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2 years ago

Help please (about projectile motion at an angle)

Physics

1 answer:

PolarNik [594]2 years ago

7 0

Answer:

Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao

Explanation:

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A parallel-plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative pla

MissTica

Answer:

The speed of proton when it emerges through the hole in the positive plate is $2.05\times 10^5\ m/s$ .

Explanation:

Given that,

A parallel-plate capacitor is held at a potential difference of 250 V.

A A proton is fired toward a small hole in the negative plate with a speed of, $u=3\times 10^5\ m/s$

We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

$qV=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2\\\\1.6\times10^{-19}\times250=\dfrac{1}{2}mv^2-\frac{1}{2}\cdot1.67\times10^{-27}\cdot(3\times10^{5})^{2}\\\\\dfrac{1}{2}mv^2=3.515\cdot10^{-17}\\\\v=\sqrt{\dfrac{3.515\cdot10^{-17}\cdot2}{1.67\times10^{-27}}}\\\\v=2.05\times 10^5\ m/s$

So, the speed of proton when it emerges through the hole in the positive plate is $2.05\times 10^5\ m/s$ .

5 0

2 years ago

If a 5kg ball is traveling at 20 m/s and it’s stopped in 4s, what’s the impulse on the ball?

Anna [14]

FT=MV

F(4)=(5)(20)

F= 100/4

F= 25 N

4 0

2 years ago

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called ____.

irina1246 [14]

Answer:

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called COMETS.

4 0

2 years ago

Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, $\overrightarrow{v_1}=-120\ \vec{i}$

Velocity in north direction is, $\overrightarrow{v_2}=120\ \vec{j}$

Now, since $v_1\ and\ v_2$ are perpendicular to each other, their resultant magnitude is given as:

$|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}$

Plug in the given values and solve for the magnitude of the resultant.This gives,

$|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h$

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

$x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)$

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

4 0

2 years ago

A student drops a rock from a bridge to the water 12.7 m below. with what speed does the rock strike the water? the acceleration

kolezko [41]

The rock strike the water with the speed of 15.78 m/sec.

The speed by which rock hit the water is calculated by the formula

v= $\sqrt{2gh}$

v= $\sqrt{2*9.8*12.7}$

v=15.78 m/sec

Hence, the rock strike the water with the speed of 15.78 m/sec.

7 0

2 years ago