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3 years ago

What is conduction? Please help. First answer will mark Brainliest

Physics

2 answers:

Kruka [31]3 years ago

7 0

Answer:

Conduction is the main method of thermal energy transfer in solids.

Hope this helps.....

ICE Princess25 [194]3 years ago

5 0

Answer:

conduction is the transfer of thermal energy through direct contact

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A shank has a moment of inertia about the knee joint of 0.18 kg∙m^2. Assume that the quadriceps muscles (vasti and the rectus fe

timofeeve [1]

Answer:

Torque will be equal to 4.176 N-m

Explanation:

It is given moment of inertia about knee joint $I=0.18kgm^2$

Angular acceleration produced $\alpha =23.2rad/sec^2$

We e have to find the torque

Torque is equal to $\tau =I\alpha$ , here I is moment of inertia and $\alpha$ is angular acceleration.

Torque will be equal to $\tau =0.18\times 23.2=4.176Nm$

So torque is equal to 4.176 N-m

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3 years ago

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mario62 [17]

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3 years ago

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VladimirAG [237]

Answer:

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3 years ago

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sergiy2304 [10]

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3 years ago

With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas

xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

$\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity$

As it just touches the 15 mm high roof, the final velocity will be zero. So,

$v_f=0\ m/s$ .

Now, the change in kinetic energy is equal to:

$\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2$

Change in gravitational potential energy = Final PE - Initial PE

So,

$\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J$ [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

$0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s$

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

4 0

3 years ago