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Because of barriers and obstructions, you can see only a tiny fraction of a standing wave on a string. You do not know, for inst

chubhunter [2.5K]

Answer:

(1) Sure, the frequency is 1000 Hz.

Explanation:

Frequency = wave speed ÷ wave distance

wave speed = 100 m/s

wave distance = 10 cm = 10/100 = 0.1 m

Frequency = 100 ÷ 0.1 = 1000 Hz

4 0

3 years ago

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A ball traveling with an initial momentum of 5.1 kgm/s bounces off a wall and comes back in the opposite direction with a moment

Elina [12.6K]

Answer: Change in momentum=9.4kgm/s

Impulse=9.4kgm/s

Explanation:

Change in momentum=5.1-(-4.3)=5.1+4.3=9.4kgm/s

Impulse=Change in momentum

There impulse=9.4kgm/s

4 0

3 years ago

If a t-shirt gun can fire t-shirts with an initial speed of 15 m/s, what is the maximum distance (along horizontal, flat ground)

kotegsom [21]

Answer:

h = 11.47 m

Explanation:

Initial speed pf the t-shirt gun is 15 m/s

We need to find the maximum distance covered by the t-shirt. It is based on the conservation of energy. The maximum distance covered is given by :

$h=\dfrac{u^2}{2g}\\\\h=\dfrac{(15)^2}{2\times 9.8}\\\\h=11.47\ m$

So, it will cover a distance of 11.47 m.

7 0

3 years ago

A gardener uses a 60-N wheelbarrow to transport two bags of fertilizer weighing W = 252-N. Determine the maximum allowable horiz

Butoxors [25]

Explanation:

Let us assume that the maximum allowable horizontal distance be represented by "d".

Therefore, torque equation about A will be as follows.

$60 \times 0.15 + 252 \times 0.15 \times 2 + 252 \times d = 2 \times 75 \times (0.7 + 0.15 + 0.15)$

d = $\frac{[2 \times 75 \times (0.7+0.15+0.15) - 60 \times 0.15 - 252 \times 0.15 \times 2]}{252}$

d = 0.409 m

Thus, we can conclude that the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm is 0.409 m.

6 0

3 years ago

A child whose weight is 235 N slides down a 4.90 m playground slide that makes an angle of 37.0° with the horizontal. The coeffi

Basile [38]

Answer:

Explanation:

a) First, let's calculate the value of the Friction force, which is given by the formula:

Ff = u*W

As the Friction force has an X component, it would be:

Ff = u*m*g*cosФ

Where m*g is the weight of the child.

Solving for Ff:

Ff = 0.051 * 235cos37 = 9.57 N

Now, to get the energy transferred to thermal energy (or heat) we need to get the Work done, so:

Wf = Ff * d

Wf = 9.57 * 4.9 = 46.9 J

b) We need to get the downslope component of the child weight, which is:

Wy = 235sin37 = 141.43 N

As you can imagine, the gravity also does work, so:

Wg = Wy * d

Wg = 141.43 * 4.9 = 693 J

Now, let's get the kinetic energy, which can be obtained by this expression:

ΔKe = Wg - Wf

ΔKe = 693 - 46.9 = 646.1 J

The formula for kinetic energy is:

Ke = 1/2 m*V^2

We have the innitial speed which is 0.355 m/s, and the mass can be obtained by m*g so:

Fw= m*g ----> m = Fw/g

m = 235 / 9.8 = 23.97 kg

so the innitial energy is:

Ke = 1/2 * 23.97 * (0.355)^2

Ke = 1.51 J

This could be Ke1, so to get Ke2:

ΔKe = Ke2 - Ke1

Ke2 = Δke + Ke1

Ke2 = 646.1 + 1.51 = 647,61 J

Finally, the speed at the bottom would be:

v = √2Ke/m

v = √2*647.61/23.97

v = 7.35 m/s

7 0

3 years ago

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