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3 years ago

Part A:

2 answers:

7 0

In a perfectly ELASTIC collision between two perfectly rigid objects <span>both the momentum and the kinetic energy of the system are conserved.hope it helps</span>

skad [1K]3 years ago

6 0

Answer:

both the momentum and the kinetic energy of the system are conserved.hope it helps

Explanation:

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Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o

sasho [114]

Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

$U = \frac{kq_{1}q_{2} }{r}$

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

$U_{1} = \frac{ke^{2} }{r_{1} }$

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

$U_{2} = \frac{ke^{2} }{r_{2} }=\frac{ke^{2} }{4.10r_{1} }$

Applying law of conservation of energy :

ΔU = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁ - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

$\frac{1}{2}mv^{2}=\frac{ke^{2} }{r_{1} }- \frac{ke^{2} }{4.10r_{1} }$

Here m and v are the mass and final speed of electron respectively.

$v^{2}=\frac{2}{m} \frac{ke^{2} }{r_{1} }(1- \frac{1 }{4.10 })$

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

$v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2} }{4.32\times10^{-10} }(1- \frac{1 }{4.10 })$

$v^{2}=8.86\times10^{11}$

v = 9.41 x 10⁵ m/s

5 0

3 years ago

a van moves with a constant speed of 79 km/h how long will it take to travel a distance of 502 kilometers

Mazyrski [523]

Answer:

6.35hours

Explanation:

s=vt

t=s/v=502/79=6.35hours

5 0

3 years ago

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×10^{4} 4 m/s^{2} 2 , and 1.85 ms (1

Paha777 [63]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 = u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

6 0

3 years ago

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

#SPJ4

6 0

2 years ago

I need help with this please

kondaur [170]

Answer:

1. The respiratory system functions when our involuntary nervous system sends impulses to the muscles in the diaphragm; thereby, causing the lungs to expand and contract.

2. The respiratory system oxygenates the blood which is vital for bodily function as oxygenated blood is carried from your lungs to the left side of your heart, to be circulated throughout the body. Furthermore deoxygenated blood is carried back to the right side of your heart to get oxygenated once more.

3. The other body systems that are crucial for the lungs to function are the nervous system and the muscular system.

4. without the raspatory system the body wouldn't receive any oxygen and the brain would slowly die. therefore, without the brain the heart would stop functioning and atrophy etc.

Explanation:

6 0

3 years ago