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Ugo [173]
3 years ago
7

Name one type of technology that utilizes magnetism to function.

Physics
1 answer:
Lubov Fominskaja [6]3 years ago
3 0

Magnetic Resonance Imaging (MRI) is one type of technology that utilizes magnetism to function. Most might know it better as an MRI scan.

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What is 541.2 mg in grams
PIT_PIT [208]

Answer:

0.5142 grams.

Explanation:

I think

5 0
3 years ago
Read 2 more answers
A simple harmonic oscillator has an amplitude of 3. 50 cm and a maximum speed of 26. 0 cm/s. What is its speed when the displace
Marianna [84]

Answer:

1.) A simple harmonic oscillator has an amplitude of 3.50 cm and a maximum speed of 26.0 cm/s. What is its speed when the displacement is 1.75 cm? 2.) Both pendulum A and B are 3.0 m long. The period of A is T. Pendulum A is twice as heavy as pendulum B. What is the period of B? 3.) The time for one cycle of a periodic process is called the _ ? 4.) In simple harmonic motion, the acceleration is proportional to? 5.) The position of a mass that is oscillating on a spring is given by x= (18.3 cm) cos [(2.35 s-1)t]. What is the frequency of this motion?

Explanation:

3 0
2 years ago
suppose a ball is thrown vertically upward. Eight seconds later it returns to its point of release. What is the initial velocity
valentinak56 [21]
The ball took half of the total time ... 4 seconds ... to reach its highest
point, where it began to fall back down to the point of release.

At its highest point, its velocity changed from upward to downward. 
At that instant, its velocity was zero.

The acceleration of gravity is 9.8 m/s².  That means that an object that's
acted on only by gravity gains 9.8 m/s of downward speed every second. 

-- If the object is falling downward, it moves 9.8 m/s faster every second.

-- If the object is tossed upward, it moves 9.8 m/s slower every second.

The ball took 4 seconds to lose all of its upward speed.  So it must have
been thrown upward at  (4 x 9.8 m/s)  =  39.2 m/s .

(That's about  87.7 mph straight up.  Somebody had an amazing pitching arm.)
6 0
3 years ago
A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its maximum displacement from
Mrac [35]

Answer:

C) True. At maximum displacement, its instantaneous velocity is zero.

Explanation:

The simple harmonic movement is given by

        x = A cos wt

Speed

        v = - A w sin wt

At the point of maximum displacement x = A

       A = A cos wt

      cos wt = 1

      wt = 0

We replace the speed

       v = -Aw sin 0 = A w

Speed ​​is maximum

Let's review the claims

A) False. Speed ​​is zero

B) False. It can be determined

C) True. Agree with our result

D) False. When one is maximum the other is minimum

4 0
3 years ago
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
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