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2 years ago

I need help getting started with number 8.

Physics

1 answer:

ratelena [41]2 years ago

4 0

In #8, the distance and the (magnitude of displacement) are equal, because he crawled in a straight line.

Displacement = (straight-line distance from start-point to end-point) in the direction from start to end, regardless of what route was actually followed.

Displacement = 5m, in the negative direction.

In #9 . . . distance will be the same. Displacement is going to be the same magnitude, but in the positive direction.

This is so simple that it's hard to talk about.

In #8, "What was the bug's distance ?". "Distance was 5 meters.". "What was the bug's displacement ?", "Displacement was 5 meters backwards."

In #9, What was the bug's distance ?". "Distance was 5 meters.". "What was the bug's displacement ?", "Displacement was 5 meters forward."

You might be interested in

How many grams of CO-60 result in 1 Millicuire of activity? How many years until the activity decays to 1 microcuire tl/2 =5.3 Y

vredina [299]

Answer:

m =8.81*10^{-6}grams

time t = 52.8 year

Explanation:

GIVEN DATA:

the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663 e+8 s

activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s

activity , $dN/dt = N* \lambda$

$\frac{dN}{dt} = N* \frac{ln2}{T_1/2}$

$N = \frac{(dN/dt )(T1/2)}{ln2}$

= ( 3.7 X 10^7 )(1.663*10^8 ) / ln2

= 8.877*10^{16}

Number of moles:

n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol

mass of the CO-60 is,

m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams

-----------------------------------------------------------------------------------------

time t = -[T1/2 / ln2]*ln[N/N0]

= - [5.3 years / ln2]*ln[1x10-6/1x10-3]

= 52.8 year

8 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

2 years ago

(b) Figure 4 shows a car travelling on a motorway.

Alik [6]

Answer:

To calculate anything - speed, acceleration, all that - we need data. The more data we have, and the more accurate that data is, the more accurate our calculations will be. To collect that data, we need to measure it somehow. To measure anything, we need tools and a method. Speed is a measure of distance over time, so we'll need tools for measuring time and distance, and a method for measuring each.

Conveniently, the lamp posts in this problem are equally spaced, and we can treat that spacing as our measuring stick. To measure speed, we'll need to bring time in somehow too, and that's where the stopwatch comes in. A good method might go like this:

Press start on the stopwatch right as you pass a lamp post
Each time you pass another lamp post, press the lap button on the stopwatch
Press stop after however many lamp posts you'd like, making sure to hit stop right as you pass the last lamp post
Record your data
Calculate the time intervals for passing each lamp post using the lap data
Calculate the average of all those invervals and divide by 40 m - this will give you an approximate average speed

Of course, you'll never find an *exact* amount, but the more data points you have, the better your approximation will become.

5 0

2 years ago

A 215-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

Misha Larkins [42]

Answer:

303.9481875 N

Explanation:

t = Time taken = 2 seconds

F = Force

r = Radius = 1.5 m

I = Moment of Inertia

$\alpha$ = Angular Acceleration

Torque

$\tau=F\times r$

$\tau=I\times \alpha$

$\\\Rightarrow F\times r=I\times \alpha\\\Rightarrow F=\frac{I\times \alpha}{r}$

Angular velocity

$\omega=rev/s\times 2\pi\\\Rightarrow \omega=0.6\times 2\pi\\\Rightarrow \omega=3.76991\ rad/s$

Angular acceleration

$\alpha=\frac{\omega}{t}\\\Rightarrow \alpha=\frac{3.76991}{2}\\\Rightarrow \alpha=1.88495\ rad/s^2$

$I=\frac{1}{2}mr^2\\\Rightarrow I=\frac{1}{2}215\times 1.5^2\\\Rightarrow I=241.875\ kgm^2$

$F=\frac{I\times \alpha}{r}\\\Rightarrow F=\frac{241.875\times 1.88495}{1.5}\\\Rightarrow F=303.9481875\ N$

The magnitude of the force to stop the merry-go-round is 303.9481875 N

3 0

3 years ago

A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x ax

alexdok [17]

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

applying conservation law of momentum

p + .444 i + .748 j = 0

p = - .444 i - .748 j

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

.024 v = .87

v = 36.25 m / s .

6 0

2 years ago