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3 years ago

How do leptons differ from hadrons?

1 answer:

4 0

I'll just give you the link for it but count it as my answer. http://www.differencebetween.com/difference-between-leptons-and-vs-hadrons/

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The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo

Jobisdone [24]

(a) 4.0 m/s

We can solve this part just by analyzing the vertical motion of the froghopper.

The initial vertical velocity of the froghopper as it jumps from the ground is given by

$u_y = u_0 sin \theta$ (1)

where

$u_0$ is the takeoff speed

$\theta=58.0^{\circ}$ is the angle of takeoff

The maximum height reached by the froghopper is

h = 58.7 cm = 0.587 m

We know that at the point of maximum height, the vertical velocity is zero:

$v_y = 0$

Since the vertical motion is an accelerated motion with constant (de)celeration $g=-9.8 m/s^2$ , we can use the following SUVAT equation:

$v_y^2 - u_y^2 = 2gh$

Solving for $u_y$ ,

$u_y = \sqrt{v_y^2-2gh}=\sqrt{-2(-9.8)(0.587)}=3.4 m/s$

And using eq.(1), we can now find the initial takeoff speed:

$u_0 = \frac{u_y}{sin \theta}=\frac{3.4}{sin 58.0^{\circ}}=4.0 m/s$

(b) 1.47 m

For this part, we have to analyze the horizontal motion of the froghopper.

The horizontal velocity of the froghopper is

$u_x = u_0 cos \theta = (4.0) cos 58.0^{\circ} =2.1 m/s$

And this horizontal velocity is constant during the entire motion.

We now have to calculate the time the froghopper takes to reach the ground: this is equal to twice the time it takes to reach the maximum height.

The time needed to reach the maximum height can be found through the equation

$v_y = u_y + gt$

Solving for t,

$t=-\frac{u_y}{g}=-\frac{3.4}{9.8}=0.35 s$

So the time the froghopper takes to reach the ground is

$T=2t=2(0.35)=0.70 s$

And since the horizontal motion is a uniform motion, we can now find the horizontal distance covered:

$d=u_x T = (2.1)(0.70)=1.47 m$

7 0

3 years ago

Lf a dropped hammer's velocity increases from 0.Omis to 15.0 m/s in 4.04s

Slav-nsk [51]

Answer: The acceleration due to gravity on the surface of the Moon is approximately 1.625 m/s2, about 16.6% that on Earth's surface or 0.166 ɡ.

Heres your answer.

6 0

3 years ago

An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first

GarryVolchara [31]

Answer:

The right answer is "1.369 m/s²".

Explanation:

The given values are:

Distance (s)

= 260 m

Initial speed (u)

= 26 m/s

Reaction time (t')

= 0.51 s

During reaction time, the distance travelled by locomotive will be:

⇒ $s'=ut'$

$=26\times 0.51$

$=13.26 \ m$

Remained distance between locomotive and car:

⇒ $x=s-s'$

$=260-13.26$

$=246.74 \ m$

Now,

The final velocity to avoid collection is, V = 0 m/s

From third equation of motion:

⇒ $V^2=u^2+2ax$

On putting the estimated values, we get

⇒ $0=(26)^2+2\times a\times 246.74$

⇒ $0=676+493.48a$

⇒ $493.48a=-676$

⇒ $a=-\frac{676}{493.48}$

⇒ $a=1.369 \ m/s^2$

3 0

2 years ago

Sunlight has its maximum intensity at a wavelength of 4.83 x 10'm; wha energy does this correspond to in e?

Lera25 [3.4K]

Answer:

E = 2,575 eV

Explanation:

For this exercise we will use the Planck equation and the relationship of the speed of light with the frequency and wavelength

E = h f

c = λ f

Where the Planck constant has a value of 6.63 10⁻³⁴ J s

Let's replace

E = h c / λ

Let's calculate for wavelengths

λ = 4.83 10-7 m (blue)

E = 6.63 10⁻³⁴ 3 10⁸ / 4.83 10⁻⁷

E = 4.12 10-19 J

The transformation from J to eV is 1 eV = 1.6 10⁻¹⁹ J

E = 4.12 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

E = 2,575 eV

5 0

3 years ago

Carbon monosulfide ionic or covalent

taurus [48]

Answer:

Covalent.

Explanation:

Carbon and Sulfur are both nonmetals.

Nonmetal+nonmetal= covalent compound.

3 0

2 years ago