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Andre45 [30]
3 years ago
13

A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg

ligible during this linear elastic collision, a)stone A will slow down but continue moving forward b) stone A will rebound and stone B will move forward c) stone a will rebound but stone b will remain stationary d) stone A will stop and stone b will move forward
Physics
1 answer:
Kitty [74]3 years ago
4 0

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

p_i=mv+0=mv

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

p_f=mv'+MV'

Equating initial and the final momenta we get

mv=mv'+MV'\\\\m(v-v')=MV'.....i

Now since the surface is frictionless thus the energy is also conserved thus

E_i=\frac{1}{2}mv^2

Similarly the final energy becomes

E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\

Equating initial and final energies we get

\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)

Solving i and ii we get

v+v'=V'

Using this in equation i we get

v'=\frac{v(m-M)}{(M-m)}=-v

Thus putting v = -v' in equation i  we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

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A mass attached to a spring vibrates back and forth. At maximum displacement, the spring force and the
Svetradugi [14.3K]

ANSWER

Velocity of the mass reaches zero

EXPLANATION

We want to identify what hapens to a mass attached toa a spring at maximum displacement.

When a mass attached to a spring is at its maximum position of displacement, the direction of the mass begins to change. This implies that the velocity of the mass will reach zero.

Hence, at maximum displacement, the velocity of the mass reaches zero.

8 0
1 year ago
After which action would the concentration of a solution remain constant?(1 point)
Alika [10]

Answer:

C. Evaporating water from the container.

Explanation:

The concentration of solution changes when solvent or solute are added/removed from a solution.

5 0
3 years ago
Find the wavelength in meters for a transverse mechanical wave with an amplitude of 10 cm and a radian frequency of 20π rad/s if
nydimaria [60]

Answer:

The wavelength of the wave is 20 m.

Explanation:

Given that,

Amplitude = 10 cm

Radial frequency \omega = 20\pi\ rad/s

Bulk modulus = 40 MPa

Density = 1000 kg/m³

We need to calculate the velocity of the wave in the medium

Using formula of velocity

v=\sqrt{\dfrac{k}{\rho}}

Put the value into the formula

v=\sqrt{\dfrac{40\times10^{6}}{10^3}}

v=200\ m/s

We need to calculate the wavelength

Using formula of wavelength

\lambda =\dfrac{v}{f}

\lambda=\dfrac{v\times2\pi}{\omega}

Put the value into the formula

\lambda=\dfrac{200\times2\pi}{20\pi}

\lambda=20\ m

Hence, The wavelength of the wave is 20 m.

5 0
3 years ago
A child is sliding down a slide at the playground. is the mechanical energy conserved. why or why not.
aleksandrvk [35]
C: the mechanical energy isn't conserved. Some energy was lost to friction. 
6 0
3 years ago
Read 2 more answers
A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
3 years ago
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