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2 years ago

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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? *

1 answer:

Ede4ka [16]2 years ago

5 0

Wheres the picture? Theres no picture below

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To determine the height of a tall building such as Sears Tower in Chicago, Illinois a ball was dropped from the top of the build

Darya [45]

Answer:

The height of Sears Tower is 1448.5 feet.

Explanation:

<h3>We apply the free fall formula to the ball: </h3><h3> $y=v_{o} *t+\frac{1}{2} *g*t^{2}$ </h3><h3>y: The vertical distance the ball moves at time t </h3><h3> $v_{o}$ i: Initial speed </h3><h3>g=Gravity acceleration= $9.8*(\frac{\frac{1ft}{0.305m} }{s^{2} } )$ </h3>

Known information

We know that the vertical distance (y) that the ball moves in 9,5s is equal to height of Sears Tower (h).

Too we know that the ball is released from rest, then, $v_{0}$ =0

Height of Sears Tower calculation:

We replace in the equation 1 the data following;

$y=h$

$v_{o} =0$

$g=32,1\frac{ft}{s^{2} }$

$t= 9,5s$

$h=0*9.5+\frac{1}{2} *32.1*9.5^{2}$

$h=1448.5 ft$

Answer: The height of Sears Tower is 1448.5 ft

6 0

3 years ago

One of the world's largest Ferris wheels, the Cosmo Clock 21 with a radius of 50.0 m is located in Yokohama City, Japan. Each of

STatiana [176]

Answer:

a = 0.55 m / s²

Explanation:

The centripetal acceleration is given by the relation

a = v² / r

angular and linear velocities are related

v = w r

we substitute

a = w² r

In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system

w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s

let's calculate

a = 0.105² 50.0

a = 0.55 m / s²

4 0

3 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

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3 years ago

What units would you use to measure a red blood cell ? (nanometers micrometers millimeters

Stels [109]

Viruses are infectious particles made of nucleic acid (DNA or RNA) and protein. Viruses are 100 times smaller than the size of bacteria.

4 0

2 years ago

Read 2 more answers

Andrews [41]

Given :

A 13.3 kg box sliding across the ground decelerates at 2.42 m/s².

To Find :

The coefficient of kinetic friction.

Solution :

Frictional force applied to the box is :

$f = ma$ ....1)

Also, force of friction is given by :

$f = \mu mg$ ....2)

Equating equation 1) and 2), we get :

$\mu mg = ma\\\\\mu = \dfrac{a}{g}\\\\\mu = \dfrac{2.42}{9.8}\\\\\mu = 0.247$

Therefore, the coefficient of kinetic friction is 0.247 .

8 0

3 years ago