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Alexandra [31]
1 year ago
13

This simulation has three parts: Intro, Compare, & Mystery.I. Explore the Intro section of this simulation.1. Based on your

exploration how will you define the density D, of a substance? Writedown an algebraic expression relating D to the mass and volume of a substance.2. Does the density of a substance change with its mass? If yes, how? If no, why not?
Physics
1 answer:
dexar [7]1 year ago
4 0

a) Density is mass of a unit volume of a material substance

b) Algebraic expression relating Density = m / v

c) Density is an intensive property, which means the density does not change as the amount of the substance present changes

a) Density is mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is the measurement of how tightly a material is packed together.

The more closely packed particles are, the more dense the object. The amount of mass in a particular space or volume. The more particles (or mass) in a given space, the more dense it is.

b) Algebraic expression relating Density = mass / volume or m/v

where d is density, M is mass, and V is volume.

c) Density (ρ) is the amount of mass (m) per unit volume (V) of a substance. Density is an intensive property, which means the density does not change as the amount of the substance present changes.

To learn more about density here

brainly.com/question/15164682?referrer=searchResults

#SPJ4

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A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),
labwork [276]

Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

<u>FOR FORCE (P)</u>:

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

W = Pd

<u>Therefore, work done by force (P) is Positive.</u>

<u></u>

<u>FOR NORMAL FORCE (Fn) AND WEIGHT (W)</u>:

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

<u>Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.</u>

<u></u>

<u>FOR KINETIC FRICTIONAL FORCE (fk)</u>:

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

<u>Therefore, work done by Kinetic Frictional Force (fk) is Negative.</u>

<u></u>

8 0
3 years ago
Please help me important will give brainliest
irinina [24]
Pretty sure it’s C. all the others are speeding up. acceleration means gradually (over time) getting faster. So it’s C.
7 0
3 years ago
A vase falls from a windowsill 50.0 m above the sidewalk. How fast is the base moving when it hits the ground?
scZoUnD [109]

Answer:

The answer is A sorry if i'm wrong

Explanation:

6 0
3 years ago
Two simple pendulum of slightly different length , are set off oscillating in step is a time of 20s has elasped , during which t
adell [148]

Answer:

Length of longer pendulum = 99.3 cm

Length of shorter pendulum = 82.2 cm

Explanation:

Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²

substituting T = 2s and g = 9.8 m/s² we have

l = T²g/4π²

= (2 s)² × 9.8 m/s² ÷ 4π²

= 39.2 m ÷ 4π²

= 0.993 m

= 99.3 cm

Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

t/T₂ = t/T₁ + x

1/T₂ = 1/T₁ + x  (1)

Also, the T₂ = t/n₂ = t/(n₁ + x)  (2)

From (1) T₂ = T₁/(T₁ + x) (3)

equating (2) and (3) we have

t/(n₁ + x) = T₁/(T₁ + x)

substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

20 s/(10 + x) = 2/(2 + x)

10/(10 + x) = 1/(2 + x)

(10 + x)/10 = (2 + x)

(10 + x) = 10(2 + x)

10 + x = 20 + 10x

collecting like terms

10x - x = 20 - 10

9x = 10

x = 10/9

x = 1.11

x ≅ 1 oscillation

substituting x into (2)

T₂ = t/n₂ = t/(n₁ + x)

= 20/(10 + 1)

= 20/11

= 1.82 s

Since length l = T²g/4π²

substituting T = 1.82 s and g = 9.8 m/s² we have

l = T²g/4π²

= (1.82 s)² × 9.8 m/s² ÷ 4π²

= 32.46 m ÷ 4π²

= 0.822 m

= 82.2 cm

6 0
3 years ago
Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

8 0
3 years ago
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