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1 year ago

5

The water hose at your home outputs 24 gallons of water a minute. how many cubic centimeters per second does the hose output? th

ere are 264.172 gal in 1 m3 and 100 cm in 1 m.

1 answer:

adoni [48]1 year ago

7 0

When we convert 24 gallon/min to cubic centimeters per second (cm³/s), the result obtained is 1514.164 cm³/s

<h3>How to convert 24 gallon/min to cm³/min</h3>

We'll begin by converting 24 gallon/min to cm³/min. This can be obtained as follow:

1 gallon/min = 3785.41 cm³/min

Therefore,

24 gallon/min = (24 gallon/min × 3785.41 cm³/min) / 1 gallon/min

24 gallon/min = 90849.84 cm³/min

<h3>How to convert 90849.84 cm³/min to cm³/s</h3>

We can convert 90849.84 cm³/min to cm³/s as follow:

60 cm³/min = 1 cm³/s

Therefore,

90849.84 cm³/min = (90849.84 cm³/min × 1 cm³/s) / 60 cm³/min

90849.84 cm³/min = 1514.164 cm³/s

Thus,

24 gallon/min = 1514.164 cm³/s

Learn more about conversion:

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A 160-N child sits on a light swing and is pulled back and held with a horizontal force of 100 N. The magnitude of the tension f

guajiro [1.7K]

Answer:

T = 190 N

Explanation:

When child is sitting on the swing then the weight of the child is vertically downwards

So it is

$F_g = 160 N$

now a force of 100 N is acting on the swing in horizontal direction

so it is given as

$F_x = 100 N$

now the net force is resultant force due to gravity and horizontal force

so it is given as

$T = \sqrt{F_x^2 + F_g^2}$

$T = \sqrt{100^2 + 160^2}$

$T = 190 N$

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2 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

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3 years ago

What happens when bromine reacts with carbon?

serious [3.7K]

D. Electrons are shared between the bromine atoms and carbon atoms

4 0

3 years ago

Read 2 more answers

Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration

eimsori [14]

The complete text of the problem is:

<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

<em />

<u>Solution:</u>

1) Acceleration: $-2336 m/s^2$ on the hardwood floor, $-382 m/s^2$ on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

$v^2 - u^2 = 2ad$

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

a = g = 9.8 m/s^2 is the acceleration of gravity

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

$v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s$

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

$d = 1.8 mm = 0.0018 m$

So we can find the acceleration by using again the equation

$v'^2 - v^2 = 2ad$

Solving for a,

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2$

For the carpeted floor instead,

$d=1.1 cm = 0.011 m$

therefore the acceleration is

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2$

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

$a=\frac{v'-v}{t}$

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

$a=-2336 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms$

For the carpeted floor,

$a=-382 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms$

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

8 0

3 years ago

a cell phone uses 3.0V battery. the circuit board needs a 0.05 A current. what size resistor is needed to generate this current

san4es73 [151]

Answer:

60 Ohms

Explanation:

Ohms law states that the voltage in the circuit is directly proportional to the current through the circuit components and expressed as

V=IR

Where V is the voltage, I is current and R is resistance

Making R the subject of the formula then

$R=\frac {V}{I}$

Substituting 3.0V for V and 0.05 A for I then

$R=\frac {3}{0.05}=60.0$

Therefore, resistance is 60.0 Ohms

7 0

3 years ago