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almond37 [142]
3 years ago
7

A small block of mass m on a horizontal frictionless surface is attached to a horizontal spring that has force constant k. The b

lock is pushed against the spring, compressing the spring a distance d. The block is released, and it moves back and forth on the end of the spring. During its motion, what is the maximum speed of the block?
Physics
1 answer:
Rashid [163]3 years ago
5 0

Answer:

v_{max} = |d|\cdot \sqrt{\frac{k}{m} }

Explanation:

The maximum speed of the block occurs when spring has no deformation, that is, there is no elastic potential energy, which can be remarked from appropriate application of the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot d^{2} = \frac{1}{2}\cdot m \cdot v^{2}

k \cdot d^{2} = m\cdot v^{2}

v_{max} = |d|\cdot \sqrt{\frac{k}{m} }

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Convection currents in air and water occur because _
Diano4ka-milaya [45]

warm fluids are less dense than cold fluids

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3 years ago
The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the
natulia [17]

Answer:

The value is   A =  2.80 *10^{-4} \  m^2

Explanation:

From the question we are told that

The  operating temperature is  T  =  2450 \  K

The emissivity is  e =  0.350

 The  power rating is  P  =  200 \  W

Generally the area is mathematically represented as

      A = \frac{P}{ e *  \sigma  *  T^2}

Where  \sigma is the Stefan Boltzmann constant  with value  

      \sigma  =  5.67 *10^{-8} \  W/m^2\cdot K^4

So

     A =  \frac{200}{0.350 *  5.67*10^{-8} *  2450^{4}}

     A =  2.80 *10^{-4} \  m^2

8 0
3 years ago
Qwertyuiop hhffhhhhf fiuefhasuhs afbauyhsbvzskhbdv
mars1129 [50]

Answer:

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4 0
2 years ago
Read 2 more answers
The length and mass of the arm are Larm = X1 = 50 cm and Marm = 0.3 kg, X2 = 15 cm, and the mass of the object is MObject = 0.25
max2010maxim [7]

Answer: 0.5N

Explanation: if the system is at equilibrium, sum of the torque will be equal to zero.

But if they are not in equilibrium.

U will find the difference in the two torque

find the attached file for solution

3 0
2 years ago
Please help
Snezhnost [94]

Answer:

Range = 22.61 m

Explanation:

We can use the formula for the Range in flat ground, given by:

Range=v_i^2\frac{sin(2\theta)}{g}

which for our case renders:

Range=15^2\frac{sin(80^o)}{9.8} \approx 22.61\,\,m

4 0
2 years ago
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