Answer:
Q=1670J
Explanation:
Mass of ice: m=5g=0.005kg
Latent heat: lambda=3.34×10⁵J/kg
Heat received by ice: Q=m×lambda
Q=0.005×3.34×10⁵=5×334=1670J
The result of the Mexican victory was that fallen defenders
became heroes to the cause of Texan independence.<span> The Battle of
the Alamo took place between February 23 and March 6, 1836 and became the
central episode of the Texas
Revolution . After this thirteen-day battle, the
Mexican troops of General President Antonio
Lopez de Santa Anna began an attack on San Antonio de
Bexar, the current San Antonio in Texas. The Battle of the Alamo fought the
army of Mexico against
a group of Texan rebels, mostly American settlers. More than four thousand
men from Santa Ana stood in front of
the Alamo Fort , the last stronghold of the rebels, which
barely reached 187. The Alamo was not a fortress prepared to withstand a siege.
It is believed that all the rebels of the Alamo died in the siege, but Santa
Anna came to lose up to about 900 men during the days that lasted the fight. However,
the worst result for Santa Ana was precisely the resistance that the Texan
rebels had in the Alamo, which fostered the fighting spirit of the Texans. A
few days later, on March 14, 1836, Texas became independent from Mexico and a
month later, Santa Ana was imprisoned.</span>
Answer:
Your answer would be letter <em><u>B</u></em><em><u>.</u></em><em><u> </u></em><em><u>Electrons</u></em><em><u> </u></em><em><u>orbit</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>energy</u></em><em><u> </u></em><em><u>level</u></em><em><u>.</u></em>
Explanation:
Hope it helps..
Just correct me if I'm wrong, okay?
But ur welcome!!
(;ŏ﹏ŏ)(◕ᴗ◕✿)
Answer:
?
Explanation:
Can u explain your question?
Answer:
K_a = 8,111 J
Explanation:
This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved
initial instant. Just before dropping the particles
p₀ = 0
final moment
p_f = m_a v_a + m_b v_b
p₀ = p_f
0 = m_a v_a + m_b v_b
tells us that
m_a = 8 m_b
0 = 8 m_b v_a + m_b v_b
v_b = - 8 v_a (1)
indicate that the transfer is complete, therefore the kinematic energy is conserved
starting point
Em₀ = K₀ = 73 J
final point. After separating the body
Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²
K₀ = K_f
73 = ½ m_a (v_a² + v_b² / 8)
we substitute equation 1
73 = ½ m_a (v_a² + 8² v_a² / 8)
73 = ½ m_a (9 v_a²)
73/9 = ½ m_a (v_a²) = K_a
K_a = 8,111 J