What condition is necessary for the sustained flow of water in a pipe?

A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

$F_k = F_{W,E}$

$kx_1 = mg$

The extension of the spring due to the weight of the object on Moon is a value of $x_2$ , then

$kx_2 = mg_m$

Recall that gravity on the moon is a sixth of Earth's gravity.

$kx_2 = m\frac{g}{6}$

$kx_2 = \frac{1}{6} mg$

$kx_2 = \frac{1}{6} kx_1$

$x_2 = \frac{1}{6} x_1$

We have that the displacement at the earth was $x_1 = 0.3m$ , then

$x_2 = \frac{1}{6} 0.3$

$x_2 = 0.05m$

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0

3 years ago

A beaker weighs 0.4N when empty and1.4N when filled with water what does ot weigh when filled with brine of density 1.2 g/cm3

PtichkaEL [24]

Answer:2.47

Explanation:

So, the beaker weighs 1.40N when filled with water, brine of density weighs about 1.7N, you add the density + water. Have a good day!

7 0

2 years ago

An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then

gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

$a=\frac{dv}{dt}=1.76$

So, the velocity can be obtained by integrating this expression:

$v=1.76t$

The velocity is, by definition: $v=\frac{dx}{dt}$ , so

$dx=1.76tdt\\x=1.76\frac{t^{2}}{2}$ .

Do x=11 in order to find the time spent.

$11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s$

At this time the velocity is: $v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}$

This velocity remains constant in the section 2, so for that section the movement equation is:

$x=v*t\\t=\frac{x}{v}$

The left distance is 89 meters, and the velocity is $6.2225\frac{m}{s}$ , so:

$t=\frac{89}{6.2225}=14.303s$

So, the total time is 14.303+3.5355s=17.838s

7 0

3 years ago

Wait so if dogs can eat human food why can't humans eat dog food?

NemiM [27]

Because animals are animals and humans hopefully don’t want to be treated like animals, also it is preferred that they don’t eat human food like gum or chocolate…

7 0

2 years ago

Read 2 more answers

You stand17.5 m from a wall holding a baseball. You throw the baseball at the wall at an angle of 20.5∘ from the ground with an

Lelechka [254]

Answer: 8.8 m

Explanation:

The movement of the baseball to the wall is a example of parabolic motion. While the baseball aproach the wall it is affected by gravity.

In this case, because the Initial Velocity of the ball is a vecotr, it can be defined using its two directionals compounds. One, on the Y-axis and another on the X-axis. This can be related, on how a hypotenuse is the product of two legs of the triangle. Because of this, each one of this, can be know using the following equation:

Vx = Vo * cos(∅)

Vy = Vo * sin(∅)

Where Vo is the initial velocity 27.5 m/s, and ∅ is the angle which is 20.5°. So we calculate:

Vx = 27.5m/s * cos(20.5)

Vx = 27.5m/s * 0.936

Vx = 25.74m/s

Vy = 27.5m/s * sin(20.5)

Vy = 9.62m/s

Now the movement is divided on two parts, one under the effect of gravity and another one with a constant velocity.

To know how tall does the baseball hit the wall, we need to know first how much time it takes the ball to reach the wall on the X-axis. The wall is 17.5m away, we velocity on Vx that is constant we can calculate as it follow:

Time (T) = Distance (D) / Velocity (V)

Where Distance is 17.5m and our Velocity is the Vx calculated before.

T = 17.5 m / 25.74m/s

T = 0.68s

This is the time it takes the ball to reach the wall.

Know with the time, we can calculate the how tall it got on that time with the following equation:

$x = (Vo*t) + (\frac{1}{2} *a*t^{2} )$