Is it okay to keep an albino gecko

In a towns water system pressure gauges in still water at street level reads 150kpa. if a pipeline connected to the system break

AnnZ [28]

Answer:

The water shoots 15.31 m high above the street level.

Explanation:

The gauge pressure drives the motion of the water to whixhever height it will attain. The expression relating the gauge pressure to the height reached by the water, is

P = ρgh

P = Gauge Pressure = 150 kPa = 150,000 Pa

ρ = density of the fluid (water) = 1000 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = Height reached by the water = ?

150,000 = 1000 × 9.8 × h

h = (150000) ÷ 9800 = 15.306 = 15.31 m

Hope this Helps!!!

5 0

4 years ago

2. Another airplane is flying in the opposite direction. It covers the same

Korolek [52]

Answer:300

Explanation:

6 0

4 years ago

In a drag race, a car takes off from rest and covers 375 meters in 5 seconds. What is the

salantis [7]

Explanation:

s = ut + 1/2 a t^2

375 = 0 * 5 + 1/2 * a * (5)^2

375 = 1/2 * a * 25

a = 375*2/25

a = 15* 2

a = 30m/sec^2

v = u + at

v = 0 + 30 * 5

v = 150 m/sec

hope it helps you

3 0

3 years ago

What information about elements can be collected from the periodic table?

Dafna1 [17]

Answer:

Using the data in the table scientists, students, and others that are familiar with the periodic table can extract information concerning individual elements. For instance, a scientist can use carbon's atomic mass to determine how many carbon atoms there are in a 1 kilogram block of carbon.

Explanation:

HOPE THIS HELPS LIKE AN RATE PLZ

3 0

3 years ago

Read 2 more answers

A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin

gogolik [260]

Answer:

a) 23 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$p_{o} = p_{f} (1)$

The initial momentum p₀, can be written as follows:

$p_{o} = m_{1} * v_{1o} + m_{2}* v_{2o} = 6.0 kg * 5.0 m/s + 2.0 kg * v_{2o} (2)$

The final momentum pf, can be written as follows:

$p_{f} = (m_{1} + m_{2} )* v_{f} = 8.0 kg* (-2.0 m/s) (3)$

Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

$v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg} = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s (4)$

This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.

6 0

3 years ago