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nataly862011 [7]
2 years ago
13

A 0.25 kg steel ball is tied to the end of a string and then whirled in a vertical circle at a constant speed v. The length of t

he string is 0.62 m, and the tension in the string when the ball is at the top of the circle is 4.0 N. What is v
Physics
1 answer:
TiliK225 [7]2 years ago
7 0

Answer:

Explanation:

Let the tension in the string be T . At the top of the circle , total force acting on them = T + mg . This will provide centripetal force

T + mg = m v² / r

4 + .25 x 9.8 = .25 x v² / .62

6.45 = .25 v² / .62

v² = 16

v = 4 m /s .

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Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0
3 years ago
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
2 years ago
A mass-spring system has k = 56.8 N/m and m = 0.46 kg.
andriy [413]

Answer:

A. \omega=11.1121\ rad.s^{-1}

B. f=1.7685\ Hz

C. T=0.5654\ s

Explanation:

Given:

  • spring constant, k=56.8\ N.m^{-1}
  • mass attached, m=0.46\ kg

A)

for a spring-mass system the frequency is given as:

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{56.8}{0.46}}

\omega=11.1121\ rad.s^{-1}

B)

frequency is given as:

f=\frac{\omega}{2\pi}

f=\frac{11.1121}{2\pi}

f=1.7685\ Hz

C)

Time period of a simple harmonic motion is given as:

T=\frac{1}{f}

T=0.5654\ s

7 0
3 years ago
Need help on this thank you
Semmy [17]

Answer:

TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction.

7 0
3 years ago
A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the
Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J

W = 3.12 J

Hope this helps!

4 0
3 years ago
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