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Licemer1 [7]
3 years ago
12

An object of mass 2.5 kg has a momentum < 3, 6, 7 > kg m/s. At this instant the object is acted on a by a force < 50, 5

0, 100 > N for 5 x 10-3 s . What is the momentum of the object at the end of this time interval?
Physics
1 answer:
rosijanka [135]3 years ago
5 0

Answer:

The momentum of the object at the end is (3.25i+6.25j+7.5k)\ kg-m/s

Explanation:

Given that,

Mass of object = 2.5 kg

Momentum p= 3i+6j+7k

Force F=50i+50j+100k

Time t=5\times10^{-3}\ s

We need to calculate the momentum of the object at the end

Using formula of impulse

J=\Delta p

J=m\Delta v...(I)

J=F\times\Delta t....(II)

From equation (I) and (II)

F\times\Delta t=m\Delta v

F\times \Delta t=m(v_{f}-v_{i})

mv_{f}=F\times \Delta t+mv_{i}

Put the value into the formula

p_{f}=(50i+50j+100k)\times5\times10^{-3}+3i+6j+7k

p_{f}=0.25i+0.25j+0.5k+3i+6j+7k

p_{f}=(3.25i+6.25j+7.5k)\ kg m/s

Hence, The momentum of the object at the end is (3.25i+6.25j+7.5k)\ kg-m/s

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Answer:

Edwin Hubble

Explanation:

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3 years ago
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
what do you think is the purpose of a periscope? Are there any other uses for periscopes besides for submarines?
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3 years ago
1. Where is the water in the pot going?
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Answer:

   

Explanation:

   

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A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t
Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

Charge of second Particle q_2=-8\ \mu C

distance between them d=4\ cm

k=9\times 10^{9}

magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}

E_1=27\times 10^7\ N/C (away from it)

Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

E_2=-18\times 10^7\ N/C(towards it)

E_{net}=E_1+E_2

E_{net}=9\times 10^7\ N/C(away from first charge)        

8 0
3 years ago
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