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skelet666 [1.2K]
1 year ago
11

A flat metal washer is heated. As the washer's temperature increases. What happens to the hole in the center?

Physics
1 answer:
Phoenix [80]1 year ago
8 0

Answer:

expands

Explanation:

It gets bigger....expands.... just like if there was a solid disk of metal there which would expand

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babymother [125]

your answer is make up artist

4 0
3 years ago
A mass m is attached to an ideal massless spring. When this system is set in motion with amplitude a, it has a period t. What is
Luden [163]

The period will be the same if the amplitude of the motion is increased to 2a

What is an Amplitude?

Amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium.

Here,

mass m is attached to the spring.

mass attached = m

time period = t

We know that,

The time period for the spring is calculated with the equation:

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

Where k is the spring constant

Now if the amplitude is doubled, it means that the distance from the equilibrium position to the displacement is doubled.

From the equation, we can say,

Time period of the spring is independent of the amplitude.

Hence,

Increasing the amplitude does not affect the period of the mass and spring system.

Learn more about time period here:

<u>brainly.com/question/13834772</u>

#SPJ4

7 0
2 years ago
In a nuclear fusion reaction, the mass of the products is _____ the mass of the reactants.
JulijaS [17]

Answer:

more than

Explanation:

In a nuclear fusion reaction, the mass of the products is more than the mass of the reactants.

7 0
3 years ago
A third baseman makes a throw to first base 40.5 m away. The ball leaves his hand with a speed of 30.0 m/s at a height of 1.4 m
Ugo [173]

Answer:

When the ball goes to first base it will be 4.23 m high.

Explanation:

Horizontal velocity = 30 cos17.3 = 28.64 m/s

   Horizontal displacement = 40.5 m

   Time  

         t=\frac{40.5}{28.64}=1.41s          

   Time to reach the goal posts 40.5 m away = 1.41 seconds

Vertical velocity = 30 sin17.3 = 8.92 m/s

    Time to reach the goal posts 40.5 m away = 1.41 seconds

    Acceleration = -9.81m/s²

    Substituting in s = ut + 0.5at²

             s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m

    Height of throw = 1.4 m

    Height traveled by ball = 2.83 m

    Total height = 2.83 + 1.4 = 4.23 m

    When the ball goes to first base it will be 4.23 m high.

8 0
3 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
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