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Sveta_85 [38]
3 years ago
8

Which has the greater change in momentum, a 50 gram clay ball that strikes a wall at 1 m/s and sticks or a 50 gram superball tha

t strikes the wall at 1 m/s and bounces away at 0.8 m/s. Explain?
Physics
1 answer:
jek_recluse [69]3 years ago
5 0

Answer: 50 gram superball that strikes the wall at 1 m/s and bounces away at 0.8 m/s has greater change in kinetic energy.

Explanation:

50 gram superball that strikes the wall at 1 m/s and bounces away at 0.8 m/s has the greater change in kinetic energy because the collision is elastic in nature that is bodies separates after collision and doesn't lose any kinetic energy.

Also for an elastic collision, both the momentum and energy of the bodies are conserved compare to inelastic collision where only momentum is conserved but not the kinetic energy(this is attributed to bodies that sticks together after collision).

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Formula for finding radius
Makovka662 [10]

r=d/2

to find radius, you divide the diameter by 2 or multiply it by one half

8 0
3 years ago
A cart on a horizontal, linear track has a fan attached to it. The cart is positioned at one end of the track, and the fan is tu
kompoz [17]

Answer:

a

  F = 0.0566 \  N  

b

   t =  6.147 \  s

Explanation:

From the question we are told that

     The distance travel in  4.22 s is  s =  1.44 \ m

     The mass of the cart plus the fan is  m  =  350 \  g =  0.35 \  kg

Generally from kinematic equation we have that

        s =  ut + \frac{1}{2}  * a * t^2

Here  u is the initial  velocity with value  u =  0 \ m/s

So  

         1.44=  0 * t + \frac{1}{2}  *  a * 4.22^2      

=>      a =  0.1617 \  m/s^2

Generally the net force is  

         F = m * a

=>      F = 0.35  *  0.1617  

=>      F = 0.0566 \  N  

Gnerally the new mass of the cart plus the fan is  M  =  656 \  g  = 0.656 \  kg

    The distance considered is s_1 = 1.63 \  m

     Generally the new acceleration of the cart is mathematically represented as

        F =  M  *  a_1

=>      a_1 =  \frac{F}{M}

=>      a_1 =  \frac{0.0566}{0.656}

=>      a_1 = 0.08628 \  m/s^2

Gnerally from kinematic equation we have

          s =  ut + \frac{1}{2} *  a_1 *  t ^2

Here u  is the initial velocity and the value is zero because it started from rest  

=>       1.63 =  0 * t + \frac{1}{2} *  0.08628*  t ^2

=>        t =  6.147 \  s

6 0
3 years ago
A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −
romanna [79]

Answer:

The speed of electron is 1.5\times10^{7}\ m/s

Explanation:

Given that,

Electric field E=1.50\times10^{3}\ N/C

Distance = -0.0200

The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy P.E=5.04\times10^{-17}\ J

Change in potential energy \Delta P.E=-9.60\times10^{-17}\ J as it goes x = 0.190 m to x = -0.210 m

We need to calculate the work done by the electric field

Using formula of work done

W=-eE\Delta x

Put the value into the formula

W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))

W=-5.04\times10^{-17}\ J

We need to calculate the initial velocity

Using change in kinetic energy,

\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2

\Delta K.E=\dfrac{-3mv^2}{8}

Now, using work energy theorem

\Delta K.E=W

\Delta K.E=\Delta U

So, \Delta U=W

Put the value in the equation

\dfrac{-3mv^2}{8}=-5.04\times10^{-17}

v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}

Put the value of m

v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}

v=1.21\times10^{7}\ m/s

We need to calculate the change in potential energy

Using given potential energy

\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})

\Delta U=-4.56\times10^{-17}\ J

We need to calculate the speed of electron

Using change in energy

\Delta U=-W=-\Delta K.E

\Delta K.E=\Delta U

\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}

Put the value into the formula

v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}

v_{f}=1.5\times10^{7}\ m/s

Hence, The speed of electron is 1.5\times10^{7}\ m/s

4 0
3 years ago
When do phase transitions occur in molecules?
jekas [21]
It occurs when energy is supplied or withdrawn :)
5 0
3 years ago
If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 1030 kg, what is the average distance between the planet
Shtirlitz [24]

The average distance between the planet and the star is:

R=9.36*10^11 m

Orbital velocity  v=√{(G*M)/R},

G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,

M = mass of the star

R =distance from the planet to the star.

v=ωR, with ω as the angular velocity and R the radius

ωR=√{(G*M)/R},

ω=2π/T,

T = orbital period of the planet

To get R we write the formula by making R the subject of the equation

(2π/T)*R=√{(G*M)/R}

{(2π/T)*R}²=[√{(G*M)/R}]²,

(4π²/T²)*R²=(G*M)/R,

(4π²/T²)*R³=G*M,

R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

As was already said, Earth is located roughly 150 million kilometres (93 million miles) from the Sun on average. It is 1 AU. Mars is on our fictitious football field's three-yard line. On average, the distance between the Sun and the red planet is around 142 million miles (228 million kilometres).

Learn more about average distance:

brainly.com/question/18366547

#SPJ4

The complete question is ''If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 10^30 kg, what is the average distance between the planet and the star? 1.43 × 10^9 m 9.36 × 10^11 m 5.42 × 10^13 m 9.06 × 10^17 m''.

4 0
1 year ago
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