The force a magnet exerts on another magnet, on iron or a similar metal, or on moving charges is

The velocity of the water in the pipe at right is given by V1 = 0.5t m/s and V2 = 1.0t m/s, where t is in seconds. Determine the

Nonamiya [84]

Answer:

A) At point 1, local acceleration = 0.5 m/s²

At point 2, local acceleration = 1.0 m/s²

B) Average Eulerian convective acceleration over the two points in the cross section shown = 0.5 m/s²

This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.

Explanation:

Local acceleration at those points is the instantaneous acceleration at those points and it is given as

a = dv/dt

At point 1, v₁ = 0.5 t

a₁ =dv₁/dt = 0.5 m/s²

At point 2, v₂ = 1.0 t

a₂ = dv₂/dt = 1.0 m/s²

b) Average Eulerian convective acceleration over the two points in the cross section shown = (change of velocity between the two points)/time

Change of velocity between the two points = v₂ - v₁ = 1.0t - 0.5t = 0.5 t

Time = t

Average acceleration = 0.5t/t = 0.5 m/s²

This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.

8 0

3 years ago

Determine W (or fuel energy) required to launch a satellite of mass m at rest from a launching pad placed at the surface earth,

jeka57 [31]

Answer:

5. 9GmM/(10R)

Explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE = $-\frac{GMm}{R}$

Therefore: W = Energy at earth surface - $\frac{GMm}{R}$

Energy at earth surface (E) at an altitude of 5R = $-\frac{GMm}{5r} +\frac{1}{2}mV^2$

But $V=\sqrt{\frac{GM}{5R} }$

Therefore: $E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2= -\frac{GMm}{5R}+\frac{GMm}{10R} = -\frac{GMm}{10R}$

W = E - PE

$W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}$

7 0

3 years ago

A teacher asks her students to jump off of the ground. Once the students complete the task, she says, "All of you just made Eart

Crazy boy [7]

Answer:

Explained below

Explanation:

A) Newton's first law of motion states that an object will remain at rest or continue in its current state of motion except it is acted upon by another force.

Now using this law, when you jump off the ground, the earth will move a tiny bit and accelerate due to the force applied by the jumping.

B) Newton's 2nd law states that the acceleration of a system is directly proportional to the net external force acting on that system, is in the same direction with it and also inversely proportional to the mass.

In this case, when one jumps, an external force is exerted on the earth and we are told it is directly proportional to the acceleration of the system which in this case it's the earth, then it means that there is some motion by the earth even though you didn't see it move.

C) Newton's third law of motion states that to every action, there is an equal and opposite reaction.

In this case the motion of the jumper will lead to an equal and opposite reaction of the earth.

8 0

2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

3 years ago

Three examples that prove newton's law of motion

s344n2d4d5 [400]

If you have a skateboard and you skate into a tree on accident the same amount of force you put onto that tree when you was on the skateboard will come back at you when you bounce back

7 0

3 years ago