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swat32
3 years ago
9

The picture below shows an experiment to find out the highest frequencies which students can hear. When the teacher turns the di

al the frequency from the loudspeaker increases. the students put their hands down when they can no longer hear the sound.
Some students put their hands up while others have put their hands down. What conclusion can you draw from this observation?

The teacher suspects that some of the students are not giving honest responses, they are keeping their hands up even when they can no longer hear the sound. How could he check if he is right? *
Physics
1 answer:
Ede4ka [16]3 years ago
5 0
Wheres the picture? Theres no picture below
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A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you
soldi70 [24.7K]

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}       (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s

The interaction time to avoid that the water balloon breaks is 0.029s

5 0
3 years ago
A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0
3 years ago
Enter an expression for the force constant for the floating raft, in terms of L, g, and the density of water, ρ.
andriy [413]

Answer:

K = ρL²g

Explanation:

Consider L as the length of the raft inside the water when the raft is displaced through additional distance y;

Then:

F = upthrust ( restoring force) = weight of the liquid displaced.

F = V_{\omega} \rho_{\omega} g= A y \rho_{\omega} g

where;

A = L²

\rho_{\omega} = \rho

F = ky.

Then,

Ay \rho g = ky

L^2y \rho g = ky

Divide both sides by y

K = ρL²g

3 0
3 years ago
What types of electromagnetic radiation has a shorter wavelength than ultraviolet?
Lunna [17]
X-rays and gamma rays are the only electromagnetci waves with a shorter wavelength, gamma rays being the smallest. Hope this helps ;)
4 0
3 years ago
How do you know which magnitude is higher or how do you compare them?
Paul [167]
Magnitudes are measured by intensity so a 3.4 earthquake is much less stronger than a 4.5 earthquake it’s very literally when measuring them the higher the number the stronger it is
7 0
3 years ago
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