If 320 J of work is done on a spring with a spring constant of 730 N/m, how far will it stretch?

A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t

levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

$F=m\omega^2 r$

where

m is the mass of the object

$\omega$ is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

$\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s$ is the angular velocity

r = 30 cm = 0.30 m is the radius of the circle

Substituting, we find the force:

$F=(0.040)(6.28)^2(0.30)=0.47 N$

3 0

2 years ago

A 60KG WOMAN IS ON A LADDER 2 M ABOVE THE GROUND. WHAT IS HER POTENTIAL ENERGY

dexar [7]

Answer:

Explanation:

PE = mgh = 60(9.8)(2.0) = 1176 J

7 0

2 years ago

The first antiparticle, the positron or antielectron, was discovered in 1932. It had been predicted by Paul Dirac in 1928, thoug

Taya2010 [7]

Answer:

Energy of Photon = 4.091 MeV

Explanation:

From the conservation of energy principle, we know that total energy of the system must remain conserved. So, the energy or particles before collision must be equal to the energy of photons after collision.

K.E OF electron + Rest Energy of electron + K.E of positron + Rest Energy of positron = 2(Energy of Photon)

where,

K.E OF electron = 3.58 MeV

Rest Energy of electron = 0.511 MeV

Rest Energy of positron = 0.511 MeV

K.E OF positron = 3.58 MeV

Energy of Photon = ?

Therefore,

3.58 MeV + 0.511 MeV + 3.58 MeV + 0.511 MeV = 2(Energy of Photon)

Energy of Photon = 8.182 MeV/2

Energy of Photon = 4.091 MeV

8 0

2 years ago

A gas, behaving ideally, has a pressure P1 and at a volume V1. The pressure of the gas is changed to P2. Using Avogadro’s, Charl

Bond [772]

Answer:

Boyle's Law

$\therefore P_1.V_1=P_2.V_2$

Explanation:

Given that:

initially:

pressure of gas, $= P_1$

volume of gas, $= V_1$

finally:

pressure of gas, $= P_2$

volume of gas, $= V_2$

To solve for final volume $V_2$

According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.

According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.

But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.

Here nothing is said about the temperature, so we consider the Boyle's Law which states that at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.

Mathematically:

$P_1\propto \frac{1}{V_1}$

$\Rightarrow P_1.V_1=k\ \rm(constant)$

$\therefore P_1.V_1=P_2.V_2$

5 0

3 years ago

A solenoid that is 78.8 cm long has a cross-sectional area of 15.9 cm2. There are 914 turns of wire carrying a current of 8.25 A

Harlamova29_29 [7]

Answer:

(a) Energy density will be equal to $57.31J/m^3$

(b) Total energy will be equal to 0.0718 J

Explanation:

It is given that length of solenoid l = 78.8 cm = 0.788 m

Cross sectional area $A=15.9cm^2=15.9\times 10^{-4}m^2$

Number of turns of the wire N = 914

Current in the solenoid i = 8.25 A

Inductance of the wire is equal to $L=\frac{\mu _0N^2A}{l}=\frac{4\pi \times 10^{-7}\times 914^2\times 15.9\times 10^{-4}}{0.788}=2.117\times 10^{-3}H$

(b) Total energy stored in magnetic field $U=\frac{1}{2}Li^2=\frac{1}{2}\times 2.11\times 10^{-3}\times 8.25^2=0.0718J$

(a) Energy density will be equal to

$U_b=\frac{0.0718}{15.9\times 10^{-4}\times 0.788}=57.31J/m^3$

7 0

2 years ago