True or false

A hungry rat is placed in a maze. It walks the following path to find a piece of cheese. 4.0m N, 7.5 m E, 6.8 m S, 3.7 m E, 3.6

storchak [24]

Answer:

Explanation:

We shall take the help of vector form of displacement . Taking east as i and north as j

4.0m N = 4 j

7.5 m E = 7.5 i

6.8 m S = - 6.8 j

3.7 m E, = 3.7 i

3.6 m S = - 3.6 j

5.3 m W = - 5.3 i

3.7 m N, = 3.7 j

5.6 m W = - 5.6 i

4.4 m S = - 4.4 j

4.9 m W = - 4.9 i

Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i

= -4.6 i -7.1 j

magnitude of displacement = $\sqrt{(4.6^2+7.1^2)}$

= 8.46 m

Direction

Tanθ = 7.1/ 4.6

θ = 57⁰ south of west .

distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9

= 49.5 m

6 0

3 years ago

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A student tests the acidity of nitric acid (HNO3) by dissolving a sample of HNO3 in a solution of liquid methane. He then uses a

DerKrebs [107]

Sorry I didn't see this before...
Okay, I see two major problems with this student's experiment:

1) Nitric acid Won't Dissolve in Methane
Nitric acid is what's called a mineral acid. That means it is inorganic (it doesn't contain carbon) and dissolves in water.
Methane is an organic molecule (it contains carbon). It literally cannot dissolve nitric acid. Here's why:
For nitric acid (HNO3) to dissolve into a solvent, that solvent must be polar. It must have a charge to pull the positively charged Hydrogen off of the Oxygen. Methane has no charge, since its carbon and hydrogens have nearly perfect covalent bonds. Thus it cannot dissolve nitric acid. There will be no solution. That leads to the next problem:

2) He's Not actually Measuring a Solution
He's picking up the pH of the pure nitric acid. Since it didn't dissolve, what's left isn't a solution—it's like mixing oil and water. He has groups of methane and groups of nitric acid. Since methane is perfectly neutral (neither acid nor base), the electronic instrument is only picking up the extremely acidic nitric acid. There's no point to what he's doing.

Does that help?

8 0

3 years ago

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Which statements accurately describe mechanical waves

AleksAgata [21]

Answer:

Explanation:

so a mechanical wave transfers energy through a medium but unlike other waves that move through very long distances

the distance of the mechanical wave is different

7 0

3 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

3 years ago

A chair of mass 15.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force of

Neko [114]

Answer: 185.5672566

Explanation: The friction is not relevant

Normal reaction is the force perpendicular to the surface.

this force resists the downwards forces applied which are gravity and a component of the applied force.

7 0

3 years ago

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