it hits the ground

$y_{o}=1 m$ is the initial height of the pencil

$y=0$ is the final height of the pencil (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity, always acting vertically downwards

Begining with (1):

$x=V_{o} cos(0\°) t$ (3)

$x=V_{o}t$ (4)

Finding $t$ from (2):

$0=1 m+ \frac{-9.8m/s^{2}}{2}t^{2}$ (5)

$t=\sqrt{\frac{-2y_{o}}{g}}$ (6)

Substituting (6) in (4):

$x=V_{o}\sqrt{\frac{-2y_{o}}{g}}$ (7)

Isolating $V_{o}$ :

$V_{o}=\frac{x}{\sqrt{\frac{-2y_{o}}{g}}}$ (8)

$V_{o}=\frac{0.25 m}{\sqrt{\frac{-2(1 m)}{-9.8m/s^{2}}}}$ (9)

Finally:

$V_{o}=0.55 m/s$

4 0

3 years ago

Create a ray diagram for eyeglasses that contain a diverging lens. Assume you are looking at a 2 cm tall object that is 4 cm fro

k0ka [10]

The ray diagram for the given object consists of 2 cm height of object, 4 cm object distance and 3 cm focal length.

<h3>Image formed by a diverging lens</h3>

Diverging lens is called a concave lens. The working of the lens is dependent on the refraction of the light rays as they pass through the lens.

Image formed by a diverging lens is always virtual, erect and diminished; smaller than the object and is located on the same side of the lens as the object.

The ray diagram for the given object is presented in the image in the diagram.

Object height = 2 cm
Focal length = 3 cm
Object distance = 4 cm

Learn more about diverging lens here: brainly.com/question/3140453

#SPJ1

8 0

2 years ago

A 12 oz can of soda is left in a car on a hot day. In the morning the soda temperature was 60oF with a gauge pressure of 40 psi.

Neko [114]

In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:

P/T = Constant

Then
$\frac{ P_{1} }{ T_{1} } = \frac{ P_{2} }{ T_{2} }$

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)

Substituting;
$P_{2} = \frac{ P_{1} T_{2} }{ T_{1} } = \frac{40*305}{289} =42.21 psi$

3 0

3 years ago

Can you help me ASAPPP please help me

AveGali [126]

Answer:it is a

Explanation hope this helps .

6 0

3 years ago