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Solnce55 [7]
3 years ago
8

Does an object in inelastic collisions exert less force than elastic collisions?

Physics
1 answer:
Montano1993 [528]3 years ago
7 0

Answer:

The object exerts <em>same</em> amount of force in elastic and inelastic collisions.

Explanation:

The force an object exerts is not different between the two types of collisions. What changes from elastic to inelastic is the amount of energy transformed from kinetic to other type during an inelastic collision.

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

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

B = Magnetic field = 0.044 T

q = Charge of electron = 1.6\times 10^{-19}\ C

The centripetal force and the magnetic forces are conserved

m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}

Velocity of first electron

v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s

Velocity of second electron

v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s

Total kinetic energy is given by

K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV

The energy of incident electron is 114.92749 keV

5 0
4 years ago
What is the speed of sound in air that has a temperature of 19.9 celcius
klasskru [66]

Answer4.4 :

Explanation:

7 0
3 years ago
Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is
NARA [144]

Answer:

E= 55.53 x 10³ V/m

Explanation:

Given that

a=  3.63 cm

Area ,A= a²

distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Value of capacitor given as

C=\dfrac{\varepsilon _oA}{d}

By putting the values

C=\dfrac{8.85\times 10^{-12}\times 3.63^2\times 10^{-4}}{0.473\times 10^{-3}}

C=2.46 x 10⁻¹¹ F

U=\dfrac{1}{2}CV^2

V=Voltage difference

V=\sqrt{\dfrac{2U}{C}}

V=\sqrt{\dfrac{2\times 8.49\times 10^{-9}}{2.46\times 10^{-11}}}

V=26.27 V

V= E d

E=Electric filed

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

7 0
3 years ago
Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r
Alborosie

Answer:

6 m/s is the missing final velocity

Explanation:

From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling P_{xi} the initial momentum of object X and P_{yi} the initial momentum of object Y, we can derive the total initial momentum of the system: P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

Final momentum of the system: M * v_f=400kg * v_f

We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}

7 0
3 years ago
Read 2 more answers
Question 1
ella [17]
I think the answer maybe C
5 0
3 years ago
Read 2 more answers
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