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1 year ago

At what height h above the ground does the projectile have a speed of 0.5v?

Physics

1 answer:

maw [93]1 year ago

6 0

Answer:

$h=\dfrac{3v^2}{8g}$

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

$v=u+at$

$v=u-gt$

Initial speed of the projectile is v and final speed is 0.5 v.

$0.5v=v-gt$

$t=\dfrac{v}{2g}$

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

$h=vt-\dfrac{1}{2}gt^2$

$h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2$

$h=\dfrac{3v^2}{8g}$

So, the height of the projectile above the ground is $\dfrac{3v^2}{8g}$ . Hence, this is the required solution.

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tekilochka [14]

This problem has three questions I believe:

> How hard does the floor push on the crate?

We have to find the net vertical (normal) Fn force which results from Fp and Fg.
We know that the normal component of Fg is just Fg, which is equal to as 1110N.
From the geometry, the normal component of Fp can be calculated:
Fpn = Fp * cos(θp)
= 1016.31 N * cos(53)
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The total normal force Fn then is:
Fn = Fg + Fpn
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> Find the friction force on the crate

We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely, so we can say that the horizontal component is zero:

Fh = Fph + Fgh
= (Fp * sin(θp)) + 0
= 1016.31 N * sin(53)
= 811.66 N

> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

μs = Fh / Fn

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8 0

2 years ago

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed?

sweet [91]

Answer:

Speed of the electron will be $v=5.896\times 10^7m/sec$

Explanation:

We have given that charge on electron $e=1.6\times 10^{-19}C$

Mass of electron $m=9.11\times 10^{-31}kg$

Potential difference = $V=9.9KV=9.9\times 10^3volt$

Now according to energy conservation $eV=\frac{1}{2}mv^2$

$1.6\times 10^{-19}\times 9.9\times 10^3=\frac{1}{2}\times 9.11\times 10^{-31}v^2$

$v=5.896\times 10^7m/sec$

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1 year ago

An airplane is delivering food to a small island. It flies 100 m above the ground at a speed of 150 m/s .

miss Akunina [59]

Answer:

The airplane should release the parcel $6.7*10^2$ m before reaching the island

Explanation:

The height of the plane is $y_0=100m$ , and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

$y=y_0 - \frac{gt^2}{2}$ [1]

And the distance X is

x = V.t [2]

Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

$X = V . \sqrt{\frac{2y_0}{g}}$

Using the given values:

$x = 150 m/s \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}$

x = $6.7*10^2$ m

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2 years ago

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2 years ago

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erica [24]

Answer:

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Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

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1 year ago