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4 years ago

A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity

Physics

1 answer:

Viktor [21]4 years ago

3 0

Answer:

-26 m/s.

Explanation:

Hello,

In this case, since the vertical initial velocity is 26 m/s and the vertical final velocity is 0 m/s at P, we compute the time to reach P:

$t=\frac{0m/s-26m/s}{-9.8m/s^2} =2.65s$

With which we compute the maximum height:

$y=26m/s*2.65s-\frac{1}{2}*9.8m/s^2*(2.65s)^2 \\\\y=34.5m$

Therefore, the final velocity until the floor, assuming P as the starting point (Voy=0m/s), turns out:

$v_f=\sqrt{0m/s-(-9.8m/s^2)*2*34.5m}\\ \\v_f=-26m/s$

Which is clearly negative since it the projectile is moving downwards the starting point.

Regards.

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3 years ago

If the particles are oxygen ions carrying a charge of −2e and the magnetic field magnitude is 0.12 T , how long does it take eac

Natali5045456 [20]

Time taken by each ion to travel the semicircular path through the magnetic field is 34 X 10⁻²⁶s.

<u>Explanation:</u>

Given:

Charge of oxygen ion, q = -2

Magnetic Field, B = 0.12 T

Mass of O₂ , m= 2.6 X 10⁻²⁶ Kg

Time, T = ?

We know,

$T = \frac{2\pi m}{qB}$

Therefore,

T = 2 X 3.14 X 2.6 X 10⁻²⁶/ 2 X 0.12

T = 68 X 10⁻²⁶s

For semicircular orbit, Time taken would be T/2

T ₓ = 68 X 10⁻²⁶/2 s

T ₓ = 34 X 10⁻²⁶s

Therefore, time taken by each ion to travel the semicircular path through the magnetic field is 34 X 10⁻²⁶s.

4 0

3 years ago

A ball is being thrown straight upward and rises to a maximum height of 12.0m above its launch point. At what height above it’s

AlexFokin [52]

Answer:

9.0 m

Explanation:

Let the initial velocity be 'u'.

Given:

Final velocity is half of initial velocity.

Maximum height reached by ball (H) = 12.0 m

Acceleration of the ball is due to gravity (g) = -9.8 m/s²(Downward)

Now, first, we will find initial velocity of the ball using equation of motion given as:

$v^2=u^2+2a(\Delta y)$

For maximum height, final velocity is 0 as the ball stops at the maximum height temporarily. So, $v=0\ m/s$

Also, $\Delta y=H=12\ m$

Now, plug in all the values and solve for 'u'.

$0^2=u^2+2(-9.8)(12)\\\\u^2=235.2\\\\u=\sqrt {235.2} =15.34\ m/s$

Now, consider the motion of the ball till the velocity reaches half of initial velocity.

So, final velocity (v) = $\frac{u}{2}=\frac{15.34}{2}=7.67\ m/s$

Now, again using the same equation and finding the new height now. Let the new height be 'h'.

So, equation of motion is given as:

$v^2=u^2+2ah\\7.67^2=15.34^2+2\times -9.8\times h\\58.83=235.2-19.6h\\\\19.6h=235.2-58.83\\\\19.6h=176.37\\\\h=\frac{176.37}{19.6}\approx9.0\ m$

Therefore, the height reached by the ball when velocity is decreased to one-half of the initial velocity is 9.0 m.

8 0

3 years ago

A proton experiences a force of 4 Newton when it enters perpendicular to the direction of the magnetic field with a speed 100 m/

____ [38]

Answer:

F = 0 N

Explanation:

Force on a moving charge in constant magnetic field is given by the formula

$F = q(\vec v\times \vec B)$

so here it depends on the speed of charge, magnetic field and the angle between velocity of charge and the magnetic field

here when charge is moving with speed 100 m/s in a given magnetic field then the force on the charge is given as

$F = 4 N$

now when charge is moving parallel to the magnetic field with different speed then in that case

$\vec v \time \vec B = 0$

so here we have

F = 0

7 0

3 years ago

Find the configuration of any tow

Yuki888 [10]

Answer:

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6 0

3 years ago