Pls answer quick with right answer !!!

With detailed explaniation

belka [17]

Ø=37°
Initial velocity=u=20m/s
g=10m/s²

#A

$\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}$

$\\ \rm\Rrightarrow H_{max}=20sin^237$

$\\ \rm\Rrightarrow H_{max}=7.2m$

#B

$\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}$

$\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}$

$\\ \rm\Rrightarrow R=40sin74$

$\\ \rm\Rrightarrow R=38.5m$

#C

$\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}$

$\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}$

$\\ \rm\Rrightarrow T=4sin37$

$\\ \rm\Rrightarrow T=2.4s$

Now

$\\ \rm\Rrightarrow v=u-gt$

$\\ \rm\Rrightarrow v=20-10(2.4)$

$\\ \rm\Rrightarrow v=20-24$

$\\ \rm\Rrightarrow v=-4m/s$

4 0

2 years ago

Need help I'll give you 30 points please.

Ksivusya [100]

Cathode Ray Tube i think?

7 0

3 years ago

What are some ways to separate a mixture?

stiks02 [169]

la separación mecánica de un sólido de grano grueso, insoluble, en un líquido; consiste en verter cuidadosamente el líquido, después de que se ha sedimentado el sólido. Por este proceso se separan dos líquidos miscibles, de diferente densidad, por ejemplola separación de un sólido disuelto en un líquido, por calentamiento, hasta que hierve y se transforma en vapor. Como no todas las sustancias se evaporan con la misma rapidez, el sólido disuelto se obtiene en forma pura, agua y aceite.

3 0

3 years ago

You fire an arrow at an angle of 45o from the horizontal and it lands a distance of 100 m from you. If you fire an identical arr

otez555 [7]

<u>Answer: </u>

Arrow projected at an angle of 46⁰ will travel 99.94 m.

<u>Explanation: </u>

The arrow fired at an angle 45⁰ travels 100 m horizontally, the horizontal movement is because of horizontal component of velocity.

Horizontal component = u cos θ, where u is the velocity of projection.

The time taken to travel 100m is depending on vertical component of velocity, time taken is calculated by dividing vertical component of velocity by acceleration due to gravity and multiplying by two.

So, time taken = 2u sin θ/g

Horizontal distance travelled by projectile, $H = u cos \theta * \frac{2u sin \theta }{g} = \frac{2u^2 sin 2\theta }{g}$

So. H ∝ sin 2θ

Here $\frac{H_1}{H_2} = \frac{sin 2\theta_1 }{sin 2\theta_2}\\ \\ \frac{100}{H_2} = \frac{sin (2*45) }{sin (2*46)}=\frac{1}{sin 92} \\ \\ H_2=99.94 m$

So an arrow projected at an angle of 46⁰ will travel 99.94 m.

4 0

3 years ago

A particle of rest energy mc2 is moving with speed v in the positive x direction. The particle decays into two particles, each o

Helga [31]

Answer:

The rest energy of the particle is $mc^2 = 498 MeV$

The velocity of the particle is $v = 1.587 *10^8 m/s$

Explanation:

From the question we are told that

The rest energy of the original particle is $E = mc^2$

The speed of the particle is $v$

The rest energy the first particle is $(mc^2)_1 = 140 MeV1$

The rest energy of the second particle is $(mc^2)_2= 140MeV$

The first particle has a kinetic energy of $KE = 282 MeV$ positive x

The second particle has a kinetic energy of $KE_2 = 25 MeV$ negative x

The total energy of the first particle is

$E_1 = (mc^2)_1 + KE_1$

Substituting values

$E_1 = 140 + 282$

$E_1 = 422 MeV$

The total energy of the first particle is

$E_1 = (mc^2)_2 + KE_2$

$E_1 = 140 + 25$

$E_1 =165MeV$

The momentum of the first particle can be mathematically evaluated using this expression

$p__{1 }}= \sqrt{E^2_1 - ((mc^2)_1) ^2}$

$p__{1 }}= \sqrt{422^2 - 140^2}$

$p__{1 }}= 398MeV$

The final energy of both particle is mathematically evaluated as

$E = E_1 + E_2$

$E = 422 + 165$

$E = 587 MeV$

The momentum of the first particle can be mathematically evaluated using this expression

$p__{2 }}= - \sqrt{E^2_2 - ((mc^2)_2) ^2}$

The negative sign shows that he direction is negative x-axis

$p__{2 }}= -\sqrt{165^2 - 140^2}$

$p__{2 }}= -87MeV$

The final momentum of both particle is mathematically evaluated as

$P = p_1 + p_2$

Substituting value

$P = 398 - 87$

$P = 311 MeV$

According to the law of conservation of momentum and the conservation of energy the final energy and momentum of both particle must be equal to the initial energy of the original particle

Then the rest energy is mathematically represented as

$mc^2 = \sqrt{E^2 - P^2}$