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shtirl [24]
2 years ago
10

define the term change and state one negative change you may encounter as a student or as an employee in the future​

Physics
1 answer:
Brums [2.3K]2 years ago
6 0

Answer:

hey iam bored pls join

ID: 724 645 3790

Password: 1111

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A water line enters a house 2.0 m below ground. A smaller diameter pipe carries water to a faucet 5.0 m above ground, on the sec
nirvana33 [79]

Answer:

 P₁- P₂ = 91.1 10³ Pa

Explanation:

For this exercise we will use Bernoulli's equation, where point 1 is at the bottom of the house and point 2 on the second floor

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

          P1-P2 = ½ ρ (v₂² - v₁²) + ρ g (y₂-y₁)

In the exercise they give us the speeds and the height of the turbid, so we can calculate the pressure difference

For heights let's set a reference system on the ground floor of the house, so we have 5m for the second floor and an entrance at -2m

 

        P₁-P₂ = ½ 1.0 10³ (7² - 2²) + 1.0 10³ 9.8 (5 + 2)

       P₁-P₂ = 22.5 10³ + 68.6 10³

       P₁- P₂ = 91.1 10³ Pa

3 0
3 years ago
How do protons and electrons determine the properties of an atom?
Nataly [62]
They determine if it has a negative or positive charge
6 0
3 years ago
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A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform abov
Rina8888 [55]

Answer:

20.96 m/s

Explanation:

Using the equations of motion

y = uᵧt + gt²/2

Since the puck slides off horizontally,

uᵧ = vertical component of the initial velocity of the puck = 0 m/s

y = vertical height of the platform = 2 m

g = 9.8 m/s²

t = time of flight of the puck = ?

2 = (0)(t) + 9.8 t²/2

4.9t² = 2

t = 0.639 s

For the horizontal component of the motion

x = uₓt + gt²/2

x = horizontal distance covered by the puck

uₓ = horizontal component of the initial velocity = 20 m/s

g = 0 m/s² as there's no acceleration component in the x-direction

t = 0.639 s

x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m

For the final velocity, we'll calculate the horizontal and vertical components

vₓ² = uₓ² + 2gx

g = 0 m/s²

vₓ = uₓ = 20 m/s

Vertical component

vᵧ² = uᵧ² + 2gy

vᵧ² = 0 + 2×9.8×2

vᵧ = 6.26 m/s

vₓ = 20 m/s, vᵧ = 6.26 m/s

Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s

4 0
3 years ago
Read 2 more answers
A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s,
Serhud [2]
By using the equation speed = distance/time we can solve for distance. The speed is 4 m/s and the time is 12 seconds. We need to rearrange the equation to Speed * Time = distance. 4(12) = 48; 48 = distance. The cliff is 48 meters high.
3 0
2 years ago
How do you graph distance and time for an object that moves at a constant speed?
LekaFEV [45]

Answer:

It would be a straight line

Explanation:

On a distance-time graph, an object that moves at constant speed would be represented by a straight line.

In fact, in a distance-time graph, the slope of the line corresponds to the speed of the object. We can demonstrate that. In fact:

- The speed of the object is equal to the ratio between the distance covered (\Delta s) and the time taken (\Delta t):

v=\frac{\Delta s}{\Delta t}

On a distance-time graph, the distance is on the y-axis while the time is on the x-axis. The slope of the line is defined as:

m=\frac{\Delta y}{\Delta x}

But the variation on the y-axis (\Delta y) is equal to the distance covered (\Delta s), while the variation on the x-axis (\Delta x) corresponds to the time taken (\Delta t), so the slope can also be rewritten as

m=\frac{\Delta s}{\Delta t}

which is equal to the speed of the object. Therefore, an object moving at constant speed would be represented by a line with constant slope, which means a straight line.

6 0
2 years ago
Read 2 more answers
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