What are the products of the double displacement reaction?
AgCINaNO3
Ag + Cl + Na + NO3
AgCINa + NO3
AgCl + NaNO3
AgCl + NaNO₃
AgNO3 + NaCl → AgCl + NaNO₃
Mass
Diameter
Height of drop
Tennis Ball
0.06 kg
6.6 cm
5 m
Baseball
0.15 kg
74 cm
5 m
Bowling Ball
5 kg
21 cm
5 m
Golf Ball
0.04 kg
4.3 cm
5 m
Statement 1: All balls hit the ground at the same time
Statement 2: All balls hit the ground with different forces
Statement 3: The bowling ball hit the ground first and the tennis and golf ball hit the ground last
Statement 4: It is not possible to tell how much force each ball hit the ground with
Answer:
statement 2
Explanation:
b. malleable and ductile.
c. gases at room temperature.
d. shiny.
Unlike metals, many non-metals are malleable and ductile. Thus, the correct option is B.
Non-metals are those metals which lack all the metallic attributes which are found in metallic elements. The non-metals are good insulators of heat and electricity. They are mostly gases and sometimes also liquid. Some of the elements then are even solid at room temperature include Carbon, sulphur and phosphorus.
A malleable material is the material in which a thin sheet can be easily formed through hammering. Gold is the most malleable metal found in nature. In contrast, ductility can be defined as the ability of a solid material to deform under the tensile stress.
Therefore, the correct option is B.
Learn more about Non-metals here:
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Calculate:
(a) the time to reach maximum height
(b) the maximum height above the base of the cliff reached by the
projectile
(c) thetotal time it is in the air
(d) the horizontal range of the projectile.
Answer:
a) 9.99 s
b) 538 m
c) 20.5 s
d) 1160 m
Explanation:
Given:
x₀ = 0 m
y₀ = 49.0 m
v₀ = 113 m/s
θ = 60.0°
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
a) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find t.
vᵧ = aᵧ t + v₀ᵧ
(0 m/s) = (-9.8 m/s²) t + (113 sin 60.0° m/s)
t ≈ 9.99 s
b) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find y.
vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)
(0 m/s)² = (113 sin 60° m/s)² + 2 (-9.8 m/s²) (y − 49.0 m)
y ≈ 538 m
c) When the projectile lands, y = 0 m. Find t.
y = y₀ + v₀ᵧ t + ½ aᵧ t²
(0 m) = (49.0 m) + (113 sin 60° m/s) t + ½ (-9.8 m/s²) t²
You'll need to solve using quadratic formula:
t ≈ -0.489, 20.5
Since negative time doesn't apply here, t ≈ 20.5 s.
d) When the projectile lands, y = 0 m. Find x. (Use answer from part c).
x = x₀ + v₀ₓ t + ½ aₓ t²
x = (0 m) + (113 cos 60° m/s) (20.5 s) + ½ (0 m/s²) (20.5 s)²
x ≈ 1160 m