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2 years ago

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1/A ball is dropped from the top of a building. After 3 seconds, its speed is measured to be 29.4 m/s. Calculate the acceleratio

n of the dropped ball.
2/ How fast will the ball be going after 4 seconds?

1 answer:

AleksAgata [21]2 years ago

7 0

Answer:

9.8

Explanation:

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A 1-kg rock is suspended by a massless string from one end of a

maxonik [38]

Answer:

The weight of measuring stick is 9.8 N

Explanation:

given information:

the mass of the rock, $m_{r}$ = 1 kg

measuring stick, x =1 m

d = 0.25 m

to find the weight of measuring stick, we can use the following equation:

τ = Fd

τ = 0

$F_{r} d$ - $F_{s}d$ = 0

F_{r} = the force of the rock

F_{s} = the force of measuring stick

$F_{s} =F_{r}$

= m g

= 1 kg x 9.8 m/s

= 9.8 N

thus, the weight of measuring stick is 9.8 N

6 0

3 years ago

Sharon the ant (Aaron’s sister) sits at the edge of a turntable of radius R that is spinning with period T. As she makes one-hal

Dmitry_Shevchenko [17]

Answer:

$a = \dfrac{4\pi^2R}{T^2}$

Explanation:

The acceleration of a circular motion is given by

$a = \omega^2 R$

where $\omega$ is the angular velocity and $R$ is the radius.

Angular velocity is related to the period, T, by

$\omega=\dfrac{2\pi}{T}$

Substitute into the previous formula.

$a = (\dfrac{2\pi}{T})^2 R$

$a = \dfrac{4\pi^2R}{T^2}$

This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.

6 0

3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago

Why does light change its velocity (or slows down) when it encounters a glass lens?

Rufina [12.5K]

Explanation:

what happens is that light slows down when it passes from the less dense air into the denser glass or water. This slowing down of the ray of light also causes the ray of light to change direction. It is the change in the speed of the light that causes refraction.

6 0

3 years ago

The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid

Karolina [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

$w=2250*2\pi *\frac{1}{60}\\ w=235.61$

$I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}$

Now using Inertia an w

$E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J$

average power= $\frac{11.4J}{0.230s}=50.2 W$

b).

power=t*w

P= $11.5465*0.25*235.61$

P=2.957 kW

8 0

2 years ago

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