A measure of how easily current will pass through a material ?

You are stowing items and come across an aerosol bottle of hairspray.what should you do?

Vlad [161]

Answer:

below

Explanation:

5 0

3 years ago

A volume of 25 milliliters is the same as a volume of ____________________ cubic centimeters.

PolarNik [594]

Answer:

25 cm³

Explanation:

In the conversion of units, we know that are one cubic centimeters (cm³) in a milliliter (mL) .

1 milliliter = 1 cubic centimeter

25 milliliters = 25 cubic centimeters

Therefore, a volume of 25 milliliters is the same as a volume of 25 cubic centimeters.

This ultimately implies that, the volume of an object in milliliters is equivalent to its volume in cubic centimeters.

6 0

3 years ago

A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s

Paraphin [41]

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$

3 0

3 years ago

Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. A

Crazy boy [7]

Answer:

W = 1884J

Explanation:

This question is incomplete. The original question was:

Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.

How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.

The amount of work done by the motor is given by:

$W=\Delta K$

$W= 1/2*I*\omega f^2-1/2*I*\omega o^2$

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

$\omega f^2=\omega o^2+2*\alpha*\theta$

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

$\tau=I*\alpha$

$\alpha=\tau/I$ => $\alpha = 0.5rad/s^2$

Now we can calculate the final velocity:

$\omega f = 8.68rad/s$

Finally, we calculate the total work:

$W= 1/2*I*\omega f^2 = 1883.56J$

Since the question asked to "Enter your answer in joules to four significant figures.":

W = 1884J

3 0

3 years ago

La ce distanta de la suprafata Pamintului intensitatea cimpului sau gravidational este egala cu 1 N/kg ?

Ulleksa [173]

Acceleration of gravity is proportional to 1/D² .
('D' is the distance from the Earth's center.)

We need (D / R)² = 9.8
('R' is the Earth's radius.)

D/R = √9.8 = 3.13

Distance from Earth's center = 3.13 R
Distance above Earth's surface = 2.13 R = about 13,570 km =====================================================

Accelerarea de gravitate este proporțională cu 1 / D².
('D este distanța de centrul Pământului.)

Avem nevoie de (D / R) ² = 9,8
('R' este raza Pământului.)

D / R = √9.8 = 3,13

Distanța de la centrul Pamantului = 3,13 R
Distanța de deasupra suprafetei Pamantului = 2.13 R = aproximativ 13,570 km

6 0

4 years ago