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2 years ago

The second hand on a clock is 3.00 cm long. What is the speed of the outermost tip of that second hand

1 answer:

4 0

In 60 minutes or 3600 seconds, the tip of the minute hand traverses the circumference of a circle with radius 3.00 cm, so it moves with a tangential speed of

(3.00 cm)/(3600 s) ≈ 0.00083 cm/s = 8.3 μm/s

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What average force is needed to accelerate a 7.00-gram pellet from rest to 155 m/s over a distance of 0.600 m along the barrel o

leonid [27]

Answer: The force needed is 140.22 Newtons.

Explanation:

The key assumption in this problem is that the acceleration is constant along the path of the barrel bringing the pellet from velocity 0 to 155 m/s. This means the velocity is linearly increasing in time.

The force exerted on the pellet is

F = m a

In order to calculate the acceleration, given the displacement d,

$d = \frac{1}{2}at^2\implies a=\frac{2d}{t^2}$

we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:

$\overline{v}=\frac{1}{2}(v_{end}-v_{start})=\frac{1}{2}(155-0)\frac{m}{s}=77.5\frac{m}{s}$

This value can be used to determine the time for the pellet through the barrel:

$t = \frac{d}{\overline{v}}=\frac{0.6m}{77.5\frac{m}{s}}\approx0.00774s$

Finally, we can use the above to calculate the force:

$F = ma = m\frac{2d}{t^2} = 0.007kg\cdot \frac{2\cdot 0.6 m}{0.00774^2 s^2}\approx 140.22N$

8 0

2 years ago

A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of

algol [13]

Answer:

12.14 cm

Explanation:

mass, m = 15.5 kg

frequency, f = 9.73 Hz

maximum amplitude, A = 14.6 cm

t = 1.25 s

The equation of the simple harmonic motion

y = A Sin ωt

y = A Sin (2 x π x f x t)

put, t = 1.25 s, A = 14.6 cm, f = 9.73 Hz

y = 14.6 Sin ( 2 x 3.14 x 9.73 x 1.25)

y = 14.6 Sin 76.38

y = 12.14 cm

Thus, the displacement of the particle from the equilibrium position is 12.14 cm.

6 0

3 years ago

Which statements describe the Mercalli scale? Check all that apply.

konstantin123 [22]

<h3><u>Full question:</u></h3>

Which statements describe the Mercalli scale? Check all that apply.

A. This scale measures seismic waves based on their size.

B. This scale rates an earthquake according to how much damage it causes.

C.This scale produces a single rating for earthquakes that reach the surface.

D. This scale uses Roman numerals to rank the damage caused by an earthquake.

E.This scale measures the magnitude of an earthquake based on the size of seismic waves.

<h3><u>Answer:</u></h3>

The Mercalli scale : This scale rates an earthquake according to how much damage it causes and This scale uses Roman numerals to rank the damage caused by an earthquake.

<h3><u>Explanation:</u></h3>

The Modified Mercalli scale is intended to illustrate the consequences of an earthquake, at a contracted station, on tangible characteristics, on modern fittings and human beings.

The Modified Mercalli Intensity value ascribed to a particular site subsequent an earthquake has an extra significant means of severity to the nonscientist than the magnitude because intensity assigns to the outcomes really encountered at that position. This scale is comprised of 12 growing levels of intensity, denoted by Roman numerals, arranging from gradual shaking to catastrophic impairment.

3 0

2 years ago

Someone please answer this, ill give you brainliest and your getting 100 points.

ss7ja [257]

Answer:

b) Asthenosphere

Explanation:

The Earths crust is broken into plates. These plates float on the Asthenosphere. This is observed in the diagram.

3 0

1 year ago

Read 2 more answers

If, as is typical, each of them breathes about 500 cm3 of air with each breath, approximately what volume of air (in cubic meter

deff fn [24]

Answer:

Explanation:

The complete question is

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm^3 of air with each breath, approximately what volume of air (in cubic meters) do these astronauts breathe in a year? (b) What would the diameter (in meters) of the space station have to be to contain all this air?

The average breathing rate is 12 breaths per minute

there are 60 minutes x 24 hours x 365 days in a year = 525600 minutes in a year

if an average human takes 12 breath per minute, then in a year an average human take 12 x 525600 = 6307200 breath

For the four astronauts, the amount of breath = 4 x 6307200 = 25228800 breath in total.

The volume of air per breath = 500 cm^3

1 cm^3 = 10^-6 m^3

500 cm^3 = $x$ m^3

$x$ = 500 x 10^-6 = 5 x 10^-4 m^3

Therefore in a year, the volume of these astronauts breath in a year = 5 x 10^-4 x 25228800 = <em>12614.4 m^3</em>

b) If the space station is spherical, the volume will be given as = $\frac{4}{3} \pi r^3$

where r is the radius of the space station

This volume of the space station must be able to contain all the volume of breath produced by the astronauts which is = 12614.4 m^3

Equating, we have

12614.4 = $\frac{4}{3} \pi r^3$

12614.4 = $\frac{4}{3}*3.142*r^3$

12614.4 = 4.1893 $r^{3}$

$r^{3}$ = 12614.4/4.1893 = 3011.1

r = $\sqrt[3]{3011.1}$ =<em> 14.4 m</em>

<em>diameter of the space station = 14.4 m x 2 = 28.8 m</em>

3 0

2 years ago