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2 years ago

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A. curious kitten pushes a ball of yarn at rest with its nose, displacing the ball of yarn 0.175 m in 2.00s. What is the acceler

ation of the ball of yarn? (Show your work.)

1 answer:

Step2247 [10]2 years ago

4 0

Answer:

Explanation:

Assuming that the acceleration is constant

0.175 = ½a2.00²

0.175/2 = a

a = 0.0875 m/s²

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To eight significant figures, Avogadro's constant is 6.0221367×10^(23)mol−1. Which of the following choices demonstrates correct

Zepler [3.9K]

Answer with Explanation:

We are given Avogadro's constant = $6.0221367\times 10^{23}mol^{-1}$

There are eight significant figures.

We have to round off.

1.If we round off to four significant figures

The ten thousandth place of Avogadro's constant is less than five therefore, digits on left side of ten thousandth place remains same and digits on right side of ten thousandth place and ten thousandth place replace by zero.

Then ,Avogadro's constant can be written as

$6.022\times 10^{23}mol^{-1}$

If we round off to 2 significant figures

Hundredth place of given number is less than 5 therefore, digits on left side of hundredth place remains same and digits on right side of hundredth place and hundredth place replace by zero.

Then,Avogadro's constant can be written as

$6.0\times 10^{23}mol^{-1}$

If we round off six significant figures

6 is greater than 5 therefore, 1 will be added to 3 and digits on right side of 6 and 6 replace by zero and digits on left side of 6 remains same except 3.

Then, the Avogadro's constant can be written as

$6.02214\times 10^{23}mol^{-1}$

3 0

2 years ago

A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.

Volgvan

Answer:

A=0.199

Explanation:

We are given that

Mass of spring=m=450 g= $=\frac{450}{1000}=0.45 kg$

Where 1 kg=1000 g

Frequency of oscillation= $\nu=1.2Hz$

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator= $E=\frac{1}{2}m\omega^2A^2$

Where $\omega=2\pi\nu$ =Angular frequency

A=Amplitude

$\pi=\frac{22}{7}$

Using the formula

$0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2$

$A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398$

$A=\sqrt{0.0398}=0.199$

Hence, the amplitude of oscillation=A=0.199

4 0

2 years ago

What is Yusef most likely finding?

Ratling [72]

Explanation:

yusef adds all of the values in his data set and then divide by the number of values in the set. the actual density of iron is 7.874 g/ml .

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1 year ago

A balloon of mass M is floating motionless in the air. A person of mass less than M is on a rope ladder hanging from the balloon

Alex73 [517]

Answer:

Down with a speed less than v

Explanation:

Let the person's mass be represented by m

The mass of the balloon = M

Total momentum is conserved thus F = external force = 0

when the person starts to climb the ladder, external force is still equal to zero.

We know that change is p = F * change in t

in this equation, F is the external force = 0

Hence change in p = 0

This means that total momentum is conserved thus F = external force = 0

(owing to the fact that exterior forces on the system are balanced)

Since all external force was zero before they began to climb the ladder, as they climb all external forces will still = 0

Thus, As the person starts moving

mv + MV = 0

or mv = MV

V = mv/M

Since M > m, m/M will definitely be a number that is less than 1,

Hence, V = (a number less than 1)v.

This means that V < v, or the balloon moves down a speed V which is less than v.

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2 years ago

The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank

Paladinen [302]

When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat

So here weight of the boat = Buoyancy force

let say boat is sunk by distance "h"

now we can say

$F_b = \rho * V * g$

$F_b = 1000*0.10 * 0.08 * h * 9.8$

now by above force balance equation we can write

$m*g = F_b$

$0.040 * 9.8 = 1000 * 0.10 * 0.08 * h * 9.8$

$0.040 = 8h$

$h = 5 * 10^{-3} m$

so boat will sunk by total 5 mm distance

8 0

2 years ago