If a mixture looks smooth and the same throughout, it is probably _________

Jaipal measures a circuit at 1.2 A and 24 . Using Ohm’s law, what can he calculate for the circuit? current (C) current (I) volt

jenyasd209 [6]

Answer: voltage (V)

Explanation:

Hi, in the answer the letter A (1.2 A) stands for Ampere, and the symbol next to the number 24 must be Ω 8 (Ohms)

So, applying Ohm's law:

Voltage (V) =current (I) x resistance (R)

V = I x R

Replacing with the values given;

V = 1.2 A x 24Ω

In conclusion, we can calculate the value of Voltage for the circuit.

Feel free to ask for more if needed or if you did not understand something.

4 0

3 years ago

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You need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling a

Drupady [299]

U gotta add like all of them bro

3 0

3 years ago

Electric field lines moves away from positive to wards negative?

Archy [21]

The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.

6 0

2 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

4 years ago

When sedimentary rock is exposed to heat and pressure what does it change into

kherson [118]

When sedimentary rock is exposed to heat and pressure it changes into METAMORPHIC ROCK.
Heat and pressure have the capacity to change a rock into a completely new type of rock. An igneous rock or a sedimentary rock can be changed into a metamorphic rock as a result of heat and pressure which the rock is subjected to. Metamorphic rocks are usually formed from already existing rocks that are exposed to pressure and heat.

8 0

3 years ago

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If a mixture looks smooth and the same throughout, it is probably __________.?