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Alenkasestr [34]
3 years ago
14

Christopher conducts an experiment in which he tests how much sugar dissolves at different temperatures of water. One step requi

res him to use water with a temperature of 32.5°C. What is this temperature in kelvins? Record your answer to the nearest tenth. K
Physics
2 answers:
sveticcg [70]3 years ago
5 0

305.5 i just took the quiz

disa [49]3 years ago
4 0

Explanation:

When we have to convert temperature from degree celsius to kelvin then we add 273 into the given degree celsius temperature.

Therefore, convert 32.5^{o}C into kelvin as follows.

         Temperature in kelvin = Temperature in degree celsius + 273

                                               = (32.5 + 273) K

                                               = 305.5 K

Thus, we can conclude that temperature in kelvins is 305.5 K.

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The drill used by most dentists today is powered by a small air-turbine that can operate at angular speeds of 350000 rpm. These
alexdok [17]

Answer:

θ  = 6.3 *10³ revolutions

Explanation:

Angular acceleration of the drill

We apply the equations of circular motion uniformly accelerated

ωf= ω₀ + α*t  Formula (1)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t :  time interval (s)

Data

ω₀ = 0

ωf = 350000 rpm = 350000 rev/min

1 rev = 2π rad

1 min= 60 s

ωf = 350000 rev/min =350000*(2π rad/60 s)

ωf = 36651.9 rad/s

t = 2.2 s

We replace data in the formula (2) :

ωf= ω₀ + α*t

36651.9 = 0 + α* (2.2)

α = 36651.9 / (2.2)

α = 17000 rad/s²

Revolutions made by the drill

We apply the equations of circular motion uniformly accelerated

ωf²= ω₀ ²+ 2α*θ Formula (2)

Where:  

θ : Angle that the body has rotated in a given time interval (rad)

We replace data in the formula (2):  

(ωf)²= ω₀²+ 2α*θ

(36651.9)²= (0)²+ 2( 17000 )*θ

θ = (36651.9)²/ (34000 )

θ  = 39510.64 rad = 39510.64 rad* (1 rev/2πrad)

θ  = 6288.31 revolutions

θ  = 6.3 *10³ revolutions

3 0
3 years ago
A 4-kilogram ball moving at 8 m/sec to the right collides with a 1-kilogram ball at rest. After the collision,
elena-14-01-66 [18.8K]
Conservation of momentum: total momentum before = total momentum after

Momentum = mass x velocity

So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32

Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32

Therefore β = 12.8 m/s
4 0
3 years ago
A plane designed for vertical takeoff has a mass of 8.0 × 10³ kg. Find the net work done by all forces on the plane as it accele
artcher [175]

Answer:

<em>the net work done after starting from rest is =  2.4 × 10⁵ J</em>

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J),  The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

<em>Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)</em>

<em>Substituting these values into equation 2</em>

<em>v² = 0² +2×1×30</em>

<em>v² = 60</em>

<em>v = √60</em>

<em>v = 7.75 m/s</em>

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

<em>Substituting these values into equation 1,</em>

<em>W = 1/2(8.0×10³)(7.75)²</em>

<em>W = (4.0×10³)(60)</em>

W = 240 × 10³ J

<em>W = 2.4 × 10⁵ J</em>

<em>Therefore the net work done after starting from rest is =  2.4 × 10⁵ J</em>

4 0
3 years ago
16 to 19 year old male and females are how much more likely to be involved in a crash?
Sedbober [7]
I say around 40% - 60%

https://www.dmv.ca.gov/portal/dmv/detail/teenweb/more_btn6/traffic/traffic
http://www.teendriversource.org/stats/support_teens/detail/57
http://www.rmiia.org/auto/teens/Teen_Driving_Statistics.asp

(I just corrected the question. Sorry if it is still incorrect.)
6 0
3 years ago
Does exists the friction force in space?If yes,tell an full example
VashaNatasha [74]
Yes, friction does exist in space. Friction has nothing to do with the earth's atmosphere. It exists everywhere in the universe. <span />
5 0
3 years ago
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