There are 2 elements:
Beryllium ( Be + 2 ), metal
Bromine ( Br - 1 ), non - metal
They form the ionic bond:
Be + Br2 → Be Br2
Naming the ionic compound:
on the 1st place is metal. On the 2nd place is non-metal. If compound is binary, use the element name and change ending to - ide .
Answer:
Beryllium Bromide.
Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
Explanation:
We want to find the statement that is proven by the fact that the balls reach the same height.
A isn't supported by the evidence. Balls can reach the same height without having the same initial speed.
B isn't supported by the evidence. Balls can reach the same height without having the same launch angle.
C is supported. Projectiles spend the same amount of time going up as they do coming down, so if two projectiles reach the same height, then they must spend the same amount of time in the air.
D isn't supported by the evidence. Balls thrown at the same speed and complementary angles have the same range but different heights.
E isn't supported by the evidence. The mass of the ball doesn't affect the height.
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Answer:
P_2 = 62.69 psi
Explanation:
given,
P₁ = 70 psia T₁ = 55° F = (55 + 459.67) R
P₂ = ? T₂ = 115° F = (115 + 459.67) R
we know,
p = ρ RT
ρ is the density which is constant
R is also constant
now,


P_2 = 62.69 psi
Hence, the increase in Pressure is equal to P_2 = 62.69 psi