Ask question

Ask question

All categories

3 years ago

11

. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim

e

1 answer:

inn [45]3 years ago

6 0

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ $\frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}$

On multiplying both sides by "100", we get

⇒ $\frac{\Delta T}{T}\times 100 =\frac{500}{24\times 60\times 60}$

⇒ $\frac{\Delta T}{T}\times 100=0.005787$ (%)

∵ $T=2\pi\sqrt{\frac{l}{g} }$

On substituting the values, we get

⇒ $\frac{\Delta T}{T}$ % = $\frac{1}{2}\times \frac{\Delta l}{l}$ %

On applying cross multiplication, we get

⇒ $\frac{\Delta l}{l}$ % = $2\times \frac{\Delta T}{T}$ %

⇒ = $2\times 0.05787$

⇒ = $0.011574$

⇒ = $0.012$ %

You might be interested in

A potter’s wheel—a thick stone disk of radius0.500 mand mass 100 kg—is freely rotating at 50.0 rev/min.The potter can stop the w

Reika [66]

Answer:

- 0.873rad/ $s^{2}$

angular velocity = -0

Explanation:

Given the mass of the disk is 100kg

the radius of the disk is 0.500m

the normal force on the disk is 700N

the normal force that acts radial inward is 70.0N

the final rotation of the wheel is 0.500rev/s and the initial rotation of wheel is 0

Formula to calculate the mass moment of inertia is

$I = \frac{1}{2} MR^{2}$

where I is the mass of inertia

R is the radius of the disk

M is the mass of the disk

substitute 100kg for M and 0.500m for R in above equation to find I

I = 1/2(100kg)(0.500m)∧2

I = 12.5kg.m∧2

thus the mass moment of inertia is 12.5kg.m∧2

formula to calculate the angular acceleration is

∝ = ωf - ωi/ t

here ∝ = angular acceleration

ωf = final angular speed

ωi = initial angular speed

t = the time taken to stop the wheel by potter.

t = 2.00s

ωf = 0.500rev/s

ωi = 0

in the equation above to find ∝

∝ $= \frac{0-5.24rev/s(\frac{x}{180} )(\frac{360}{60s} )}{2.00s}$

∝ = -0.873rad/ $s^{2}$

thus the angular acceleration is -0

4 0

4 years ago

Can a car moving with a negative velocity moves faster than a car moving with a positive velocity? explain.

Nesterboy [21]

Answer:

Yes.

Explanation:

This is because "negative velocity" just means it is in the negative in relation to the point of 0. Negative velocity doesn't equal a decrease in velocity. For example lets say you were parked next to a cone (this cone represents zero) if you accelerate forwards then that would be positive acceleration. If you were to accelerate backwards, this would be in the negative direction, aka negative velocity.

SUMMARY:

A negative velocity means that the object which has the negative velocity is moving in the opposite direction of an object moving at a positive velocity. This is a question of frame of reference. The possibility for the velocity is what makes it different to the speed. Speed is only positive.

6 0

4 years ago

A 24-V battery is powering a light bulb with a resistance of 3.0 ohms. What is the current flowing through the bulb? A) 7.20 A B

Usimov [2.4K]

According to Ohm's law for a portion of the circuit we have:

U=RI=>I=U/R=24/3=8 A

The correct answer is B

3 0

3 years ago

The Electron Cloud Model shows shaded regions of probabilty where the electons most likely are at a given point in time. True Fa

melisa1 [442]

Answer:

True

Explanation:

The statement is true because According to Schrödinger it's not possible to get the exact position of where an electron is located. However, he showed from an equation that we can at determine the most likely area where an electron will be and he said they are most likely to be in the most dense area of the clouds.

7 0

3 years ago

Read 2 more answers

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago