Ask question

Ask question

All categories

3 years ago

10

An ideal pendulum is often discussed when studying simple harmonic motion. What are the conditions needed for an ideal pendulum?

Check all that apply.

2 answers:

tamaranim1 [39]3 years ago

8 0

Answer:

Gravitational force acts to pull the pendulum’s mass down.

The pivot on which the pendulum swings is frictionless.

The pendulum swings through its equilibrium position as a result of inertia.

Explanation:

Just took the test

vlada-n [284]3 years ago

6 0

Complete question is;

An ideal pendulum is often discussed when studying simple harmonic motion.

What are the conditions needed for an ideal pendulum? Check all that apply.

1) Gravitational force acts to pull the pendulum’s mass down.

2) The mass of the rod of the pendulum is evenly distributed along the rod.

3) The pivot on which the pendulum swings is frictionless.

4) The pendulum swings through its equilibrium position as a result of inertia.

5) Tension acts downward on the pendulum’s mass

Answer:

Conditions 2,3 & 4 are the ones that apply

Explanation:

The conditions required for an ideal pendulum are;

- The mass of the rod of the pendulum has to be evenly distributed along the rod.

- Ensure that the pivot on which the pendulum is swinging is frictionless

-The pendulum has to swing through it's equilibrium position as a result of inertia.

Thus,the correct options are 2, 3, 4.

You might be interested in

What is a major contributor of greenhouse gases in the atmosphere?

ASHA 777 [7]

Answer:

a. burning of fossil fuel

Explanation:

Greenhouse effect is the trapping of the sun infrared rays in the outermost layer of the earths atmosphere due to the accumulation of some harmful gasses. This gases depletes the ozone layer

The major contributor of greenhouse gases is the burning of fossil fuels. Carbon dioxides are released into the atmosphere and leads to global warming and climatic changes per time

6 0

3 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

4 years ago

A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes.

Semenov [28]

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

$\epsilon=NA\dfrac{dB}{dt}$

For N = 1

$\epsilon=A\dfrac{dB}{dt}$

We need to calculate the current

Using formula of current

$i=\dfrac{\epsilon}{R}$

Put the value of emf

$i=\dfrac{A\dfrac{dB}{dt}}{R}$

Now, if the number of turn is 22 , then induced emf would be

$\epsilon'=NA\dfrac{dB}{dt}$

Then the current would be

$i'=\dfrac{\epsilon'}{NR}$

$i'=\dfrac{NA\dfrac{dB}{dt}}{NR}$

$i'=\dfrac{A\dfrac{dB}{dt}}{R}$

$i'=i$

Hence, The current would be same in both situation.

4 0

3 years ago

A. Describe one meal that you typically order when eating out.

marta [7]

A. “I usually get 10 piece nugget meal with medium fries at McDonald’s with a McFlurry

B. I could replace the 10 piece nuggets with a grilled chicken wrap and replace the medium fries with a small fries then replace the mcflurry with either water or unsweetened tea.

3 0

3 years ago

Read 2 more answers

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago