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3 years ago

6

Mrs. Smith can walk 1.4 m/s. If it takes her 8.5 seconds to get to the teacher lounge, how far is the teacher lounge from her ro

om ?

1 answer:

AlexFokin [52]3 years ago

5 0

Answer:

6 meters away

Explanation:

6*1.4= 8.4 which is pretty close

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A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s

Natasha2012 [34]

Explanation:

There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

6 0

3 years ago

If gas particles start colliding with the walls of their metallic container with increased force, what is their direct effect?.

aliya0001 [1]

<span>So we want to know what is the direct effect of the gas particles if the force of their collision with the walls of the container is increased. Pressure, microscopically, is defined as the number of collisions per unit area. If we increase the force of the collision, the pressure increases so that is the direct effect. </span>

3 0

3 years ago

Read 2 more answers

What will be the change in velocity of a 850kg car if a force of 50,000 N

yKpoI14uk [10]

Answer:

29.412m/s

Explanation:

$F=ma$ where F= force, m= mass, and a=acceleration

we also know that,

a = Δv / t where Δv = change in velocity and t = time

thus F = m ( Δv / t)

$50000=850(\frac{v}{0.5})$

$\frac{50000}{1700}=$ Δv

29.412m/s=Δv

8 0

3 years ago

Maxwell’s theory of Electromagnetism in 1865 was the first "unified field theory" _________ no further theory has united the ele

Andrei [34K]

Answer: E. and Electroweak in 1961 the only other, as

Explanation:

This is more an English grammar question than a physics question, so taking that perspective, one should look for the answer that best completes the sentence.

Based on the word "first" in the sentence, implying the need for a conjunction to join the two theories and the last part of the sentence does not give a reason but further supports the determination of the theory being the first unified theory.

So, to complete the sentence, the best option is;

Maxwell’s theory of Electromagnetism in 1865 was the first "unified field theory" and Electroweak in 1961 the only other, as no further theory has united the electroweak field with either the Strong (Hadronic) force or Gravity.

6 0

3 years ago

What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc

Liono4ka [1.6K]

The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

$Q= mc\Delta \theta$

where;

c is the specific heat capacity
Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

$Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J$

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

$C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC$

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

$C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC$

Specific heat capacity of the metal for the third trial;

$C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC$

Specific heat capacity of the metal for the fourth trial;

$C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC$

Specific heat capacity of the metal for the fifth trial;

$C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC$

Average specific heat capacity

$C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC$

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

8 0

2 years ago