What is a good example of netwon 3rd law

If you had an unlimited amount of mass to hang, what would be the range of possible accelerations for the system?

LenaWriter [7]

Answer:

The entire cart/hanging mass system follows the same law, ΣF = ma. This means that plotting force vs. acceleration yields a linear relationship (of the form y = mx).

3 0

2 years ago

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50

PSYCHO15rus [73]

Answer:

(a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Explanation:

Given that,

Length = 50.00 cm

Time period = 2.50 s

Time period of pendulum is defined as the time for one complete cycle.

The period depends on the length of the pendulum.

Using formula of time period

$T=2\pi\sqrt{\dfrac{I}{mgh}}$

Where, I = moment of inertia

We need to calculate the value of d

Using parallel theorem of moment of inertia

$I=I_{cm}+md^2$

For a meter stick mass m , the rotational inertia about it's center of mass

$I_{cm}-\dfrac{mL^2}{12}$

Where, L = 1 m

Put the value into the formula of time period

$T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}$

$T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}$

$T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})$

Multiplying both sides by d

tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]

$(\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0$

Put the value of T, L and g into the formula

$4.028d^2-6.25d+0.336=0$

$d = 0.056\ m, 1.496\ m$

The value of d is 0.056 cm and 1.496 cm.

(b). Given that,

L = 50-5 = 45 cm

We need to calculate the time period

Using formula of period

$T=2\pi\sqrt{\dfrac{l}{g}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}$

$T=1.35\ sec$

Hence, (a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

7 0

3 years ago

If you double the voltage in a circuit and reduce the resistance by a factor of four, what will happen to the current?

Kamila [148]

<span>doubling or tripling the voltage will cause the current to be doubled or tripled. On the other hand, any alteration in the resistance will result in the opposite or inverse alteration of the current. So doubling or tripling the resistance will cause the current to be one-half or one-third the original value.</span>

5 0

3 years ago

The geese fly 23 m/s to the south when they migrate for the winter. Identify if this example is speed or velocity.

kari74 [83]

Answer:

Its velocity.

7 0

2 years ago

Two small charged spheres are located on the y-axis. One is at y = 1.00 m, the other is at y = −1.00 m, and they both have a cha

yuradex [85]

Answer:

(a) 23.946 kV

(b) -0.077 J

Explanation:

(a) The electric potential is given by the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$ (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = q2 = 1.60*10^{-6}C

r1 and r2 are the distance from the charges to the point in which electric potential is evaluated.

Firs, you calculate the distance r1 and r2 by taking into account the position of the charges

$r_1=\sqrt{(1.00)^2+(0.670)^2}m=1.20m\\\\r_2=\sqrt{(-1.00)^2+(0.670)^2}m=1.20m$

Next, you replace the values of the parameters to calculate V:

$V=(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}+(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}\\\\V=23946.66\ V=23.946\ kV$

(b) The potential electric energy is given by:

$U_T=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}\\\\r_{1,2}=2.00m\\\\r_{1,3}=1.20m\\\\r_{2,3}=1.20m\\\\U_T=(8.98*10^9)[\frac{(1.6*10^{-6})^2}{2.00m}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}]J\\\\U_T=-0.077J$

5 0

3 years ago

What is a good example of netwon 3rd law​

What is a good example of netwon 3rd law