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3 years ago

If an object moving at 12 m/s has a kinetic energy of 72 j what is the mass

1 answer:

5 0

Simply, apply the formula $E = \frac{1}{2}mv^{2}$ and insert the values of m = mass, v = velocity and E = Energy.
The result will be $72J = \frac{1}{2}m(12m/s)^{2}$ , m = 1 kg

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. A 2.0-m wire carries a current of 15 A directed along the positive x axis in a region where the magnetic field is uniform and

Lady bird [3.3K]

Answer:

The resulting magnetic force on the wire is -1.2kN

Explanation:

The magnetic force on a current carrying wire of length 'L' with current 'I' in a magnetic field B is

F = I (L*B)

Finding (L * B) , where L = (2, 0, 0)m , B = (30, -40, 0)

L x B = $\left[\begin{array}{ccc}i&j&k\\2&0&0\\30&-40&0\end{array}\right]$ = (0, 0, -80)

we can now solve

F = I (L x B) = I (-80)

F = -1200 kmN

F = -1200 kN * 10⁻³

F = -1.2kN

6 0

3 years ago

) Suppose a particle travels along a straight line with velocity v(t) = t 2 e −3t meters per second after t seconds. How far doe

pshichka [43]

Answer:

x(t=3s) = 0.07 m to the nearest hundredth

Explanation:

v(t) = t² e⁻³ᵗ

Find displacement after t = 3 s.

Recall, velocity, v = (dx/dt)

v = (dx/dt) = t² e⁻³ᵗ

dx = t² e⁻³ᵗ dt

∫ dx = ∫ t² e⁻³ᵗ dt

This integration will be done using the integration by parts method.

Integration by parts is done this way...

∫ u dv = uv - ∫ v du

Comparing ∫ t² e⁻³ᵗ dt to ∫ u dv

u = t²

∫ dv = ∫ e⁻³ᵗ dt

u = t²

(du/dt) = 2t

du = 2t dt

∫ dv = ∫ e⁻³ᵗ dt

v = (-e⁻³ᵗ/3)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ (-e⁻³ᵗ/3) 2t dt

= (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt

But the integral (∫ 2t (-e⁻³ᵗ/3) dt) is another integration by parts problem.

∫ u dv = uv - ∫ v du

u = 2t

∫ dv = ∫ (-e⁻³ᵗ/3) dt

u = 2t

(du/dt) = 2

du = 2 dt

∫ dv = ∫ (-e⁻³ᵗ/3) dt

v = (e⁻³ᵗ/9)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ 2t (-e⁻³ᵗ/3) dt = 2t (e⁻³ᵗ/9) - ∫ 2 (e⁻³ᵗ/9) dt = 2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)

Putting this back into the main integration by parts equation

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt = (-t²e⁻³ᵗ/3) - [2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)]

x(t) = ∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k (k = constant of integration)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k

At t = 0 s, v(0) = 0, hence, x(0) = 0

0 = 0 - 0 - (2/27) + k

k = (2/27)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + (2/27)

At t = 3 s

x(3) = (-9e⁻⁹/3) - (6e⁻⁹/9) - (2e⁻⁹/27) + (2/27)

x(3) = -0.0003702294 - 0.0000822732 - 0.0000091415 + 0.0740740741 = 0.07361243 m = 0.07 m to the nearest hundredth.

7 0

3 years ago

Read 2 more answers

A paragraph that can serve the specific purpose of creating a special effect within the essay is called?

Katarina [22]

Its b.functional paragraph because writers use this for interest presents and special effects

7 0

3 years ago

Read 2 more answers

What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable

VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: $1.92\cdot 10^{-18}J$

Explanation:

When a particle with charge $q$ travels across a potential difference $\Delta V$ , then its change in electric potential energy is

$\Delta U = q \Delta V$

In this problem, we know that:

The particle is an electron, so its charge is

$q=-1.60\cdot 10^{-19}C$

We also know that the positive terminal is at potential

$V_+=0V$

While the negative terminal is at potential

$V_-=-12 V$

Therefore, the potential difference (final minus initial) is

$\Delta V = -12-0 = -12 V$

So, the change in potential energy of the electron is

$\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J$

This means that the electron when it is at the negative terminal has $1.92\cdot 10^{-18}J$ of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

$1.92\cdot 10^{-18}J$

3 0

3 years ago

Two trumpet players are trying to tune their instruments. When they are in tune, they will both be playing the same note and no

marusya05 [52]

Answer:

The second trumpeter will be playing at frequency = 515 Hz

Explanation: Given that the note sounds lower and they can hear 20 beats in 4.0 s.

Beat frequency = 20/4 = 5 Hz

Beat frequency = F2 - F1

5 = 520 - F1

F1 = 520 - 5

F1 = 515 Hz

Since the note sound lower, the second trumpeter will be playing at 515 Hz frequency

8 0

3 years ago