The coefficient of expansion is 13 * 10^-6 m per meter length.per oK
The temperature difference = 42 - - 8 = 50 oC
delta T = (42 + 273) - (-8 + 273) = 50 oK
delta L = L * 13* 10^6 m/oK
oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m
So we need to find the length and it is computed by:
0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m
Answer:
Newton's Second Law tells us that the more mass an object has, the more force is needed to move it. A larger rocket will need stronger forces (eg. more fuel) to make it accelerate. The space shuttles required seven pounds of fuel for every pound of payload they carry.
Explanation:
Answer: 245 meters
Explanation: 49 ? Meters
over over over
1 5 seconds
Cross multiply- 49 times 5 divided by 1 which equals 245
Answer:
In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.
Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.
Explanation:
In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.
Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.
