They all have segmented limbs, a hard exoskeleton, a pair of antennae and a segmented body.
S=125km
t=2h
v=s/t=125/2=62,5km/h
or 62,5/3,6=17,36m/s
The gravitational constant was experimentally measured by W Cavendish using the attraction between big and small lead balls. is true
The correct answer is true
<h3>How do you define gravitational constant?</h3>
the strength of gravity. a factor in use in Newton's gravity law to relate the strength of the gravitational pull between two bodies with their masses and distance from one another. 6.67259 X 10-11 newtons per square kilogram is roughly the gravitational constant. G is its identifier.
<h3> where is the strongest gravity is?</h3>
The gravitational pull of the earth is greatest near sea level, normally, and weakens as you get further from the center, such as to the summit of Mt. Everest. Because the obloid earth was slightly wider, but only by a minor ratio, the gravity just at poles is stronger than that at the equator.
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Answer:
a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
b) T = 295.37 K
Explanation:
Given;
Initial temperature of tea T1 = 31 C
Initial temperature of ice T2 = 0 C
Mass of tea m1 = 0.89 kg
Mass of ice m2 = 0.075kg
The heat capacity of both water and tea c = 4186 J/(kg⋅K)
the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg
And T = the final temperature of the mixture
Heat loss by tea = heat gained by ice
m1c∆T1 = m2c∆T2 + m2Lf
m1c(T1-T) = m2c(T-T2) + m2Lf
m1cT1 - m1cT = m2cT - m2cT2 + m2Lf
m1cT + m2cT = m1cT1 + m2cT2 - m2Lf
T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
Substituting the values;
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)
T = 22.37 °C
T = 273 + 22.37 K
T = 295.37 K
Answer:
a. 1/2 of the radius of earth
Explanation:
<h2>g= GM/r²</h2>
thus g is inversely proportional to radius
