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3 years ago

How does increasing the tension of a spring affect a wave on the spring?

2 answers:

4 0

Increasing the tension of a spring affects a wave on the spring because it increases the frequency. When the tension rises, so does the frequency.

mezya [45]3 years ago

3 0

Answer: Increases the frequency of the wave

Explanation:

Increasing the tension of a spring means increasing its stiffness. The wave produced thereafter has greater frequency.

Higher tension causes the spring to have greater restoring force causing it to quickly come back to its equilibrium position. Thus, the speed of the wave also increases.

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Jail

kykrilka [37]

Answer:

c.o Kg • ms 08

Explanation:

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7 0

3 years ago

A sample of radium-226 will decay to ¼ of its original amount after 3200 years. What is the half-life of radium-226?

lesantik [10]

<h3><u>Answer</u>;</h3>

1600 years

<h3><u>Explanation</u>;</h3>

Half life is the time taken for a radioactive isotope to decay by half of its original amount.
We can use the formula; N = O × (1/2)^n ; where N is the new mass, O is the original amount and n is the number of half lives.
A sample of radium-226 takes 3200 years to decay to 1/4 of its original amount.

Therefore;

Thus; <em>3200 years is equivalent to 2 half lives.</em>

<em>Hence, the half life of radium-226 is 1600 years</em>

3 0

3 years ago

vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses

dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

$v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}$

(This is correct because the horizontal motion has acceleration zero). Then:

$v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s$

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

$E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}$

Then, plugging in the given values, we obtain:

$k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m$

Finally, the effective spring constant of the firing mechanism is 1808N/m.

3 0

3 years ago

What best describes gravity?

DanielleElmas [232]

F = G*((m sub 1*m sub 2)/r^2)

8 0

2 years ago

The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f

zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

584 = Ha + Lg

an also this mean that torques acting on the body is balanced

So, 0.410 Ha = 0.840 Lg

Making Lg the subject of formula in the equation above we

Lg = 0.4881 Ha

Considering the first equation and replacing Lg with this recent equation we have

584 = Ha + 0.4881 Ha

Therefore Ha = 392.44 N

This value obtained is for both hands for each hand we divide by 2

Therefore we have for each hand $= 392.44/2$ =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

Lg = 0.4881 × 392.44

= 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

191.22/2 = 95.8 N.

8 0

3 years ago