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3 years ago

Which gives the correct relationship for kinetic energy?

Physics

2 answers:

Karolina [17]3 years ago

6 0

Answer: K = (m/2)*v^2

Explanation: The kinetic energy is the enrgy that an object has because of the movement of the object, and as the mass of the moving object increases, also should increase the energy (more mass requires more energy to ve moved)

So the equation of the kinetic energy is:

K = (m/2)*v^2

where m is mass and v is velocity squared, this also cna be written as:

half the product of the mass and the squared velocity

lisabon 2012 [21]3 years ago

5 0

Answer:

MV/2

Explanation:

Because that is what I've been taught

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A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,

egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0

3 years ago

What is the equation for the enthalpy of vaporization?

Natalka [10]

Answer: Use the formula q = m·ΔHv in which q = heat energy, m = mass, and ΔHv = heat of vaporization.

4 0

2 years ago

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

3 years ago

The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 4.6 1018 electrons/s are drifting through a co

Vesna [10]

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density $= 8.4\times10^{28}\ m^3$

Current $i = 4.6\times10^{18}\ electron/s$

Diameter of wire = 1.2 mm

Length = 31 cm

We need to calculate the drift velocity

Using formula of drift velocity

$v_{d}=\dfrac{I}{neA}$

$v_{d}=\dfrac{Ne}{tne\times\pi r^2}$

Put the value into the formula

$v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}$

$v_{d}=4.842\times10^{-5}\ m/s$

We need to calculate the time

Using formula for time

$v_{d}=\dfrac{l}{t}$

$t=\dfrac{l}{v_{d}}$

Where, l = length

$v_{d}$ = drift velocity

Put the value into the formula

$t=\dfrac{31\times10^{-2}}{4.842\times10^{-5}}$

$t=6402.31\ sec$

$t=106.7\ minute$

Hence, The time is 106.7 minute.

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3 years ago

RP COME AND GET IT!!!!!!!!

Ahat [919]

I need someone to help me with my finals!!!!

6 0

2 years ago

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