Answer:
We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved
Explanation:
Answer:
233.1 miles per hours
Explanation:
Speed: This is defined as the ratio of distance to time. The S.I unit of speed is m/s. speed is a vector quantity because it can only be represented by magnitude only. Mathematically, speed can be expressed as,
S = d/t ....................... Equation 1
Where S = speed of the runner, d = distance covered, t = time.
Given: d = 100 meter , t = 9.580 seconds
Conversion:
If, 1 meter = 0.00062 miles
Then, 100 meters = (0.00062×100) miles = 0.62 miles.
Also
If, 3600 s = 1 h
Then, 9.580 s = (1×9.580)/3600 = 0.00266 hours.
Substitute into equation 1
S = 0.62/0.00266
S = 233.1 miles per hours.
Hence the runner speed is 233.1 miles per hours
Mass divided by volume is density, while mass times density is volume. you cannot calculate density without volume and you cannot calculate volume without density.
i believe that's the answer.
oh gosh i didn't realize my middle school education involved high school stuff.
Answer:
a) B = 1.99 x 10⁻⁴ Tesla
b) B = 0.88 x 10⁻⁴ Tesla
Explanation:
According to Biot - Savart Law, the magnetic field due to a currnt carrying straight wire is given as:
B = μ₀ I L/4πr²
where,
μ₀ = permebility of free space = 1.25 x 10⁻⁶ H m⁻¹
I = current = 2 A
L = Length of wire = 40 cm = 0.4 m
a)
r = radius of magnetic field = 2 cm = 0.02 m
Therefore,
B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.02 m)²
<u>B = 1.99 x 10⁻⁴ Tesla</u>
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b)
r = radius of magnetic field = 3 cm = 0.03 m
Therefore,
B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.03 m)²
<u>B = 0.88 x 10⁻⁴ Tesla</u>
Answer:
A book on its side exerts a greater force.
Explanation:
Pressure = Force / Area
Assuming that 1kg = 10N
2kg = 20N
Area of book lying flat = 0.3m × 0.2m
= 0.6m²
Pressure of book lying flat = 20N / 0.6m²
= 30Pa (1 s.f.)
Area of book on its side = 0.2m × 0.05m
= 0.01m²
Pressure of book on its side = 20N / 0.01m²
= 2000Pa (1 s.f.)
Since 2000Pa (1 s.f.) > 30Pa (1 s.f.), a book on its side applies greater pressure than lying flat.
