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3 years ago

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A motionless mass, hung on the end of a spring and pulled 5 cm downward, is a height of 1 meter above the classroom floor

1 answer:

Travka [436]3 years ago

3 0

Answer:

A

Explanation:

I TOOK IT

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Two 2.0kg bodies A and B collide The velocities before the collision are U1=15i+30j and U2=-10j+5.0j After the collision V1=-5.0

Svetllana [295]

Use the law of conservation of momentum. Since the momentum is a linear measure, we can treat each of the dimension separately:

i-direction:

$m_1u_{1i}+m_2u_{2i}=m_1v_{1i}+m_2v_{2i}\\v_{2i} = \frac{m_1u_{1i}+m_2u_{2i}-m_1v_{1i}}{m_2}=\frac{(2\cdot 15-2\cdot10+2\cdot5)kg\frac{m}{s}}{2kg}=10\frac{m}{s}$

j-direction:

$m_1u_{1j}+m_2u_{2j}=m_1v_{1j}+m_2v_{2j}\\v_{2j} = \frac{m_1u_{1j}+m_2u_{2j}-m_1v_{1j}}{m_2}=\frac{(2\cdot 30+2\cdot5-2\cdot20)kg\frac{m}{s}}{2kg}=15\frac{m}{s}$

Answer: Final velocity is: (10i + 15j) m/s

Change in the kinetic energy:

$\Delta E_k = E_{ku}-E_{kv} = \frac{1}{2}m(u_1^2+u_2^2-v_1^2-v_2^2)=\\=\frac{1}{2}2kg(1125+125-425-325)\frac{m^2}{s^2}=500J$

Answer: The system lost 500J worth of kinetic energy in the collision

4 0

3 years ago

The electric motor of a model train accelerates the train from rest to 0.540 m/s in 27.0 ms. The total mass of the train is 610

Gnesinka [82]

Answer:

The value is $P =3.294 \ W$

Explanation:

From the question we are that

The velocity $v = 0.540 \ m/s$

The time taken is $t = 27.0 ms = 27.0 *10^{-3} \ s$

The total mass of the train is $m = 610 \ g = 0.610 \ kg$

Generally the average power delivered is mathematically represented as

$P =\frac{KE }{t}$

$P =\frac{ \frac{1}{2} * m * v^2 }{t}$

=> $P =\frac{ \frac{1}{2} * 0.610 * 0.540 ^2 }{ 27.0 *10^{-3}}$

=> $P =3.294 \ W$

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3 years ago

Read 2 more answers

What is the particles instantaneous speed at t=16 sec

Ira Lisetskai [31]

It's 3.6 meters per second less than my speed was
at 4:19 PM last Tuesday.

Does that tell you anything ?
Why not ?

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3 years ago

If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds

BabaBlast [244]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

According to above question ~

Current (I) = 4 Amperes

Time (t) = 3 seconds

Charge (q) = ?

Let's find the charge (q) by using formula ~

$I = \dfrac{q}{t}$

$4 = \dfrac{q}{3}$

$q = 4 \times 3$

$q = 12 \: \: coulombs$

Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds

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3 years ago

What instrument measures air temperature?

zavuch27 [327]

Its a thermometer . i hope this helps you

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3 years ago