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irina1246 [14]
3 years ago
15

1. If an object of mass m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, th

en the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor of ___. The new velocity will be ___v.
2. If an object of mass m collides and velocity v collides inelastically with an object of mass 4m that is initially at rest, then the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor of ___. The new velocity will be ___v.

3. If an object of mass 3m collides and velocity v collides inelastically with an object of mass 4m that is initially at rest, then the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor if ___. The new velocity will be ___v.

4. If an object of mass 5m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, then the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor if ___. The new velocity will be ___v.
Physics
1 answer:
Archy [21]3 years ago
3 0

Answer:

1a). 4 times. 1b) 4. 1c) 1/4.

2a) 5 times. 1b) 5 1c) 1/5

3a) 7/3 times 3b) 7/3 3c) 3/7

4a) 8/5 times 4b) 8/5 4c) 5/8

Explanation:

1) Assuming no external forces acting during the collision, total momentum must be conserved, so the following general expression applies:

m₁*v₁₀ + m₂*v₂₀ = m₁*v₁f  + m₂*v₂f (1)

If we assume that the collision is perfectly inelastic, this means that both masses stick together after the collision, so v₁f = v₂f.

If m₂ is initially at rest, ⇒ v₂₀ = 0.

Replacing in (1) we get the expression of vf as a function of v₁₀, as follows:

vf = v₁₀*(m₁/(m₁+m₂)

So, for the four cases we have the following:

1) initial mass = m

  final mass = m+3m = 4 m

⇒final mass / initial mass = 4

vf = v₀* (m/4m) = v₀/4  ⇒v₀/vf = 4

So, the velocity of the system will decrease by a factor of 4. The new velocity will be vf= v₀/4.

2) Applying the same considerations, we get:

2a)  final mass / initial mass = 5

2b) vf = v₀* (m/5m) = v₀/5  ⇒v₀/vf = 5

2c) vf = v₀/5

3) Applying the same considerations, we get:

3a)  final mass / initial mass = 7/3

3b) vf = v₀* (3m/7m) =3/7* v₀  ⇒v₀/vf = 7/3

3c) vf = 3/7*v₀

4) Applying the same considerations, we get:

4a)  final mass / initial mass = 8/5

4b) vf = v₀* (5m/8m) = 5/8*v₀ ⇒v₀/vf = 8/5

4c) vf = 5/8*v₀

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If we use to construct the latches on the windows and doors, then the magnetism will keep thee latches secure.
Daniel [21]

Answer:

<em>a. True</em>

<em></em>

Explanation:

I'll assume the question is about magnetic latches and locks.

Magnetic door locks use an electromagnetic force to stop doors from opening, so they are ideal for security. There are two main types of electric locking devices. Locking devices can either be a fail-secure locking device that remains locked when power is lost, or a fail-safe locking device that is unlocked when de-energized. An electromagnetic lock creates a magnetic field when energized or powered up, this causes an electromagnet and armature plate to become attracted to each other strongly enough to keep a door from opening.

6 0
4 years ago
In 1977 off the coast of Australia, the fastest speed by a vessel on the water
fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration a_{ave} is the variation of velocity  \Delta V over a specified period of time  \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

a_{ave}=1.80 m/s^{2}

\Delta V=V_{f}-V_{o} being V_{o}=0 the initial velocity and V_{f} the final velocity

\Delta t=85.6 s

Then:

a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}

Since V_{o}=0:

a_{ave}=\frac{V_{f}}{\Delta t}}

Finding V_{f}:

V_{f}=a_{ave} \Delta t

V_{f}=(1.80 m/s^{2})(85.6 s)

Finally:

V_{f}=154.08 m/s

8 0
3 years ago
A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl
Taya2010 [7]

Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given that,

Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

F-f=ma

Put the value into the formula

340-f=60\times2.0

f=340-60\times2.0

f=220\ N

We need to calculate the coefficient of friction

Using formula of friction force

f= \mu mg

\mu=\dfrac{f}{mg}

\mu=\dfrac{220}{60\times9.8}

\mu =0.37

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

F = ma

a=\dfrac{F}{m}

a=\dfrac{220}{100}

a=2.2\ m/s^2

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

3 0
4 years ago
A 3 kg stone is dropped from a height of 4 m what is its momentum as it strikes the ground
Vitek1552 [10]

Okay first you have to recognize that the maximum Gravitational potential energy will equal the maximum kinetic energy. The maximum GPE will equal when the stone is at its highest point and the max KE will be right before the stone hits the ground. So

GPE max = KE max

mgh = 1/2mv^2

Mass cancels as it is on both sides

gh = 1/2v^2

Multiply by 2

2gh = v^2

Square root

v = √2gh

Now plug in

v = √2(9.8)(4)

v = 8.85 m/s

Now use the mass to calculate the momentum just before it hits the ground as this is the speed right before it hits the ground

p = mv

p = (3)(8.85)

p = 26.55 kgm/s

7 0
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Sonbull [250]
Heat can be reduced but it cannot be eliminated completely there is some amount if heat energy present even if it is absolute zero(-273degrees Celsius)
3 0
4 years ago
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