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just olya [345]

3 years ago

7

A person leaves a house and walks three blocks north, then walks three blocks west. In what direction is the displacement of the

person, relative to the starting point? A North B Northwest C South D Southwest

1 answer:

Sergeeva-Olga [200]3 years ago

4 0

B Northwest hope this helps

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Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw

Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0

3 years ago

Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. Exp

VARVARA [1.3K]

The unbalanced reaction is

<em>a</em> NH₄NO₃ ⇒ <em>b</em> N₂ + <em>c</em> O₂ + <em>d</em> H₂O

where <em>a</em>, <em>b</em>, <em>c</em>, and <em>d</em> are unknown constants.

Count how many times each element appears on either side of the reaction.

• reactants:

N = 2<em>a</em>, H = 4<em>a</em>, O = 3<em>a</em>

• products

N = 2<em>b</em>, O = 2<em>c</em> + <em>d</em>, H = 2<em>d</em>

<em />

Now we solve the system of equations,

2<em>a</em> = 2<em>b</em> … … … [1]

4<em>a</em> = 2<em>d</em> … … … [2]

3<em>a</em> = 2<em>c</em> + <em>d</em> … … … [3]

<em />

From [1] we immediately have

<em>a</em> = <em>b</em>

In [2], we get

2<em>a</em> = <em>d</em>

and substituting for <em>d</em> in [3] gives

3<em>a</em> = 2<em>c</em> + 2<em>a</em>

<em>a</em> = 2<em>c</em>

<em />

Let <em>c</em> = 1; then

• <em>a</em> = 2×1 = 2

• <em>b</em> = 2

• <em>d</em> = 2×2 = 4

So, the balanced reaction is

2 NH₄NO₃ ⇒ 2 N₂ + O₂ + 4 H₂O

8 0

3 years ago

Simora [160]

the father has to sit 0.5meter away from the kid because he is a 3/4 heavier that the kid

6 0

2 years ago

A car travels from point A to Be in 3 hours and returns back to point A in 5 hours .Points A and B are 150 miles apart along str

Rainbow [258]

Explanation:

Given that,

The time taken from A to B is 3 hours

The time from B to A = 5 hours

The total distance between A and B is 150 miles

The average velocity from A to B,

$v=\dfrac{150}{3}=50\ mph$

The average velocity from B to A,

$v=\dfrac{150}{5}=30\ mph$

As the velocity of an object is a vector quantity. It means when the car reaches the initial point, the displacement is 0. So, the average velocity is equal to 0.

3 0

3 years ago

A commercial jet liner takes off with an average acceleration of 3 g. How long does it take to reach the end of its runway which

ANEK [815]

Answer:

The time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

Explanation:

Given;

average acceleration of the commercial Jet liner, a = 3g = 3 x 9.8 m/s² = 29.4 m/s²

distance traveled by the commercial Jet liner, s = 1542 m

The time taken for the commercial Jet liner to reach the end of its runway is calculated as follows;

s = ut + ¹/₂at²

where;

u is the initial velocity of the commercial Jet liner = 0

s = 0 + ¹/₂at²

s = ¹/₂at²

2s = at²

$t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2 \times 1524}{29.4} } \\\\t = 10.18 \ s$

Therefore, the time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

6 0

3 years ago