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How much work is done if a 10kg mass is moved 5 meters?
The work done is the force multiplied by the distance. You don't have the force in this case; if there is no friction, you would need zero work. If the mass moves is on a flat surface, multiply by the coefficient of friction to get the force required. The coefficient of friction varies for different combinations of materials.The work done is the force multiplied by the distance. You don't have the force in this case; if there is no friction, you would need zero work. If the mass moves is on a flat surface, multiply by the coefficient of friction to get the force required. The coefficient of friction varies for different combinations of materials.The work done is the force multiplied by the distance. You don't have the force in this case; if there is no friction, you would need zero work. If the mass moves is on a flat surface, multiply by the coefficient of friction to get the force required. The coefficient of friction varies for different combinations of materials.The work done is the force multiplied by the distance. You don't have the force in this case; if there is no friction, you would need zero work. If the mass moves is on a flat surface, multiply by the coefficient of friction to get the force required. The coefficient of friction varies for different combinations of materials.
What does µ stand for in physics?
It could be the friction-constant or the prefix micro, which is 10^-6
A material has a coefficient of volume expansion of 60×10^-6/degree celsius. What is its coefficient of expansion?
The coefficient of volume expansion is the triple of the linear expansion coefficient. So with a volume expansion coefficient of 60×10^-6/°C, the linear expansion coefficient would be 20×10^-6/°C.
If you know the coefficient of friction how do you find the maximum velocity to maintain a circular path at a given radius?
say mass(m) = 10 kg, radius(r) = 10 m, say friction coefficient = 0.5 force to break friction = 10 * 0.5 = 5 kgf = say 50 n to find acceleration required to produce this force use f=m*a, shuffle to a = f / m so a = 50 / 10 = 5 (m/s)/s, install in a = v^2 / r, so 5 = v^2 / 10, so 10 * 5 = v^2, so sq. root 50 = v, so v = 7.07 metres / second if friction coefficient and radius remain the same, altering the mass wont alter the velocity at breakaway point
What number is greater 10 to the power 390?
If you mean "greater than", you can raise it to a higher power.
If your pulling 30 lb tire on a lawn what is the estimated weight youre actually pulling because of friction?
the equation for static friction coefficient is:static friction coefficient = force required to break bond / weight of object (tire)you need the friction coefficient between rubber and grass, say its 0.5, this means the force you have to apply to equal the friction force is:0.5 = x / 30x = 0.5 * 30x = 15 lbs fanything greater than 15 lbs f will break the bond and accelerate the tire.notes :1 / moving friction coefficient is usually less than static friction coefficient, so youve less drag once its moving.2 / friction coefficients are never greater than 1.0actually , some friction coefficients do exceed 1.0 , see google / friction coefficients table
If two numbers are written in scientific notation how can you tell which is the greater number?
The number with the greater coefficient and/or exponent is the greater number. n x 10^E where "n" is the coefficient, and "E" is the exponent. examples: 1.23 x 10^21 > 1.23 x 10^20 1.22 x 10^21 > 1.21 x 10^21 1.22 x 10^21 > 1.22 x 10^-21 see http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson25.htm for reference.
How much work is done if a 10kg mass is moved 5 meters?
The work done is the force multiplied by the distance. You don't have the force in this case; if there is no friction, you would need zero work. If the mass moves is on a flat surface, multiply by the coefficient of friction to get the force required. The coefficient of friction varies for different combinations of materials.The work done is the force multiplied by the distance. You don't have the force in this case; if there is no friction, you would need zero work. If the mass moves is on a flat surface, multiply by the coefficient of friction to get the force required. The coefficient of friction varies for different combinations of materials.The work done is the force multiplied by the distance. You don't have the force in this case; if there is no friction, you would need zero work. If the mass moves is on a flat surface, multiply by the coefficient of friction to get the force required. The coefficient of friction varies for different combinations of materials.The work done is the force multiplied by the distance. You don't have the force in this case; if there is no friction, you would need zero work. If the mass moves is on a flat surface, multiply by the coefficient of friction to get the force required. The coefficient of friction varies for different combinations of materials.
What does µ stand for in physics?
It could be the friction-constant or the prefix micro, which is 10^-6
A material has a coefficient of volume expansion of 60×10^-6/degree celsius. What is its coefficient of expansion?
The coefficient of volume expansion is the triple of the linear expansion coefficient. So with a volume expansion coefficient of 60×10^-6/°C, the linear expansion coefficient would be 20×10^-6/°C.
If you know the coefficient of friction how do you find the maximum velocity to maintain a circular path at a given radius?
say mass(m) = 10 kg, radius(r) = 10 m, say friction coefficient = 0.5 force to break friction = 10 * 0.5 = 5 kgf = say 50 n to find acceleration required to produce this force use f=m*a, shuffle to a = f / m so a = 50 / 10 = 5 (m/s)/s, install in a = v^2 / r, so 5 = v^2 / 10, so 10 * 5 = v^2, so sq. root 50 = v, so v = 7.07 metres / second if friction coefficient and radius remain the same, altering the mass wont alter the velocity at breakaway point
What number is greater 10 to the power 390?
If you mean "greater than", you can raise it to a higher power.
Does a rollercaoster have greater kinetic energy if it fall from 25 meters instead of 10 meters?
yes as long as friction is negligible.
Does 0.1 mean 10?
No. 10 is 100 times greater than .1
When using scientific notation the the coefficient is less than what?
When using scientific notation, the coefficient must be between greater than or equal to 1, and less than 10. For example, 5.17, 9.63, and 1.49 are all correct coefficients, while 0.12 and 10.87 are not.
What will be the acceleration of a block of mass 10kg a force of 50Newton is acting on it coefficient of friction is 0.5Newton and the block is at rest initially?
Coefficent of friction isn't measured in Newtons. F=COF.R R=mg therefore F=49.05N F=ma a=(50-49.05)/10 a=0.095m.s-2
What is the coefficient in the expression 10x2-7?
In the expression (10x^2 - 7), the coefficient is the numerical factor that multiplies the variable. Here, the coefficient of (x^2) is 10, while the term -7 is a constant and does not have a variable associated with it. Thus, the coefficient in this expression is 10.
