What is the BEST example of a purebred

A liquid of density 1200kg/m³ is filled in a beaker upto the depth of 50 cm. calculate the pressure at bottom of the breaker.(g=

sweet-ann [11.9K]

Density=1200kg/m^3=p
Height=50cm=0.5m=h
Acceleration due to Gravity=9.8m/s^2=g

$\boxed{\sf Pressure=pgh}$

$\\ \qquad\quad\sf{:}\dashrightarrow Pressure=1200(0.5)(9.8)$

$\\ \qquad\quad\sf{:}\dashrightarrow Pressure=600(9.8)$

$\\ \qquad\quad\sf{:}\dashrightarrow Pressure=5880Pa$

6 0

3 years ago

The isomerization of cyclopropane to form propene is a first-order reaction. At 760 K, 85% of a sample of cyclopropane changes t

Nataliya [291]

Answer:

so rate constant is 4.00 x 10^-4 $s^{-1}$

Explanation:

Given data

first-order reactions

85% of a sample

changes to propene t = 79.0 min

to find out

rate constant

solution

we know that

first order reaction are

ln [A]/[A]0 = -kt

here [A]0 = 1 and (85%) = 0.85 has change to propene

so that [A] = 1 - 0.85 = 0.15.

that why

[A] / [A]0= 0.15 / 1

[A] / [A]0 = 0.15

here t = (79) × (60s/min) = 4740 s

so

k = - {ln[A]/[A]0} / t

k = -ln 0.15 / 4740

k = 4.00 x 10^-4 $s^{-1}$

so rate constant is 4.00 x 10^-4 $s^{-1}$

3 0

2 years ago

A runner is jogging in a straight line at a

soldi70 [24.7K]

Answer:

13 km

Explanation:

The bird flies from the runner, to the finish line, and back to the runner. We can write two equations for the distance it travels:

d = 7.8 km + 7.8 km − 4.9 km/hr × t

d = 24.5 km/hr × t

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

t = d / 24.5

d = 7.8 + 7.8 − 4.9 (d / 24.5)

d = 15.6 − 0.2 d

1.2 d = 15.6

d = 13

The bird flies a cumulative distance of 13 km.

4 0

2 years ago

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

sergiy2304 [10]

Answer:

10.53m/s²

Explanation:

Centripetal acceleration is the acceleration of an object about a circle. The formula for calculating centripetal acceleration is expressed by:

$a = \frac{v^2}{r}$

v is the velocity of the car = 24.5m/s

r is the radius of the track = 57.0m

Substitute the given values into the formula:

$a = \frac{24.5^2}{57} \\\\a = \frac{600.25}{57}\\ \\a = 10.53m/s^{2}$

Hence the centripetal acceleration of the race car is 10.53m/s²

6 0

2 years ago

Velocity is intensive how?

White raven [17]

It is an intensive property as it varies with time and position within the system.

4 0

3 years ago