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geniusboy [140]

3 years ago

7

Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially

stationary and rebounds with 70% of its initial kinetic energy. What is the magnitude of the change in momentum of object A?

1 answer:

AnnZ [28]3 years ago

6 0

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

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A 5.0 kg wooden block Is 100 meters from the ground. What is the gravitational potential energy for the block?

iren [92.7K]

Answer:

<h2>4900 J</h2>

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

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7 0

3 years ago

How far from a -7.80 μC point charge must a 2.40 μC point charge be placed in order for the electric potential energy of the pai

Naddik [55]

Answer:

Distance between two point charges, r = 0.336 meters

Explanation:

Given that,

Charge 1, $q_1=-7.8\ \mu C=-7.8\times 10^{-6}\ C$

Charge 2, $q_2=2.4\ \mu C=2.4\times 10^{-6}\ C$

Electric potential energy, U = -0.5 J

The electric potential energy at a point r is given by :

$U=k\dfrac{q_1q_2}{r}$

$r=k\dfrac{q_1q_2}{U}$

$r=9\times 10^9\times \dfrac{-7.8\times 10^{-6}\times 2.4\times 10^{-6}}{-0.5}$

r = 0.336 meters

So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.

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3 years ago

An electron travels west to east with a kinetic energy of 10 keV. The Earth's magnetic field in Pittsburgh is 19,911.5 nT in the

ch4aika [34]

Explanation:

It is given that,

Kinetic energy of the electron, $E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J$

Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.

The magnetic field in north direction, $B_y=19911.5\ nT$

The magnetic field in west direction, $B_x=-3257.1\ nT$

The magnetic field in vertical direction, $B_z=48381.8 \ nT$

Magnetic field, $B=(-3257.1i+19911.5j+48381.8k)\ nT$

Firstly calculating the velocity of the electron using the kinetic energy formulas as :

$E_k=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E_k}{m}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}$

$v=5.92\times 10^7 i\ m/s$ (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

$F=q(v\times B)$

$F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})$

Since, $i\times j=k\ \\j\times k=i\\k\times i=j$

And, $i\times i=j\times j=k\times k=0$

$F=1.6\times 10^{-19}\times [1178 k-2864.20j]$

$|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}$

$F=4.95\times 10^{-16}\ N$

(b) Let a is the acceleration of the electron. It can be calculated as :

$a=\dfrac{F}{m}$

$a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}$

$a=5.43\times 10^{14}\ m/s^2$

Hence, this is the required solution.

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