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3 years ago

Quarks

Physics

1 answer:

Basile [38]3 years ago

4 0

C. T,B,U,D,C, and S. D is a tempting answer but it's actually the gluons that transmit the strong force.

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Transverse waves are sent along a 5.00-m-long string with a speed of 30.00 m/s. The string is under a tension of 10.00 N. What i

Kipish [7]

Answer:

0.055 kg

Explanation:

Given that

Length of the string, l = 5 m

Speed of the wave, v = 30 m/s

Tension on the string, F(t) = 10N

From the formula written in the attachment, we have

v = velocity of the wave, in m/s

F(t) = Tension on the string, in N

U = Mass per length of the string, in kg/m

m = Mass of the string, in kg

l = Length of the string, in m

See attachment for the calculation

3 0

3 years ago

The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian

Llana [10]

Complete question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v = c* \sqrt{1 - \frac{1}{n^2} }$

Explanation:

From the question we are told that

a = nb

The length of the minor axis of the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the Empire's symbol, (an ellipse)

Now this length seen by the observer can be mathematically represented as

$h = t \sqrt{1 - \frac{v^2}{c^2} }$

Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)

So t = a = nb

and b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the Empire's symbol, (an ellipse)

i.e h = b

$b = nb [\sqrt{1 - \frac{v^2}{c^2} } ]$

$[\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}$

$v^2 =c^2 [1- \frac{1}{n^2} ]$

$v^2 =c^2 [\frac{n^2 -1}{n^2} ]$

$v = c* \sqrt{1 - \frac{1}{n^2} }$

6 0

3 years ago

A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl

Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

4 0

3 years ago

A ball is dropped from a window and takes two seconds to reach the ground .it starts from rest and reaches a final speed of 20m/

Akimi4 [234]

The ball can't reach the speed of 20 m/s in two seconds, unless you THROW it down from the window with a little bit of initial speed. If you just drop it, then the highest speed it can have after two seconds is 19.6 m/s .

If an object starts from rest and its speed after 2 seconds is 20 m/s, then its acceleration is 20/2 = 10 m/s^2 .

(Gravity on Earth is only 9.8 m/s^2.)

3 0

3 years ago

If a builder of mass 75kg climbs a vertical ladder of 25m how much energy has she gained ?

erica [24]

So,

GPE (graviational potential energy) = mass x g x height

GPE is depends on where zero height is defined. In this situation, we define h = 0 as the initial height.

$GPE = 75 \ kg*9.8 \ \frac{m}{s^2}*25 \ m$

$GPE = 18,375 \frac{kg*m^2}{s^2}$

$GPE = 18,375 \ joules(J) \ or \ 18.375 \ kilojoules(kJ)$

The builder has gained 18.375 kJ of PE.

4 0

3 years ago