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Hooke's law if a 3 kg mass stretches a spring 40 cm how far will a 5 kg mass stretch the spring?
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.
Can you measure mass using spring balance?
Yes
How do you measure the mass of a cow?
Put her on a truck and have the truck weighed on a balance scale (NOT a spring scale). From this you can get true mass.
What unit of measure is spring scale to?
Mass (Newtons(N)) and Weight (grams(g))
Can the speed and frequency of the waves in a spring be changed?
All violinists believe so. The springs (= violin string) properties of frequency and timbre can be altered by the pressure of the bow against the string. Even in a simple coiled spring, you'll find the period of the pluck waves will change as the spring is elongated. But the purist will point out (correctly) that it is now a different spring. My favourite demonstration is of a rubber band about 200 mm long, with a mass at the lower end. This spring+mass apparatus has at least three resonant periods! First is that of a simple pendulum. Second is that of a torsional resonance (much slower). Third is that of the vertical oscillation of a spring-mass system.
What mass would be needed to stretch the spring to a length of 60 cm?
1,500 grams2,500 grams500 grams2,000 grams
A mass of 1.7kg caused a vertical spring to scretch 6m what's the spring constant?
A mass of 1.7kg caused a vertical spring to stretch 6m so the spring constant is 2.78.
Suggest suitable Experiments to find spring constant of a helical spring?
Connect a mass to the bottom of the spring. (depending on the spring size, the mass will vary, the larger the spring the greater the mass u can use) Suppose you use a 100 g mass on a spring, measure the amount by which it stretches and record the data. Use hooke law to figure out the constant of the spring. K = m.g/x m = mass, g =gravity, x = stretch
Predict how many centimeters the spring will stretch if a total mass of 700 grams were attached?
hello how are u
What is the density in 55cm?
Density is the measure of mass divided by volume. To know the density of whatever you're referring to, we would need a lot more information.
If a 40kg weight is hanging on a spring with spring constant 50Nm how far will the spring stretch?
The net force acting in the stretch direction on the spring is proportional to its deformation (e.g., its stretch). In math talk that's F = kdX where k = 50 N/m and F = mg = 40*9.81 is the weight of the m mass. NOTE: 40 kg is not...not...a weight, it's a measure of mass. To get the weight, the force of gravity, we must multiply by g = 9.81 m/sec^2 which is a typical average acceleration due to gravity on the Earth's surface. So the stretch dX = F/k = mg/k = 40*9.81/50 = 7.848 meters. ANS.
A 200 g mass is attached to a spring causing it to stretch 5 cm If another 200 g mass is added to the spring the potential energy of the spring will be how many times as much?
do your homework yourself. You may acually learn something interesting :) For this problem, think of the ratio between the two masses to find the answer
What forces are acting on the Mass a mass is hanging from a spring?
A mass is hanging from a spring experiences the force of gravity.
Which states the relationship between the mass attached to the end of the spring and the length the spring is stretched?
more mass the longer the spring
A mass is hanging from a spring what forces are acting on the mass?
I do believe that it is either kinetic or potential energy. From jellyfish
Hooke's law if a 3 kg mass stretches a spring 40 cm how far will a 5 kg mass stretch the spring?
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.
What happens to the length of a spring when you add a mass to it?
Usually it gets bigger, if you wanted to do an experement, you would- Measure the spring, put it in water, take it out then measure it again and it would be bigger!
