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3 years ago

What are two different ways to understand time? Explain and give examples.

1 answer:

6 0

There are two different ways to understand time and these are:
A. What time is it?
B. How much time?
The examples of these two different ways are:
A. What time is it? The best example that would help us understand and know what time are the clock and the calendar. This gives us the exact hour, minutes and seconds. The calendar tells us the exact day, month and year.
B. How much time? This makes us understand how much time did it take from the starting time. An example for this would be a stopwatch.

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Which climate does Western California have

maxonik [38]

Answer:

Along the western side of the Coast Range the climate is dominated by the Pacific Ocean. Warm winters, cool summers, small daily and seasonal temperature ranges, and high relative humidities are characteristic of this area.

Explanation:

3 0

4 years ago

How would you obtain a mean value for the specific heat<br> capacity of a material?

lidiya [134]

The heat capacity and the specific heat are related by C=cm or c=C/m. The mass m, specific heat c, change in temperature ΔT, and heat added (or subtracted) Q are related by the equation: Q=mcΔT. Values of specific heat are dependent on the properties and phase of a given substance.

8 0

3 years ago

A 0.0434-m3 container is initially evacuated. Then, 4.19 g of water is placed in the container, and, after some time, all of the

adell [148]

Answer:

P =18760.5 Pa

Explanation:

Given that

Volume ,V= 0.0434 m³

Mass ,m= 4.19 g = 0.00419 kg

T= 417 K

If we assume that water vapor is behaving like a ideal gas ,then we can use ideal gas equation

Ideal gas equation P V = m R T

p=Pressure ,V = Volume ,m =mass

T=Temperature ,R=Universal gas constant

Now by putting the values

P V = m R T

For water R= 0.466 KJ/kgK

P x 0.0434 = 0.00419 x 0.466 x 417

P =18.7605 KPa

P =18760.5 Pa

Therefore the answer is 18760.5 Pa

6 0

4 years ago

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 5.0 km/h

Ratling [72]

Answer

given,

time = 10 s

ship's speed = 5 Km/h

F = m a

a is the acceleration and m is mass.

In the first case

F₁=m x a₁

where a₁ = difference in velocity / time

F₁ is constant acceleration is also a constant.

Δv₁ = 5 x 0.278

Δv₁ = 1.39 m/s

$a_1=\dfrac{1.39}{10}$

a₁ = 0.139 m/s²

F₂ =m x a₂

F₃ = F₂ + F₁

Δv₃ = 19 x 0.278

Δv₃ = 5.282 m/s

a₃=Δv₂ / t

$a_3=\dfrac{5.282}{10}$

a₃ = 0.5282 m²/s

m a₃=m a₁ + m a₂

a₃ = a₂ + a₁

0.5282 = a₂ + 0.139

a₂=0.3892 m²/s

F₂ = m x 0.3892...........(1)

F₁ = m x 0.139...............(2)

F₂/F₁

ratio = $\dfrac{0.3892}{0.139}$

ratio = 2.8

6 0

4 years ago

A 1,450 kg car drives toward a 60 kg shopping cart that has a velocity of -1.2 m/s toward the car. The two objects collide, givi

Y_Kistochka [10]

Answer:

A) v₁ = 5.66 [m/s]

Explanation:

To solve this problem we must use the definition of linear momentum conservation, which tells us that momentum is conservation before and after a collision.

The linear momentum is equal to the mass by the product of the Velocity.

P = m*v

where:

P = lineal momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

Now, to the right side of the equal sign will take the linear momentum before the collision and to the left side of the equal sign as after the collision.

Pbefore = Pafter

(m₁*v₁) - (m₂*v₂) = (m₁*v₃) + (m₂*v₄)

where:

m₁ = mass of the car = 1450 [kg]

v₁ = velocity of the car before the collision [m/s]

m₂ = mass of the shopping cart = 60 [kg]

v₂ = velocity of the shopping cart before the collision = -1.2 [m/s]

v₃ = velocity of the car after the collision = 5.13 [m/s]

v₄ = velocity of the shopping cart after the collision = 11.75 [m/s]

Now replacing:

(1450*v₁) - (60*1.2) = (1450*5.13) + (60*11.75)

1450*v₁ - 72 = 7438.5 + 705

1450*v₁ = 7438.5 + 705 + 72

1450*v₁ = 8215.5

v₁ = 5.66 [m/s]

4 0

3 years ago

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