A jet airliner, moving initially at 300 mi/h to the east, suddenly enters a region where the wind is blowing at 100 mi/h

If an ideal fluid element moves from a region of lower pressure to a region of higher pressure, is the work done on it by the pr

nikklg [1K]

Answer:

Positive

Explanation:

An ideal fluid is on which is incompressible and non-viscous.

In an ideal fluid element when a fluid element moves from a region of high pressure to region of low pressure then the work done on the fluid element is positive because the force acts in the direction from high pressure towards low pressure and displaces the fluid element, so the work-done is positive.

6 0

3 years ago

What would it mean if neither the red or blue litmus paper changes colors?

Anna35 [415]

if the color changes, it is neutral but if it stays the same, it is an acid.

8 0

3 years ago

In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor

leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

$W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}$ .....(I)

Where, $F(x)$ = net force

We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

$W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

We need to find the value of $\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

Using newton's second law

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}$ ...(II)

We know that,

Acceleration is rate of change of velocity.

$a=\dfrac{dv}{dt}$

Put the value of a in equation (II)

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

Now, the work done by the net force on the block is,

$W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0

3 years ago

What is the mass of an object that is 6kg on a planet with an acceleration due to gravity

goldfiish [28.3K]

Answer:

Mass of the body will be = 6 kg

Explanation:

Given:

Mass of an object = 6 kg

Acceleration due to gravity of the planet = 10 $m/s^2$

To find the mass of the body on that planet.

Solution:

Mass of the body is defined as the total amount of matter contained in the body.

Thus, mass of a body will always remain constant irrespective of the acceleration due to gravity. This is because it is an independent quantity and does not vary with acceleration due to gravity.

<em>It is the weight of the body that changes with the change in the acceleration due to gravity as it is given by:</em>

$W=mg$

where $m$ represents mass of the body and $g$ represents the accelration due to gravity.

Hence, the mass of the given body will remain = 6 kg.

6 0

3 years ago

A horizontal disk with a radius of 23 m rotates about a vertical axis through its center. The disk starts from rest and has a co

lys-0071 [83]

Answer:

time is 0.42 sec

Explanation:

Given data

radius = 23 m

angular acceleration = 5.7 rad/s²

to find out

time

solution

we know that radius is constant so that

tangential acceleration At = angular acceleration × radius ............. 1

tangential acceleration = 5.7 × 23 = 131.1 m/s²

and

radial acceleration Ar = (angular velocity)² × radius ........................2

we consider angular velocity = ω

this is acting toward center

so

compare 1 and 2

At = Ar

5.7 r =ω³ r

ω = √5.7 = 2.38746 rad/s

so

ω = 5.7 t

2.387 = 5.7 t

t = 2.387 / 5.7

t = 0.4187

time is 0.42 sec

8 0

3 years ago