Answer:
a) 578.0 cm²
b) 25.18 km
Explanation:
We're given the density and mass, so first calculate the volume.
D = M / V
V = M / D
V = (6.740 g) / (19.32 g/cm³)
V = 0.3489 cm³
a) The volume of any uniform flat shape (prism) is the area of the base times the thickness.
V = Ah
A = V / h
A = (0.3489 cm³) / (6.036×10⁻⁴ cm)
A = 578.0 cm²
b) The volume of a cylinder is pi times the square of the radius times the length.
V = πr²h
h = V / (πr²)
h = (0.3489 cm³) / (π (2.100×10⁻⁴ cm)²)
h = 2.518×10⁶ cm
h = 25.18 km
Answer:
(a) m = 1.6 x 10²¹ kg
(b) K.E = 2.536 x 10¹¹ J
(c) v = 7.12 x 10⁵ m/s
Explanation:
(a)
First we find the volume of the continent:
V = L*W*H
where,
V = Volume of Slab = ?
L = Length of Slab = 4450 km = 4.45 x 10⁶ m
W = Width of Slab = 4450 km = 4.45 x 10⁶ m
H = Height of Slab = 31 km = 3.1 x 10⁴ m
Therefore,
V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)
V = 6.138 x 10¹⁷ m³
Now, we find the mass:
m = density*V
m = (2620 kg/m³)(6.138 x 10¹⁷ m³)
<u>m = 1.6 x 10²¹ kg</u>
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(b)
The kinetic energy will be:
K.E = (1/2)mv²
where,
v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)
v = 3.17 x 10⁻¹⁰ m/s
Therefore,
K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²
<u>K.E = 2.536 x 10¹¹ J</u>
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(c)
For the same kinetic energy but mass = 77 kg:
K.E = (1/2)mv²
2.536 x 10¹¹ J = (1/2)(77 kg)v²
v = √(2)(2.536 x 10¹¹ J)
<u>v = 7.12 x 10⁵ m/s</u>
Answer:
the spring be displaced by 25.0 cm
Explanation:
The computation is shown below:
As we know that
F= -K × x
So,
Now
= -0.250m
= 25.0 cm
Hence, the spring be displaced by 25.0 cm
Answer:
C
Explanation:
An element is a pure substance that can not be broken down into anything simpler
A compound in also a pure substance held together in fixed proportion through chemical bonds
Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
