All categories

3 years ago

Which is easier for a beginner? (This is for archery) O long bow O recurve bow

Physics

2 answers:

AlexFokin [52]3 years ago

6 0

Answer:

The recurve bows

Explanation:Vote brainliest plz

Zepler [3.9K]3 years ago

4 0

The recurve bow is smaller and easier to handle while the longbow is used more for competitions. So I would say the recurve bow is easier for a beginner

You might be interested in

A student stretches out a rubber band. What primary form of energy does the stretched-out rubber band possess? A. gravitational

hammer [34]

Answer : )b

Explanation:

bc i said its b sznankana

3 0

3 years ago

Read 2 more answers

George determines the mass of his evaporating dish to be 3.375 g. He adds a solid sample to the evaporating dish, and the mass o

zhannawk [14.2K]

Explanation:

The given data is as follows.

Mass of evaporating dish = 3.375 g

Total mass = Mass of solid sample + evaporating dish

That is, Mass of solid sample + evaporating dish = 26.719 g

Therefore, we will calculate the mass of solid sample as follows.

Mass of solid sample = (Mass of solid sample + evaporating dish) - mass of evaporating dish

= 26.719 g – 3.375 g

= 23.344 g

Thus, we can conclude that mass of his solid sample must be 23.344 g.

3 0

4 years ago

A high-speed K0 meson is traveling at β = 0.90 when it decays into a π + and a π − meson. What are the greatest and least speeds

san4es73 [151]

Answer:

greatest speed=0.99c

least speed=0.283c

Explanation:

To solve this problem, we have to go to frame of center of mass.

Total available energy fo π + and π - mesons will be difference in their rest energy:

$E_{0,K_{0} }-2E_{0,\pi } =497Mev-2*139.5Mev\\$

=218 Mev

now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,

K=(γ-1) $E_{0,\pi }$

$K=E_{0,\pi}(\frac{1}{\sqrt{1-\beta ^{2} } } -1)\\\frac{1}\sqrt{1-\beta ^{2} }}=\frac{K}{E_{0,\pi}}+1\\ {1-\beta ^{2}=\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta ^{2}=1-\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta = +-\sqrt{\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\\\\beta =+-\sqrt{1-\frac{1}{(\frac{109Mev}{139.5Mev+1)^2}}$

$u'=+-0.283c$

note +-=±

To find speed least and greatest speed of meson we would use relativistic velocity addition equations:

$u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c$

6 0

3 years ago

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

3 years ago

Across the sky from the great Orion,

mars1129 [50]

Answer:

Explanation:

Across the sky from the great Orion,

I killed him who hunted the lion!

My alpha star, so very large and red,

Rivals the red planet, truthfully said!

Near the center of the Milky Way

My star clusters send many a ray

To Earth, revealing sights so wondrous!

I am the <u>Scorpious</u>, <u>The scorpian</u>.

The above mentioned poem is the greek mythdology that is related to the scorpian ( one of the constellations).

8 0

3 years ago