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3 years ago

Which numbers indicate the phases during which the moon is waxing?

Physics

1 answer:

wolverine [178]3 years ago

6 0

what the answer i need it please

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Anon25 [30]

In several of the questions you've posted during the past day, we've already said that a wave with larger amplitude carries more energy. That idea is easy to apply to this question.

8 0

3 years ago

Read 2 more answers

A bullet glider and a target glider both have a mass of 0.200 kg. The bullet glider is moving 0.450 m/s

Romashka [77]

Answer:

the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Explanation:

When the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Changes can occur that the gliders unite and move with a cosecant speed less than the initial one.

The whole process must be analyzed using conservation of the moment.

p₀ = m v₀

celestines que clash case

p_f = (m + M) v

po = pf

m v₀ = (n + M) v

v = $\frac{m}{m+M}$

calculemos

v= $\frac{0.200}{0.200+M} 0.450$

v= 0.09 m/s

elastic shock case

p₀ = m v₀

p_f = m v₁ +M v₂

p₀ = p_f

m v₀ = m v₁ + m v₂

6 0

3 years ago

A projectile is fired from the ground at a velocity of 30.0 m/s, 35.0 º from the horizontal. What is the maximum height the proj

Norma-Jean [14]

Answer:

Vy = V sin theta = 30 * ,574 = 17.2 m/s

t1 = 17.2 / 9.8 = 1.76 sec to reach max height

Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m

H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m

Time to fall from zero speed to ground = rise time = 1.76 sec

Vx = V cos 35 = 24.6 m / sec horizontal speed

Time in air = 1.76 * 2 = 3.52 sec before returning to ground

S = 24.6 * 3.52 = 86.6 m

8 0

2 years ago

Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.

andreev551 [17]

Answer:

the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = $m_A$

then, the mass of object B = $m_B = \frac{m_A}{4}$

work done on object A = -18 J

work done on object B = -18 J

let $v_i$ be the initial speed

let $v_f$ be the final speed

For object A;

$K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\$

$v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\$

Thus, the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

To obtain the change in the final speed of object B, apply the following equations.

$K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B = \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\$

$\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i$

Thus, the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

3 0

2 years ago

Apakah yang dimaksud dengan gaya

tangare [24]

Nah gaya seperti titik fashion seperti apa yang Anda mana dan hal-hal seperti itu. (tell me if you cant understand)

8 0

3 years ago

Read 2 more answers