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3 years ago

11

When force is perpendicular to the direction of motion, no work is done. please select the best answer from the choices provided

t f?

2 answers:

DerKrebs [107]3 years ago

3 0

True if that is all the question is asking

Nady [450]3 years ago

3 0

The answer is true if all is asking

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A spring with a mass on the end of it hangs in equilibrium a distance of 0.420 m above the floor. The mass is pulled down a dist

yanalaym [24]

Answer:

the highest point above the ground = 0.48 m

Explanation:

lets ignore air friction, spring details,

The rest mass hanges 0.42 m above the floor and you pull 0.06 m below the original position, this is done by putting energy into the stratched spring so that when released the mass will compress the spring and cover a distance 0.06 m above the rest height.

then the highest point the mass will attain is 0.42 m + 0.06 m = 0.48 m above the ground.

4 0

4 years ago

An airplane is heading due south at a speed of 430 km/h . A wind begins blowing from the northwest at a speed of 85.0 km/h (aver

Igoryamba

Answer:

It should fly 8° to west of south at 430km/h

Explanation:

According to the diagram. X components for both velocities must have the same magnitude in order to get the resultant velocity due south.

$V_{w}*cos(45) = V_{A}*sin(\alpha )$ Solving for α:

α = 8.03°

5 0

3 years ago

3.An object that begins at rest has an acceleration of 2 m/s/s What is its instantaneous speed after 3 seconds?

neonofarm [45]

The speed will be 6m/s i believe

6 0

4 years ago

Find out the speed of the car from home to school in meters per second.

julia-pushkina [17]

1.875

Explanation:

speed = distance moved ÷ time taken

7 0

3 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the

Rama09 [41]

1) $1.11\cdot 10^{-7} J$

The capacitance of a parallel-plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

$r=\frac{2.0 cm}{2}=1.0 cm=0.01 m$

so the area is

$A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2$

While the separation between the plates is

$d=0.50 mm=5\cdot 10^{-4} m$

So the capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F$

And now we can find the energy stored,which is given by:

$U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J$

2) 0.71 J/m^3

The magnitude of the electric field is given by

$E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m$

and the energy density of the electric field is given by

$u=\frac{1}{2}\epsilon_0 E^2$

and using

$E=4\cdot 10^5 V/m$ , we find

$u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3$

7 0

4 years ago