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3 years ago

Random kinetic energy possessed by objects in a material at finite temperature. An object that feels hot has a lot of this.

Physics

2 answers:

gregori [183]3 years ago

3 0

Internal energy or thermal energy.

KATRIN_1 [288]3 years ago

3 0

Answer:

Thermal Energy

Explanation:

This is an example of kinetic energy. It results in an object having a temperature that can be measured. Also, thermal energy can be transferred from one object to another in the form of heat.

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A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west

just olya [345]

Answer:

5 m/s

Explanation:

Given that,

A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west.

It is assumed to find the resultant velocity of the vehicle. Let east side is positive and west is negative. So,

$v=v_1+v_2\\\\=20+(-15)\\\\=5\ m/s$

Hence, the resultant velocity of the vehicle is equal to 5 m/s.

6 0

2 years ago

3- For given three vectors a, b and c, c = a x b, then the vector c is:

o-na [289]

Answer:

Explanation:

8 0

2 years ago

A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh

daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is $B=2.958\times10^{-5}T$

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

$\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \ \overrightarrow B_s\perp \overrightarrow B_w$

Angle of net magnetic field from axial direction is given by:

$tan\ \theta=\frac{B_w}{B_s}$ ,

Field due to solenoid:

$B_s=\mu_onI_s, \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m$

Field due to wire:

$B_w=\frac{\mu_oI_w}{2\pi r}$

Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

$B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T$