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Artyom0805 [142]
4 years ago
15

You hold a small metal ball of mass a height above the floor. You let go, and the ball falls to the floor. Choose the origin of

the coordinate system to be on the floor where the ball hits, with up as usual.
Just after release, what are yᵢ and vᵢᵧ? Express all results in terms of m, g, and h.
Physics
1 answer:
baherus [9]4 years ago
5 0

Answer:

yᵢ = h

vᵢᵧ = 0

Explanation:

Let initial velocity = u

     final velocity = v

     height = h

     acceleration due to gravity = g

Therefore fro equation of motion,

v^{2}=u^{2}+2gh

Here initial velocity is zero.

Hence, v^{2}=2gh

v = \sqrt{2gh}

but since it points downward and you positive is taken in the up direction

v = -\sqrt{2gh}

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9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku
Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately <u>33.23 m/s</u> in a direction from the North of

approximately <u>9.18°</u>.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

  • v₁ = -(\frac{\sqrt{2} }{2} × 3.5)·i + (\frac{\sqrt{2} }{2} × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

  • v₂ =  (\frac{\sqrt{2} }{2} × 4.0)·i - (\frac{\sqrt{2} }{2} × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-(\frac{\sqrt{2} }{2} × 3.5) - (\frac{\sqrt{2} }{2} × 4.0))·i + ((\frac{\sqrt{2} }{2} × 3.5 + 12.5) + (\frac{\sqrt{2} }{2} × 4.0 + 15))·j

  • v₁₂ = -\frac{ 15 \cdot \sqrt{2} }{4}·i + \frac{ 110 + 15 \cdot \sqrt{2} }{4}·j

Red Skull's velocity relative to Captain America,  v₁₂ = -\frac{ 15 \cdot \sqrt{2} }{4}·i + \frac{ 110 + 15 \cdot \sqrt{2} }{4}·j

  • v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

  • Red Skull appears to be moving West at <u>5.3 m/s</u> and North at <u>32.8 m/s</u>

  • The direction is arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}

Therefore;

  • Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

  • |v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ <u>33.23 m/s</u>

Learn more here:

brainly.com/question/24430414

6 0
3 years ago
A moon of mass 1×10^20kg is in a circular orbit around a planet. The planet exerts a gravitational force of 2×10^21n on the moon
vagabundo [1.1K]

Hi there!

In this instance, the centripetal force experienced by the moon is equivalent to the gravitational force.

Thus:
\large\boxed{F_c = F_g}

Centripetal acceleration, according to Newton's Second Law:


\Sigma F = ma \\\\F_c = m * a_c\\\\a_c = \frac{F_c}{m}

Therefore:

a_c = \frac{F_g}{m_m} = \frac{2 * 10^{21}N}{1 * 10^{20}kg} = \boxed{20 N/kg}

6 0
2 years ago
The more domains that are aligned in a magnet ___________.
I am Lyosha [343]
Hey, Sissy05pooh!
The more domains that are aligned in a magnet it is stronger, and the magnetic field is larger.
6 0
3 years ago
The combined focal length of two thin lens is 24 cm and the focal length of one converging lens is 8
qwelly [4]

Answer: f = -12 cm

Explanation: <u>Combined</u> <u>lenses</u> is an array of  simple lenses with a common axis. The combination is useful for correction of optical aberrations which cannot be corrected by simple lenses.

When two lenses are in contact and are thin, focal lengths are related as:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}

If there is a distance between the lenses, the focal length will be:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}} -\frac{d}{f_{1}f_{2}}

Since the lenses in the question above are thin and in contact, the focal length of one of them will be:

\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}

\frac{1}{f_{2}} =\frac{1}{f_{1}} -\frac{1}{F}

\frac{1}{f_{2}} =\frac{1}{8} -\frac{1}{24}

\frac{1}{f_{2}} =\frac{-2}{24}

f_{2}= -12

The focal length of the other lens is -12 cm, with the negative sign meaning it's a converging lens.

7 0
3 years ago
A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force)
salantis [7]

Answer:

<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>

Explanation:

The Non conservative force is defined as a force which do not store energy or get he energy dissipate the energy from the system as the system progress with the motion.

Given are

   <em>  mass of the student 73 kg</em>

<em>      height of water glide 11.8 m</em>

<em>      work done as -5.5*10³ J</em>

Have to find speed at which the student goes down the glide.

According to<em> Law of Conservation of energy</em>,

          K.E =P.E+Work Done

 mv²/2=mgh +W

Rearranging the above eqn for v

v = √2(gh+W/m)

Substituting values,

V =  12.48 m/s.

<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>

 

3 0
3 years ago
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