Ask question

Ask question

All categories

3 years ago

11

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen

t: The centripetal acceleration of the ball can be increased by a factor of 4 by Entry field with incorrect answer keeping the radius fixed and increasing the period by a factor of 4. keeping the radius fixed and increasing the speed by a factor of 4. keeping the radius fixed and decreasing the period by a factor of 4. keeping the speed fixed and decreasing the radius by a factor of 4. keeping the speed fixed and increasing the radius by a factor of 4.

1 answer:

Eva8 [605]3 years ago

7 0

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

$a=\dfrac{v^2}{R}$

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

$a'=\dfrac{v^2}{R'}$

$a'=\dfrac{v^2}{R/4}$

$a'=4\times \dfrac{v^2}{R}$

$a'=4\times a$

So, the centripetal acceleration of the ball can be increased by a factor of 4.

You might be interested in

"a horn emits a frequency of 1000 HZ. A 14 m/s wind is blowing toward a listener. What is the frequency of the sound heard by th

Semenov [28]

Answer:

$f_L = 1000 Hz$

Explanation:

given,

speed of wind, = 14 m/s

frequency of horn,f_s = 1000 Hz

speed of sound,V = 344 m/s

frequency heard by the listener

using Doppler effect

$f_L = \dfrac{v+v_L}{v+v_s}f_s$

f_L is the frequency of the sound heard by the listener

f_s is the frequency of sound emitted by the listener

V is the speed of sound

v_L is the speed of listener

v_s is the speed of source

now,

considering the frame of reference in which wind is at rest now, both listener and the source will be moving at 14 m/s

$f_L = \dfrac{v+14}{v+14}\times 1000$

now on solving we will get

$f_L = 1000 Hz$

hence, the frequency heard by the listener is equal to 1000 Hz

7 0

3 years ago

2. Harry is pushing a car down a level road at 2.0 m/s with a force of 243 N. The total force

Arte-miy333 [17]

b) Equal to 243 N.

Explanation:

The total force acting on the car in the opposite direction including the road friction and air resistance is equal to 243 N.

This is in conformity with newton's third law of motion.

Newton's third law of motion states that "action and reaction are equal and opposite in direction. "

The action force is that of the pull by Harry acting on the car.
The reaction force is in the opposite direction.
Both action and reaction force equal and opposite and magnitude and direction

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0

3 years ago

Which sport requires the least amount of agility? HELLLLPPP

ZanzabumX [31]

Answer:

golf

Explanation:

it's less physical strength in more of you hitting the ball with whatever the stick is called

5 0

3 years ago

14. The inner transition metals are made up two series known as:

o-na [289]

Answer:

lanthanide and actinide

Explanation:

An inner transition metal (ITM) of chemical elements on the periodic table. They are normally shown in two rows below all the other elements. They include elements 57-71, or lanthanides, and 89-103, or actinides.

8 0

3 years ago

Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the o

julsineya [31]

Answer:

1) $k=-\frac{1}{46}\approx -0.02$

2) $\textit{Limiting value} = 24$

3) $T(10)\approx \frac{494611944}{6436343}\approx 76.8$

Explanation:

First of all note that

$T_s=24$ is the surroundings temperature, the temperature of the room where the cup of coffee is. Then, the differential equation is:

$\frac{dT}{dt}=k(T-24)$

Also, note that all units are in degrees celsius and minutes. Then, we don't have to convert units. Let's not write units explicitly from now on.

Explanation

1) We have that

$\textit{rate of cooling}=\frac{dT}{dt}=1,\quad T=70$

at some point - the exact time at which this is true doesn't really play any role because the equation doesn't have t on the right hand side. Then, from the equation we get

$1=-(70-24)=46k\Rightarrow k=-\frac{1}{46}\approx -0.02$

The minus comes from considering the temperature must decrease. With this value we can write the equation more explicitely:

$\frac{dT}{dt}=-\frac{1}{46}(T-24)$

2) The coffee is cooling off as time goes by, and it won't get any cooler than 24 degrees celsius because that's the temperature of the room. Then, in the long run, the temperature of the coffee is 24 degrees celsius.

3) Remember that Euler's method consists of using an initial exact measurement to predict what will happen in the future, approximately. There is a formula to make those predictions an it depends on the time step they gave us. Let's compute things first and then I tell you the equations we used.

In this case we know that we start with a 90 degrees celsius cup of coffee, or, in terms of math,

$T(0)=90$

Then, we can predict:

$T(2)\approx 90+2\left[-\frac{1}{46}(90-24)\right]=\frac{2004}{23}\approx 87.1$

Let's use fractions so we don't lose accuracy from now. With this number we can make an approximation of the temperature after 2 more seconds:

$T(4)\approx \frac{2004}{23}+2\left[-\frac{1}{46}\left(\frac{2004}{23}-24\right)\right]=\frac{44640}{529}\approx 84.4$

and then

$T(6)\approx \frac{44640}{529}+2\left[-\frac{1}{46}\left(\frac{44640}{529}-24\right)\right]=\frac{994776}{12167}\approx 81.8$

and then

$T(8)\approx \frac{994776}{12167}+2\left[-\frac{1}{46}\left(\frac{994776}{12167}-24\right)\right]=\frac{22177080}{279841}\approx 79.2$

and finally, the number we wanted to find:

$T(10)\approx \frac{22177080}{279841}+2\left[-\frac{1}{46}\left(\frac{22177080}{279841}-24\right)\right]=\frac{494611944}{6436343}\approx 76.8$

I hope you noticed the pattern to compute the next prediction:

$\textit{next prediction} = \textit{previous one (or exact value if it's the first step)}\\+ h\ast(\textit{right hand side of the differential equation at the previous one})$

5 0

3 years ago