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4 years ago

A ball rolls around a circular track with an angular velocity of 4π rad/s. what is the period of the motion

1 answer:

4 0

The period, T, is the time required for one revolution.

The angular velocity is 4π rad/s.
Because one revolution = 2π radians, the period is
T = (2π rad)/(4π rad/s) = 0.05 s.

Answer: 0.5 s

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The drawing shows a tire of radius R on a moving car

masha68 [24]

Answer:

H / R = 2/3

Explanation:

Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.

Initial. Lowest point

Em₀ = K = 1/2 m v²

Final. In the sought height

$Em_{f}$ = U = mg h

Energy is conserved

Em₀ = $Em_{f}$

½ m v² = m g h

v² = 2 gh

Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass

Initial. Lower

Em₀ = K = ½ I w²

Final. Heights sought

Emf = U = m g R

Em₀ = $Em_{f}$

½ I w² = m g R

The moment of inertial of a cylinder is

I = $I_{cm}$ + ½ m R²

I= ½ $I_{cm}$ + ½ m R²

Linear and rotational speed are related

v = w / R

w = v / R

We replace

½ $I_{cm}$ w² + ½ m R² w² = m g R

moment of inertia of the center of mass

$I_{cm}$ = ½ m R²

½ ½ m R² (v²/R²) + ½ m v² = m gR

m v² ( ¼ + ½ ) = m g R

v² = 4/3 g R

As they indicate that the linear velocity of the two points is equal, we equate the two equations

2 g H = 4/3 g R

H / R = 2/3

7 0

3 years ago

I NEED THE ANSWER TO PASS SO I CAN GET MY PHONE BACK HELPPPPPPPPPPPPPP

lianna [129]

Answer:

it would be c.

Explanation:

starts at zero and isn't a straight shot up

6 0

3 years ago

An aluminum sphere is 8.55 cm in diameter. What will be it's % change in volume if it is heated from 30 C to 155 C?

Ghella [55]

We know that

• The sphere diameter is 8.55 cm.

• The temperature change is from 30 C to 155 C.

First, we have to find the radius of the sphere. The radius is the half diameter.

$\begin{gathered} r=\frac{d}{2}=\frac{8.55\operatorname{cm}}{2} \\ r=4.275\operatorname{cm} \end{gathered}$

Now we have to find the volume of the sphere using the following formula.

$V=\frac{4}{3}\pi(r)^3$

Where r = 4.275 cm.

$\begin{gathered} V=\frac{4}{3}\pi(4.275\operatorname{cm})^3 \\ V\approx327.26(cm)^3 \end{gathered}$

Then, we use the following formula

$V_2-V_1=BV_0(T_2-T_1)$

Where the initial volume is 327.26 cubic cm, B is a constant about thermal expansion for aluminum, and we have to find the final volume to then calculate the percentage change.

$\begin{gathered} \Delta V=75\times10^{-6}\cdot327.26(155-30) \\ \Delta V=75\times10^{-6}\cdot327.26(125) \\ \Delta V\approx3.07(cm)^3 \end{gathered}$

This means that the volume change is 3.07 cubic centimeters.

At last, we have to divide the volume change by the initial volume, and then we have to multiply it by 100% to express it as a percentage.

$\frac{\Delta V}{V_1}\times100=\frac{3.07}{327.26}\times100\approx0.938$ <h2>Therefore, the percentage change is 0.938%.</h2>

5 0

1 year ago

An electron is accelerated from rest by a potential difference of 412 V. It then enters a uniform magnetic field of magnitude 18

Salsk061 [2.6K]

Explanation:

Given that,

Potential difference, V = 412 V

Magnitude of magnetic field, B = 188 mT

(a) The potential energy of electron is balanced by its kinetic energy as :

$eV=\dfrac{1}{2}mv^2$

v is speed of the electron

$v=\sqrt{\dfrac{2eV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 412}{9.1\times 10^{-31}}} \\\\v=1.2\times 10^7\ m/s$

(b) When the charged particle moves in magnetic field, it will move in circular path. The radius of the circular path is given by :

$r=\dfrac{mv}{eB}\\\\r=\dfrac{9.1\times 10^{-31}\times 1.2\times 10^7}{1.6\times 10^{-19}\times 188\times 10^{-3}}\\\\r=3.63\times 10^{-4}\ m$

Hence, this is the required solution.

7 0

3 years ago

The insulating solid sphere in the previous Example 24.5 has the same total charge Q and radius a as the thin shell in the examp

BaLLatris [955]

Answer and Explanation:

Using Gauss's law,

If r>a

then charge enclosed in all the three cases is same as Q.

So Electric field for all three is same.

So {1,1,1}.

(b) r<a,

Charge enclosed in case of shell is zero since all charge is present on the surface. So E = 0.

Charge enclosed by incase of point charge is Q.

Charge enclosed in case of sphere is Qr3/a3 which is less than Q.

So ranking {2,3,1}

6 0

4 years ago