Add the following displacement vectors: <br> 12 m south and 15 m 55° E of N??

A camel is living in an oasis(origin), but wants to travel to a new oasis. To do this the camel travels 3,000 meters east, and t

Alex17521 [72]

Answer:

7,000 meters

Explanation:

5 0

3 years ago

A person walks at a speed of 6 km/h from point A to point B. If he improves his pace by 1.5 km/h, he will arrive 1 hour earlier.

Fofino [41]

Answer:

a) The distance is 30 km

The time duration is 5 hours

b) s₁ is approximately 28.142 km or s₁ is approximately 1.505 km

Explanation:

The initial speed with of the person, v₁ = 6 km/h

The distance the walked by the person, d = From point A to point B

The rate at which the person increases the speed, Δv = 1.5 km/h

The time it takes for the person to arrive at point B from point A at the new speed, t₂ = 1 hour earlier than when walking at 6 km/h

a) Let t₁ represent the time it takes the person walking from point A to point B at 6 km/h, we have;

t₂ = t₁ - 1...(1)

d/t₁ = 6...(2)

d/t₂ = 6 + 1.5 = 7.5

∴ d/t₂ = 7.5...(3)

From equation (2), we have;

d = 6 × t₁ = 6·t₁

Plugging in d = 6·t₁, and t₂ = t₁ - 1 in equation (3) gives;

d/t₂ = 7.5

∴ 6·t₁/t₁ - 1 = 7.5

6·t₁ = 7.5 × (t₁ - 1) = 7.5·t₁ - 7.5

7.5·t₁ - 6·t₁ = 7.5

1.5·t₁ = 7.5

t₁ = 7.5/1.5 = 5

t₁ = 5

The time it takes the person walking from point A to point B at 6 km/h, t₁ = 5 hours

The distance from point A to point B, d = 6 km/h × 5 hours = 30 km

b) The distance the person travels at the initial speed, v₁ (6 km/h) = s₁

The duration the person pauses for a rest = 15 minutes = 1/4 hours

The speed with which he walks the rest of the journey, v₂ = 7.5 km/h

The time earlier than expected that he arrives, Δt = 30 minutes = 0.5 hours

We note that the total distance, d = 30 km

The expected time, t₁ = 5 hours

Therefore, we have;

s₁ + s₂ = 30 km

s₂ = 30 - s₁

v₁/s₁ + 1/4 + v₂/s₂ = t₁ - 0.5

Therefore;

6/s₁ + 1/4 + 7.5/(30 - s₁) = 5 - 0.5 = 4.5

6/28.142+ 1/4 + 7.5/(30 - 28.142) = 5 - 0.5 = 4.5

6/s₁ + 7.5/(30 - s₁) = 4.5 - 1/4 = 4.25

-(3·s₁ + 360)/(2·s₁²- 60·s₁) = 4.25

2·s₁²- 60·s₁) × 4.25 + 3·s₁ + 360 = 0

17·s₁²- 504·s₁ + 720 = 0

s₁ = (504 ± √((-504)² - 4 × 17 × 720))/(2 × 17)

s₁ ≈ 28.142 or s₁ = 1.505

The distance the individual travels at v₁ = 6 km/h, s₁ ≈ 28.142 km or 1.505 km

4 0

3 years ago

Help please!

kipiarov [429]

<h2>Answer: Diamond</h2>

Explanation:

This described situation is known as Refraction, a phenomenon in which the light bends or changes it direction when passing through a medium with a index of refraction different from the other medium.

In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.

According to Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (1)

Where:

$n_{1}=1$ is the first medium index of refraction (air)

$n_{2}$ is the second medium index of refraction (the value we want to know)

$\theta_{1}=27\°$ is the angle of the incident ray

$\theta_{2}=11\°$ is the angle of the refracted ray

Now, let's find $n_{2}$ from (1):

$n_{2}=n_{1}\frac{sin(\theta_{1}}{sin(\theta_{2}}$ (2)

Substituting the known values:

$n_{2}=(1)\frac{sin(27\°)}{sin(11\°)}}$

Finally:

$n_{2}=2.379\approx 2.4$

If we compare this result with the given table, the index of refraction value that is close to this number is diamond's index of refraction.

Therefore, the correct option is A: the material is diamond.

3 0

3 years ago

Acceleration due to gravity on the moon is 1.6m/s^2 or about 16% of the value of gg on Earth. If an astronaut on the moon threw

xxTIMURxx [149]

To solve this problem it is necessary to apply the concepts related to the conservation of Energy. Mathematically the conservation of kinetic energy must be paid in the increase of potential energy or vice versa. This expressed in algebraic terms is equivalent to

Kinetic Energy = Potential Energy

$\frac{1}{2}mv^2 = mgh$

Where

m = Mass

v = Velocity

g = Gravity

h = Height

As the mass is the same then we have to

$\frac{1}{2} v^2 = gh$

Rearrange to find v,

$v = \sqrt{2gh}$

Our values are given as

$g = 1.6m/s^2$

$h = 7.8m$

Therefore replacing we have

$v = \sqrt{2(1.6)(7.8)}$

$v = 4.99m/s$

Hence the velocity at the moon would be 4.99m/s

The only direct affectation is that concerning the Resistance or drag force generated by a fluid - such as air in the ground - that can diminish / sharpen the direct effects of gravity. Disregarding the resistance of the air, as we can see in the equation previously given, there should be no affectation because the speed depends on the gravity and height.

5 0

3 years ago

How many atoms are in a molecule of C6H12O6?

Andrews [41]

Answer:

The answer is 24

Explanation:

Its made up of 6 carbon atoms

6 oxygen atoms

12 hydrogen atoms

5 0

3 years ago

Add the following displacement vectors: 12 m south and 15 m 55° E of N??