44.2 in scientific notation

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates

Novosadov [1.4K]

Answer:

4334.4 J

Explanation:

Work done equals to kinetic energy change

KE=½mv²

Change in KE is given by

∆KE=½m(v²-u²)

Where m is mass of water-skier, KE is kinetic energy, ∆KE is the change in kinetic energy, v is final velocity and u is initial velocity.

Substituting 72 kg for m, 12.1 m/s for v and 5.10 m/s for u then

∆KE=½*72(12.1²-5.10²)=4334.4J

Therefore, the work done by the net external force acting on the skier is equal to 4334.4 J

6 0

3 years ago

What is science and physics? yeha avo.

Dafna1 [17]

Answer:

Science is the pursuit and application of knowledge and understanding of the natural and social world following a systematic methodology based on evidence. Scientific methodology includes the following: Objective observation: Measurement and data (possibly although not necessarily using mathematics as a tool)

Explanation:

HOPE IT HELPS 

HAVE A NICE DAY :) 

IT'S SRESHTHA77 

GORGEOUS DIDI 

8 0

3 years ago

The potential energy between two atoms in a particular molecule has the form U(x) = 2.1 x 8 − 5.2 x4 where the units of x are le

a_sh-v [17]

Answer:

$x\approx 0.948$

Explanation:

The correct formula for the potential energy between two atoms in a particular molecule is:

$U(x) = \frac{2.1}{x^{8}}-\frac{5.2}{x^{4}}$

Where $x$ is the distance.

According to the definitions of potential energy and work, as well as the Work-Energy Theorem and the Principle of Energy Conservation. The relation between that and related force is:

$F = -\frac{dU}{dx}$

The function is derived in terms of distance:

$F (x) = \frac{84}{5\cdot x^{9}} -\frac{104}{5\cdot x^{5}}$

Then, it is needed to find at least of x so that F(x) equals to 0.

$\frac{84}{5\cdot x^{9}}-\frac{104}{5\cdot x^{5}}=0$

$\frac{84}{x^{4}}-104 = 0$

$84-104\cdot x^{4} = 0$

$x=\sqrt[4]{\frac{84}{104} }$

$x\approx 0.948$

7 0

3 years ago

To construct an oscillating LC system, you can choose from a 11 mH inductor, a 6.0 μF capacitor, and a 4.2 μF capacitor. What ar

Free_Kalibri [48]

Answer:

a. 475.14 Hz

b. 1959 Hz

c. 2341.53 Hz , 3053.34 Hz

Explanation:

$f = \frac{1}{2\pi*\sqrt{C*L}}$

a. smallest use the capacitive 4.2 uF + 6.0 uF = 10.2uF replacing:

$f = \frac{1}{2\pi*\sqrt{C*L}}f=\frac{1}{2\pi*\sqrt{10.2uF*11mH}}$

$f = 475.14 Hz$

b. second smallest use the capacitive 6 uF so:

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{6uF*11mH}}$

$f = 1959Hz$

c. second largest and largest oscillation first combination so:

Use 4.2 uF

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{4.2uF*11mH}}$

$f = 2341.53 Hz$

And finally largest oscillation cap in serie so:

$C=\frac{c_1*c_2}{c_1+c_2}=\frac{4.2uF*6.0uf}{4.2uf+6.0uF}$

$C=2.47 uF$

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{2.47uF*11mH}}$

$f = 3053.34 Hz$

5 0

3 years ago

A test charge of +4 µC is placed halfway between a charge of +6 µC and another of +2 µC separated by 20 cm. (a) What is the magn

S_A_V [24]

Answer:

(a) Magnitude: 14.4 N

(b) Away from the +6 µC charge

Explanation:

As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:

$F_e = K\frac{qq_{test}}{r^2}$

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.

Let's say that a force that goes toward the +6 µC charge is positive, then:

$F_e_1 = K\frac{q_1q_{test}}{r^2}=-9*10^9 \frac{Nm^2}{C^2} \frac{6*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =-21.6 N$

$F_e_2 = K\frac{q_2q_{test}}{r^2}=9*10^9 \frac{Nm^2}{C^2} \frac{2*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =7.2 N$

The magnitude will be:

$F_e = -21.6 + 7.2 = -14.4 N$ , away from the +6 µC charge

3 0

4 years ago