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3 years ago

What is a Kelvin(K)?

Physics

2 answers:

Arturiano [62]3 years ago

6 0

<span>Kelvin is an SI base unit of thermodynamic temperature, equal in magnitude to the degree Celsius.</span>

sukhopar [10]3 years ago

5 0

It is a form of Measurement of temperature
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If the mass of each ball were the same, but the velocity of ball A were twice as much as ball B, what do you think would happen

Margaret [11]

Given momentum is conserved and is mv

if
mass is the same and va=2vb
then momentum is m(2vb + vb)

if collision is elastic (bounces off)
then m(2vb + vb) = m(vfa + vfb)
3vb = vfa + vfb
meaning final velocities would be different

if collision is inelastic (sticks together)
then m(2vb + vb) = (m+m)vf
meaning final velocity would be same

is this what you wanted?

5 0

3 years ago

Sully is riding a snowmobile on a flat, snow-covered surface with a constant velocity of 10 meters/second. The total mass of the

Gelneren [198K]

Answer: 168N

16 - 10 = 6
6 / 10 = .6
F = 280 x .6 = 168

7 0

3 years ago

Read 2 more answers

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

5 0

3 years ago

A migrating bird flew across a lake at an average speed of 18 meters per second. Was the distance that the bird flew across the

Natasha_Volkova [10]

Answer:

It cannot be said with 100% guarantee.

Explanation:

Since bird is travelling with 18 meters per second, that means that it covers a distance of 18 meters in 1 second.

for part 1 we convert 20 minutes into seconds, thus, 20 * 60 = 1200 seconds.

Distance covered in 1200 seconds will be:

S = v * t (S is distance , v is speed, t is time)

18 * 1200 = 21600 meters = 21.6 Kilometers

If the bird took less than 20 minutes than distance covered would be less than 21.6 Kilometer which could be greater than or lesser than 21 Kilometers. On average we can say that the distance bird travelled will be less than 21 Kilometer. But it is not definite.

For part B convert 0.4 hours into seconds to find the distance it travelled.

0.4 * 60 * 60 = 1440 seconds

Distance covered will be

18 * 1440 = 25920 meters = 25.9 Kilo meters.

Again, if the bird takes less than 0.4 hours to cover the lake, the lake will be less than 25.9 kilometers long. This means that it could be greater than or lesser than 21 kilometers. It cannot be definitive

7 0

4 years ago

A mosquito of mass 0.15 mg is found to be flying at a speed of 50 cm/s with an uncertainty of 0.5 mm/s. (a) How precisely may it

amid [387]

Answer:

a) Δx = t 0.05 + 0.5

, Δx = 0.5 cm, b) Do not present any problem

Explanation:

The kinematic equation for constant speed is

v = x / t

x = v t

a) the uncertainty can be calculated with

Δx = dx /dv Δv + dx /dt Δt

Δx = t Δv + v Δt

Speed is

v = (50.00 ± 0.05) cm / s

The most common uncertainty for the time of Δt = 0.01 s

We replace

Δx = t 0.05 + 50 0.01

Δx = t 0.05 + 0.5

We must know the time to have an explicit value, if we assume that the measure was t = 1s

Δx = 0.5 cm

Do not present any problem since its value is not very small, we must take as soon as the quantum effects and the velocity are not so high that we must take into account the relativistic effects

5 0

3 years ago