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scoray [572]
3 years ago
9

When is the magnitude of the acceleration of a mass on a spring at its maximum value?

Physics
1 answer:
DiKsa [7]3 years ago
6 0

Answer:

D

Explanation:

Now we know that Force is the rate of change of momentum meaning

F= mv/t

But

mv/t = Ke

v/t = ke/m

a= ke/m

Where a is acceleration

K is constant of proportionality of tension on a spring

e is the extension substended by a string.

From the formula of acceleration we can see that as mass decreases acceleration increases so we can see that m = 1

We would have a maximum value of acceleration.

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Answer:

29,032′

Explanation:

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2 years ago
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An ocean wave has an amplitude of 2.5 m. Weather conditions suddenly change such that the wave has an
Dovator [93]

Answer:

2.5

Explanation:2.5 +2.5 = 5.0

8 0
3 years ago
Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 2 h, and Car B traveled the dista
lions [1.4K]
The answer is 60 mph.

The speed (v) is distance (d) per time (t): v = d/t

Car A:
v1 = ?
t1 = 2 h
d1 = ?
___
v1 = d1/t1
d1 = v1 * t1

Car B:
v2 = ?
t2 = 1.5 h
d2 = ?
___
v2 = d2/t2
d2 = v2 * t2

<span>Two cars traveled equal distances:
d1 = d2
</span>v1 * t1 = v2 * t2

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v2 = v1 + 15


</span>v1 * t1 = v2 * t2
v2 = v1 + 15
________
v1 * 2 = (v1 + 15) * 1.5
2v1 = 1.5v1 + 22.5
2v1 - 1.5v1 = 22.5
0.5v1 = 22.5
v1 = 22.5/0.5
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8 0
3 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
3 years ago
Phase change requires...
amm1812
<span>Evaporation involves a liquid becoming a gas and sublimation is the change of a solid directly to a gas.Phase changes require either the addition of heat energy (melting, evaporation, and sublimation) or subtraction of heat energy (condensation and freezing.</span>
3 0
3 years ago
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