Asked by Marie
A beaker weighs 0.4N when empty and 1.4N when filled with water what does it weigh when filled with brine of density 1.2g/cm^3
Answers
Answered by
oobleck
the volume is (1.4-0.4) = 1.0N = 1000/9.81 = 101.936 cm^3
so the weight will be
1.2 * 101.936 * 9.81 + 0.4 N = 1.6N
or, with less work, 1.2 times the weight of the water, which weight 1.0N
so the weight will be
1.2 * 101.936 * 9.81 + 0.4 N = 1.6N
or, with less work, 1.2 times the weight of the water, which weight 1.0N
Answered by
Anonymous
the water weighed 1.4 - 0.4 = 1 Newton
mass of the water then is 1 N / 9.81 m/s^2 = (1/9.81) kg
water density is about 1 gram / cm^3 or 1000 kg / m^3
so the volume of the water is (1/9.81) kg / 1000 kg / m^3 = (10^-3/9.81)m^3
1 m^3 = 100^3 cm^3 = 10^6 cm^3
so vol of water = 1000/9.81 cm^3
1.2 g /cm^3 * 1000/9.81 = 1,200/9.81 grams mass of brine = 1.2/9.81 kg
weight = m g = 1.2/9.81 * 9.81 = 1.2 Newtons
or just 1.2/1 * 1 = 1.2
the brine density is 1.2 times water density immediately.
mass of the water then is 1 N / 9.81 m/s^2 = (1/9.81) kg
water density is about 1 gram / cm^3 or 1000 kg / m^3
so the volume of the water is (1/9.81) kg / 1000 kg / m^3 = (10^-3/9.81)m^3
1 m^3 = 100^3 cm^3 = 10^6 cm^3
so vol of water = 1000/9.81 cm^3
1.2 g /cm^3 * 1000/9.81 = 1,200/9.81 grams mass of brine = 1.2/9.81 kg
weight = m g = 1.2/9.81 * 9.81 = 1.2 Newtons
or just 1.2/1 * 1 = 1.2
the brine density is 1.2 times water density immediately.
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