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2 years ago

Determine the work done to raise a mass of 8.0 kg through a height of 2.5 m on this planet.

Physics

1 answer:

ololo11 [35]2 years ago

6 0

Answer:

196 J

Explanation:

Formula

Work done(W) = Mass (m) x gravity (g) x height (h)

Here, we are given :

m = 8.0 kg
g = 9.8 m/s² (on Earth)
h = 2.5 m

Solving

W = 8 × 9.8 x 2.5
W = 20 × 9.8
W = 196 J

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A ball moves along a table at a constant velocity and then rolls off the edge of the table. The forces that should be included i

Len [333]

Answer:

Gravity

Explanation:

When the ball is falling to the ground, it is already detached from the table, so the table does not exert any force on it.

Gravity is always present, therefore it is acting on the ball (acting downward), so it must be included into the free-body diagram. Apart from that, there are no other forces acting on the ball (if we neglect air resistance, which is negligible, and it is not mentioned in the options given), therefore the only force which has to be included in the diagram is gravity.

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3 years ago

Electrons are accelerated through a voltage difference of 230 kV inside a high voltage accelerator tube. What is the final kinet

Aliun [14]

Answer:

kinetic energy is 3.68 × $10^{-14}$ J

speed of these electrons is 0.5565 c

Explanation:

Given data

voltage difference = 230 kV

to find out

kinetic energy and speed of these electrons

solution

we know kinetic energy formula that is

kinetic energy = qv

kinetic energy = 1.6 × $10^{-19}$ × 230 × $10^{3}$

kinetic energy = 3.68 × $10^{-14}$ J

and

speed of electron that is

kinetic energy = m1c² - m2 c²

here m1 = m / √(1-v/c)²

KE/mc² = ( 1 / √(1-(v/c)² ) - 1

put value here

3.68 × $10^{-14}$ / (3× 10^8)²×9.11× $10^{-31}$ = ( 1 / √(1-v/c)² ) - 1

0.448835223 = ( 1 / √(1-(v/c)² ) - 1

( 1 / √(1-(v/c)² ) = 0.448835223 + 1 = 1.448835223

√(1-(v/c)² = 1 / 1.448835223 = 0.6902096

(v/c)² = 1 - 0.6902096 = 0.309790

v = 0.5565 c

speed of these electrons is 0.5565 c

5 0

3 years ago

Which parts of The Action Potential Are Represented On The ECG?

Aloiza [94]

<h3>Which parts of The Action Potential Are Represented On The ECG?</h3>

The QRS complex of the electrocardiogram corresponds to the action potential depolarization, while the T wave is associated with ventricular repolarization. Torsades de pointes is associated with the twisting of the QRS complex around the isoelectric line on the electrocardiogram.

3 0

3 years ago

Read 2 more answers

An electric field of intensity 4.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.

Anestetic [448]

Answer:

Electric flux will be equal to $1041.25Nm^2/C$

Explanation:

We have given electric field $E=4.25KN/C=4.25\times 10^3N/C=4250N/C$

Sides of rectangle is given as 0.350 m wide and 0.700 m long

So length of rectangle l = 0.7 m and width w = 0.35 m

So area of rectangle $A=0.7\times 0.35=0.245m^2$

It is given that electric field is along the x axis

So angle between plane and electric filed $\Theta =0^{\circ}$

So electric flux will be equal to $\Phi =EAcos\Theta =4250\times 0.245\times cos0^{\circ}=1041.25Nm^2/C$

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3 years ago

Suppose that the dipole moment associated with an iron atom of an iron bar is 2.8 × 10-23 J/T. Assume that all the atoms in the

masya89 [10]

To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.

The diple moment associated with an iron bar is given by,

$\mu = \alpha *N$