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jek_recluse [69]
3 years ago
7

A leaky faucet drips once each second. at this rate, it takes 1 hour to fill a 1-liter graduated cylinder to the 200-cm3 mark. w

hat unit conversion do you need in order to calculate the average volume of a single drop in cm3?
Physics
1 answer:
UNO [17]3 years ago
8 0
Define
v = volume of a drop per second, cm³/s

The time taken to fill 200 cm³ is 1 hour.
Let V = 200 cm³, the filled volume.
Let t = 1 h = 3600 s, the time required to fill the volume.

Therefore,
\frac{200 \,  cm^{3}}{v \, cm^{3}/s}  = 3600 \, s \\ v =  \frac{200}{3600} =0.0556 \, cm^{3}

The average volume of a single drop is approximately 0.0556 cm³.

Answer: 0.0556 cm³

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The Burj Khalifa in Dubai is the world's tallest building. The structure is 828 m (2,716.5 feet) and has more than 160 stories.
Elena L [17]

Answer:

 h = 599.5 m

Explanation:

Given,

height of structure = 828 m

weight of the tourist = 184 lb

                                 = 184 x 0.45359 = 83.43 Kg

Potential energy = 187000 J

PE = m gd

d = \frac{PE}{mg}

d = \frac{187000}{83.43\times 9.81}

h = 228.5 m

Height of the room above the ground.

 h = 828 - 228.5

 h = 599.5 m

Height of the floor above ground is equal to 599.5 m.

4 0
3 years ago
A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis
Natali5045456 [20]

Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

M_{before} = M_{after}

where:

M = momentum [kg*m/s]

M = m*v

where:

m = mass [kg]

v = velocity [m/s]

(m_{1} *v_{1} )-(m_{2} *v_{2})=(m_{1} *v_{3} )+(m_{2} *v_{4})

where:

m1 = mass of the basketball = 0.5 [kg]

v1 = velocity of the basketball before the collision = 5 [m/s]

m2 = mass of the tennis ball = 0.05 [kg]

v2 = velocity of the tennis ball before the collision = - 30 [m/s]

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

v4 = velocity of the tennis ball after the collision = 34 [m/s]

Now replacing and solving:

(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)

1 - (0.05*34) = 0.5*v3

- 0.7 = 0.5*v

v = - 1.4 [m/s]

The negative sign means that the movement is towards left

3 0
3 years ago
Consider the plot below describing motion along a straight line with an initial position of 10 m. −4 −3 −2 −1 0 1 2 3 4 5 6 1 2
Ugo [173]

Answer:

+1 m/s^2

Explanation:

Acceleration is given by

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

where

\Delta v is the change in velocity

\Delta t is the time interval in which the change in velocity occurs

To find the acceleration at 1 second, we can take the data at t = 1 s and t = 2. We find:

\Delta t = -3 -(-4) = 1 s

\Delta v = 2 - (1) = +1 m/s

So, the acceleration is

a=\frac{+1}{1}=+1 m/s^2

4 0
3 years ago
Gravity is defined as: Select one: a. an apple falling from a tree toward the ground. b. ability to accomplish a job with the le
Scorpion4ik [409]

Answer:

c. natural force or pull toward the earth

Explanation:

Gravity of the earth is the force of attraction that it naturally possesses to attract any mass.

An apple falls on the earth due to this force of gravity.

The force of gravity near to the surface of the earth is given as:

F=m.g

where:

m= mass of the body

g= acceleration due to gravity

The variation of the gravitational force with height is given as:

F'=m.g'

where:

g'=g\times (1+\frac{h}{R} )^{-2}

where

R = radius of the earth \approx64000\ km

4 0
3 years ago
PLEASE HELP ME (stop putting links ) Two objects m1 and m2, each with a mass of 5 kg and 6 kg separated by a distance. A third o
BabaBlast [244]

Answer:

Explanation:

Newton's Gravitation Law

\displaystyle \frac{GmM}{d^2}

where G is a constant, M and M the masses e d the distance betwen masses.

\displaystyle G\frac{5\cdot2}{x^2}=G\frac{6\cdot 2}{(2x+1)^2} \quad \sqrt{6}x=(2x+1)\sqrt {5} \quad x=\frac{\sqrt{5}}{\sqrt{6}-2\sqrt{5}}

7 0
3 years ago
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