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3 years ago

A moving object must have which type of energy

Physics

1 answer:

xxMikexx [17]3 years ago

7 0

Answer:

Kinetic Energy

Explanation:

Kinetic energy is the energy an object has because of its motion. If we want to accelerate an object, then we must apply a force. ... Kinetic energy can be transferred between objects and transformed into other kinds of energy. For example, a flying squirrel might collide with a stationary chipmunk.

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Co (5. 00 g) and co2 (5. 00 g) were placed in a 750. 0 ml container at 50. 0 °c. The partial pressure of co in the container was

konstantin123 [22]

Use PV = nRT and solve for P.
n = grams/molar mass = 5.00/molar mass CO2
Ignore the SO2.

But I think the answer it 6.74 atm

But I’m not saying but is but I’m just thinking this is the answer.

Goodluck!

6 0

2 years ago

The current through a certain heater wire is found to be fairly independent of its temperature. If the current through the heate

irina [24]

Answer:

Explanation:

energy = power x time, and power = resistance x current ^2. 2^2 = 4.

5 0

3 years ago

What are the 2 forces acting on a projectile

nasty-shy [4]

Answer:

gravity and normal force I think

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3 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

3 years ago

Why does an ice cube float on water?

Dmitry [639]

Answer:

Ice cubes float on water because they are less dense.

Explanation:

Ice is about 9% less dense than water. Water is heavier, so ice is lighter, causing it to float. It is less dense because the density of water decreases along with a decrease in temperature. Therefore, causing ice to be less dense than water

6 0

2 years ago