Can someone please help?

Water ______________ easier in the mountains because there is less air pressure.

m_a_m_a [10]

Answer:

water flows easier in the mountains because there is less air pressure.

6 0

3 years ago

Una manguera de agua de 1.3 cm de diametro es utilizada para llenar una cubeta de 24 Litros. Si la cubeta se llena en 48 s. A) ¿

Viefleur [7K]

Answer:

We must translate this:

a 1.3 cm diameter water hose is used to fill a 24-liter bucket. If the bucket is filled in 48 s.

A) What is the speed with which the water leaves the hose?

B) if the diameter of the hose is reduced to 0.63 cm and assuming the same flow, what will be the speed of the water leaving the hose?

A) If the velocity of the water is Xcm/s

and the radius of the hose is equal to half its diameter, so it is 1.3cm/2

Then in one second we can considerate that a cylinder of:

V = pi*(1.3cm/2)^2*X cm^3 of water.

So we have that quantity in one second of flow.

where pi = 3.14

then in 48 seconds, the amount of water in the bucket is:

V = 48*pi*(1.3/2)^2*X = 24 L = 24,000 cm^3

Now we need to solve this for X.

48*3.14*(1.3/2)^2*X = 24000

63.679*x = 24000

x = 24000/63.679 = 376.89

So the velocity of the water is 376 cm per second.

B) if the diameter is 0.64cm, we have the equation:

48*3.14*(0.63/2)^2*x = 24000

14.955*X = 24000

X = 24000/14.955 = 1604.814 cm/s

6 0

3 years ago

Q 24, 25, 26 i dont get them and need answers for it

Alexandra [31]

Answer:

24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)

Explanation:

24) This problem can be solved by means of the following equation.

$DU = Q-W\\$

where:

DU = internal energy difference [J]

Q = Heat transfer [J]

W = work [J]

Since there are no temperature changes the internal energy change is equal to zero

DU = 0

therefore:

$Q = W\\$

The work is equal to the heat transfered, W = 75 [J].

25) The heat transfer can be calculated by means of the following equation.

$Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]$

Q = 0.4*897*5 = 1794[J]

Work is equal to heat transfer, W = 1794[J]

26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.

$E_{p}=Q\\\\E_{p}=m*g*h$

where:

m = mass = 0.5[kg]

g = gravity = 9.81[m/s^2]

h = 1.5 [m]

$E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]$

The heat developed can be calculated by means of the following equation.

$Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]$

The number of times will be calculated as follows

n = 65/7.36

n = 8.8 (times) or 9 (times)

7 0

4 years ago

A jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction

Elis [28]

Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

$v=\frac{10}{2}=5m/s\;\cdots(i)$

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, $d=vt$

From equation (i)

$\Rightarrow d=5t\;\cdots(ii)$

As the jogger starts from origin, so, the distance, $d$ , also represents the position of the jogger at the time $t$ s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

$d=5\times 5=25 m$

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

$d=5\times 15=75 m$

So, the jogger is at a distance of 75 m from the origin.

8 0

3 years ago

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

7 0

3 years ago