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3 years ago

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A mass attached to a spring oscillates with a period of 22 sec. After 22 kg are added, the period becomes 33 sec. Assuming that

we can neglect any damping or external forces, determine how much mass was originally attached to the spring.

1 answer:

polet [3.4K]3 years ago

7 0

Answer:

Mass of 17.854 kg is only attached to the spring

Explanation:

We have given time period in first case is 22 sec

Let initially mass is m

After 22 kg are added the period becomes 33 sec

Time period of spring mass system is

$T=2\pi \sqrt{\frac{m}{k}}$ , here m is mass and k is spring constant

From the relation we can see that

$\frac{T_1}{T_2}=\sqrt{\frac{m}{m+22}}$

$\frac{22}{33}=\sqrt{\frac{m}{m+22}}$

Squaring both side

$0.444={\frac{m}{m+22}}$

$0.444m+9.777=m$

m = 17.584 kg

So mass of 17.854 kg is only attached to the spring

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An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro

Mashutka [201]

1) The distance travelled by the electron is $1\cdot 10^{17} m$

2) The time taken is $4.0\cdot 10^{10}s$

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

$u=5\cdot 10^6 m/s$ is the initial velocity

v = 0 is the final velocity (it comes to a stop)

$a=-1.25\cdot 10^{-4} m/s^2$ is the acceleration

Solving for s, we find the distance travelled:

$s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m$

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

$u=5\cdot 10^6 m/s$

v = 0

$a=-1.25\cdot 10^{-4} m/s^2$

And solving for t, we find the time taken:

$t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s$

Learn more about accelerated motion:

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If I wanted to generate a maximum emf of 20 V, what angular velocity (radians/sec, aka Hz) would be required given a circular co

sergij07 [2.7K]

Answer:

Angular velocity, $\omega=35.36\ rad/s$

Explanation:

It is given that,

Maximum emf generated in the coil, $\epsilon=20\ V$

Diameter of the coil, d = 40 cm

Radius of the coil, r = 20 cm = 0.2 m

Number of turns in the coil, N = 500

Magnetic field in the coil, $B=9\times 10^{-3}\ T$

The angle between the area vector and the magnet field vector varies from 0 to 2 π radians. The formula for the maximum emf generated in the coil is given by :

$\epsilon=NBA\omega$

$\omega=\dfrac{\epsilon}{NBA}$

$\omega=\dfrac{20\ V}{500\times 9\times 10^{-3}\ T\times \pi (0.2\ m)^2}$

$\omega=35.36\ rad/s$

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vladimir1956 [14]

Answer:

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Explanation:

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Answer:

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Since we know that electric potential is a scalar quantity

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$V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_3 = 63 Volts$

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