From tables, the density of mercury is
13545 kg/m^3 at 20°C,
13472 kg/m^3 at 50°C.
Because mass = density * volume, the mass of mercury at 20°C is
m = (13545 kg/m^3)*(0.002 m^3) = 27.09 kg
Let V = volume of mercury at 50°C.
Because the mass of mercury does not change, therefore at 50°C,
(13472 kg/m^3)*(V m^3) = 27.09
V = 27.09/13472 = 0.0020108 m^3
Answer: B. 0.002010812 m³
Answer:
2.29 m/s
Explanation:
given,
Speed of the stream to north,x = 3.3 m/s
speed of the ship,z = 8 m/s
Speed of the stream is in north direction, to move the ship in west direction ship should move in south direction.
To solve this a right angle triangle is constructed.
where speed of the stream is base and speed of cruise is hypotenuse.
now, speed of ship with respect to ground
z² = x² + y²
y² = z² - x²
y² = 8² - 3.3²
y = 2.29 m/s
Hence, the speed of the ship with respect to ground is equal to 2.29 m/s.
I’m not entirely certain, but usually, spring symbolizes rebirth.
Answer:
8.46E+1
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = 39 C
Charge 2 (q₂) = –53 C
Force (F) of attraction = 26×10⁸ N
Electrical constant K) = 9×10⁹ Nm²/C²
Distance apart (r) =?
The distance between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
26×10⁸ = 9×10⁹ × 39 × 53 / r²
26×10⁸ = 1.8603×10¹³ / r²
Cross multiply
26×10⁸ × r² = 1.8603×10¹³
Divide both side by 26×10⁸
r² = 1.8603×10¹³ / 26×10⁸
r² = 7155
Take the square root of both side
r = √7155
r = 84.6 m
r = 8.46E+1 m
Answer:
Bottom of the circle.
Explanation:
At the top of the circle the tension and the weight contribute on being the centripetal force, at the middle of the circle only the tension contributes on being the centripetal force (the weight being perpendicular to it), while <u>at the bottom</u> of the circle the tension contributes on being the centripetal force (as always) <em>but the weight against to it</em>, so here is where the tension must be greater to allow the same centripetal force as the other cases, thus here is where the string will break.
