It would help if there were more of the question!
F = m A = (3 kg) (4 m/s2) = 12 kg-m/s2 = 12 newtons
The first step here is to find the spring constant. Use Hooke's law and the information given about the 3 kg mass: F=kx F=w3kg-mass=m3kg-massg=(3 kg)(9.8 m/s2) (3 kg)(9.8 m/s2)=(.40 m)k k=[(3 kg)(9.8 m/s2)/(.40 m)] Now plug that in (I don't have a calculator handy, and some of the units and numbers there will cancel, so I didn't bother to calculate it out) to Hooke's law for the 5 kg mass: F=kx F=w5kg-mass=m3kg-massg=(5 kg)(9.8 m/s2) (5 kg)(9.8 m/s2)=kx (5 kg)(9.8 m/s2)=[(3 kg)(9.8 m/s2)/(.40 m)]x x=2/3 m The 5 kg mass would stretch the spring two thirds of a meter.
It mass mass per unit volume which is unit of density in (kg/m^3) Hope it helped M. Arsalan
With a temperature of T = 293 K. Lead: 11.3 103 kg m-3 Silver: 10.50 103 kg m-3
Four and half
Yes engine oil has density but the temperature varies. For 0 degrees Celsius it is 899.0 kg/m^3. For 20 degrees it is 888.1 kg/m^3. For 40 degrees it is 876.0 kg/m^3. For 60 degrees it is 863.9 kg/m^3. For 80 degrees it is 852.0 kg/m^3. For 100 degrees it is 840.0 kg/m^3. For 120 degrees it is 828.9 kg/m^3. For 140 degrees it is 816.8 kg/m^3. For 150 degrees it is 810.3 kg/m^3.
From the Steam Tables I get the following: T v sub f d ( kg / m^3 ] 15.0 C 0.001001 m^3/ kg 999.0 kg / m^3 20.0 C 0.001002 m^3/ kg 998.0 kg/ m^3 25.0 C 0.001003 m^3 / kg 1001 kg / L^3 30.0 C 0 .001004 m^3 / kg 996.0 kg/m^3 -------------------- Search also NISTIR 6969, table 9.8 (it's free on Internet) or a density calculator.
mass [kg] = volume [m^3] * density [kg / m^3] From Newton's 2nd law of motion, weight [N] = mass [kg] * 9.8 [m/s^2] = volume [m^3] * density [kg / m^3] * 9.8 [m/s^2]. Rearranging, we get volume = weight / (density * 9.8) [m^3] Find the weight in [N], then you will know the volume in [m^3].
F = m A = (3 kg) (4 m/s2) = 12 kg-m/s2 = 12 newtons
0.4 kg(3 m/s) + 0.8kg(-2 m/s) = 1.2 kg m/s -1.6 kg m/s = -0.4 kg m/s -0.4 kg m/s = 1.2 kg(v) = (-0.4 kg m/s)/(1.2 kg) = v = -0.33 m/s the cars are traveling at 0.33 m/s in the direction of the second car.
The first step here is to find the spring constant. Use Hooke's law and the information given about the 3 kg mass: F=kx F=w3kg-mass=m3kg-massg=(3 kg)(9.8 m/s2) (3 kg)(9.8 m/s2)=(.40 m)k k=[(3 kg)(9.8 m/s2)/(.40 m)] Now plug that in (I don't have a calculator handy, and some of the units and numbers there will cancel, so I didn't bother to calculate it out) to Hooke's law for the 5 kg mass: F=kx F=w5kg-mass=m3kg-massg=(5 kg)(9.8 m/s2) (5 kg)(9.8 m/s2)=kx (5 kg)(9.8 m/s2)=[(3 kg)(9.8 m/s2)/(.40 m)]x x=2/3 m The 5 kg mass would stretch the spring two thirds of a meter.
I don't think there is any equivalency.... A watt isW = J/s = (N*m)/s = ((kg*(m/s^2))*m)/s = kg*(m^2))/(s^3)Therefore: W*kg = (kg^2)*((m^2)/(s^3)) which is not equal to N = kg*(m/s^2)I guess the closest answer would be :1 kg*W = (1 kg*m/s) N or1 kg*W = (1 N*s) N
It depends on the temperature. For 0 degrees Celsius it is 899.0 kg/m^3. For 20 degrees it is 888.1 kg/m^3. For 40 degrees it is 876.0 kg/m^3. For 60 degrees it is 863.9 kg/m^3. For 80 degrees it is 852.0 kg/m^3. For 100 degrees it is 840.0 kg/m^3. For 120 degrees it is 828.9 kg/m^3. For 140 degrees it is 816.8 kg/m^3. For 150 degrees it is 810.3 kg/m^3.
(1000 kg ) / (1.29 kg/m^3) = 775 m^3
Saturn is less dense...Water is 1000 kg/m^3 and Saturn is 700 kg/m^3
density of copper = 8930 kg/m^3 & density of brass = 8700 kg/m^3 copper is heavier
About 7848 kg/m3