All categories

3 years ago

A common technique used to measure the

Physics

1 answer:

vodomira [7]3 years ago

3 0

Answer:

$K=6\ N/m$

Explanation:

<u>Force Constant of a Spring</u>

The force F applied to a spring produces a stretching distance x. These variables are linearly related as expressed by Hook's law:

$F=K.x$

If we could measure the distance a spring stretches when applying a known force, we'd be able to find the value of K. The experiment stated in the question places some mass of a known weight of 96 N that produced a stretching distance of 16 m. Knowing both variables, we can solve the above equation for K

$\displaystyle K= \frac{F}{x}=\frac{96\ N}{16\ m}=6\ N/m$

$\boxed{K=6\ N/m}$

You might be interested in

An ideal gas is brought through an isothermal compression process. The 3.00 mol 3.00 mol of gas goes from an initial volume of 2

ozzi

Answer:

The answers can be found by considering the isothermal expansion equation as well as the ideal gas equation from where we have

The temperature T = 602.64K and the final pressure P = 110.24MPa

Explanation:

Numbeer of moles of gas = 3.00 mol

initial volume = 230.8×10−6 m3

final volume = 133.4×10−6 m3 .

released energy = 8240 J

Temperature = Constant = T

Pressure = $p_{f}$ =unknown

From the relation the combined ideal gas law, PV = nRT

Where R = 8.314 4621.JK−1mol−1

we have The release energy from compression P1V1 -P2V2

-qrev = -nRTln( $\frac{V_{2} }{V_{1} }$ ) = 8240J

n = 3

Hence -nRTln( $\frac{V_{2} }{V_{1} }$ ) = 3×8.314 462×ln $(\frac{133.4}{230.8})$ × T= -8240 J

or -13.67×T = -8240J, thus T = -8240/-13.67 = 602.64K

The Final pressure is given by

PV = n×R×T from where we have V = final volume thus

P = (n×R×T)/V = (3×8.134×602.64)÷(133.4× $10^{-6}$ ) = 110237041.1 N/ $m^{2}$ = 110.237MPa

8 0

3 years ago

A horizontal spring required a force of 1.0 N to compress it 0.1 m. How much work is required to stretch the spring 0.4 m?

Crank

Answer:

Explanation:

given data

Force applied F= 1N

extension e= 0.1m

let us find the spring constant first

applying

F=ke

k=F/e

k=1/0.1

k=10N/m

Step two:

Required is the work done

we know that the expression/formula for the work done by a spring is given as

Wd=1/2kx^2

x=0.4m

substitute

Wd= 1/2*10*0.4^2

Wd=0.5*10*0.16

Wd=0.5J

3 0

3 years ago

In Fraunhofer diffraction wave front used is __________. A. Spherical B. Circular C. Plane D. Conical

Zepler [3.9K]

I am pretty sure the answer is C.

4 0

3 years ago

If there is 8 g of a substance before a physical change , how much will there be afterwards?

Dimas [21]

A substance undergoing a physical change will still weigh the same even after the change. This is in accordance to the law of conservation of mass which states that mass is neither created nor destroyed. so an 8 g substance remains of the same weight even after undergoing a physical change.

8 0

4 years ago

Read 2 more answers

A high school physics student is sitting in a seat read-

Nataly_w [17]

The equilibrium condition allows finding the result for the force that the chair exerts on the student is:

The reaction force that the chair exerts on the student's support is equal to the student's weight.

Newton's second law gives the relationship between force, mass and acceleration of bodies, in the special case that the acceleration is is zero equilibrium condition.

∑ F = 0

Where F is the external force.

The free body diagram is a diagram of the forces on bodies without the details of the shape of the body, in the attached we can see a diagram of the forces.

Let's analyze the force on the chair.

$N_{chair} - W_{chair} - W_{student} = 0 \\ \\N_{chair} = W_{chair} + W_{student}$

Let's analyze the forces on the student.

$N_{student} - W_{student} = 0 \\N_{student} = W _{student}$

In conclusion using the equilibrium condition we can find the result for the force that the chair exerts on the student is:

The reaction force that the chair exerts on the student's support is equal to the student's weight.

Learn more here: brainly.com/question/18117041

7 0

3 years ago