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3 years ago

Find the current through a 12 ohm resistive circuit when 24 volts is applied

Physics

1 answer:

marusya05 [52]3 years ago

5 0

Answer:

Explanation:

Voltage = Current * Resistance

V= I*R

I=V/R

I = 24/12 = 2A.

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A car hits another and the two bumpers lock together during the collision. is this an elastic or inelastic collision?

valkas [14]

Inelastic.
If it was elastic, they'd bump right off each other. But since they've been locked, or stuck together, this is inelastic.

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3 years ago

A student is transmitting sound waves through various materials. Through which metal in the table will the sound waves travel th

MariettaO [177]

Answer:

Aluminum

Explanation:

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2 years ago

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For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mas

inn [45]

Answer:

Yes is large enough

Explanation:

We need to apply the second Newton's Law to find the solution.

We know that,

$F= ma$

And we know as well that

$a= \frac{v}{t}$

Replacing the aceleration value in the equation force we have,

$F= \frac{mv}{t}$

Substituting our values we have,

$F= \frac{(0.060)(55)}{4*10^{-3}}$

$F=825N$

The weight of the person is then,

$W = mg$

$W = (60)(9.8) = 558N$

<em>We can conclude that force on the ball is large to lift the ball</em>

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2 years ago

Which will increase the kinetic energy of an object more, increasing its mass or increasing its speed? Why?

agasfer [191]

Increasing its velocity will add to the kinetic energy more as the formula for kinetic energy is 0.5*m*v^2. (The speed will be squared making it greater)

4 0

3 years ago

A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur

morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

$\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\$

$\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\$

Substitute the given parameters and solve for the angular speed;

$\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s$

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0

2 years ago