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3 years ago

Evaluate the scenarios and select the one that demonstrates the training principle overload.

Physics

2 answers:

Irina18 [472]3 years ago

8 0

Answer:

The answer is A.

Explanation:

The answer is A. because overload means to do more and Kai swims more and more every day.

matrenka [14]3 years ago

5 0

Answer:

B :)

Explanation:

:) JUST TRUST ME I GOT IT CORRECT

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A cyclotron is used to produce a beam of high-energy deuterons that then collide with a target to produce radioactive isotopes f

serg [7]

Answer:

19080667.0818 m/s

0.637294 m

$2.1875\times 10^{15}$

Explanation:

m = Mass of deuterons = $3.34\times 10^{-27}\ kg$

v = Velocity

K = Kinetic energy = 3.8 MeV

d = Diameter

B = Magnetic field = 1.25 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

t = Time = 1 s

i = Current = 350 μA

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 3.8\times 10^6\times 1.6\times 10^{-19}}{3.34\times 10^{-27}}}\\\Rightarrow v=19080667.0818\ m/s$

The speed of the deuterons when they exit is 19080667.0818 m/s

In this system the centripetal and magnetic force will balance each other

$\dfrac{mv^2}{r}=qvB\\\Rightarrow \dfrac{mv^2}{\dfrac{d}{2}}=qvB\\\Rightarrow d=\dfrac{2mv}{qB}\\\Rightarrow d=\dfrac{2\times 3.34\times 10^{-27}\times 19080667.0818}{1.6\times 10^{-19}\times 1.25}\\\Rightarrow d=0.637294\ m$

The diameter is 0.637294 m

Current is given by

$i=\dfrac{nq}{t}\\\Rightarrow n=\dfrac{it}{q}\\\Rightarrow n=\dfrac{350\times 10^{-6}\times 1}{1.6\times 10^{-19}}\\\Rightarrow n=2.1875\times 10^{15}$

The number of deuterons is $2.1875\times 10^{15}$

8 0

3 years ago

The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the

tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.

A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.

B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

6 0

4 years ago

Atoms in a solid are in continuous vibrational motion due to thermal energy. At room temperature, the amplitude of these atomic

Bingel [31]

Answer:

10 m/s

Explanation:

Given:

Amplitude of atomic vibrations (λ) = 10⁻⁹ cm = 10⁻⁹ × 10⁻² m = 10⁻¹¹ m [1 cm = 10⁻² m]

Frequency of the vibrations (f) = 10¹² Hz

In order to find the atom's maximum speed, we need to make use of the formula that relates speed, frequency and wavelength of the vibration.

Therefore, the formula for maximum speed is given as:

$v=f\lambda$

Now, plug in the values given and solve for speed 'v'. This gives,

$v=(10^{12}\ Hz)(10^{-11}\ m)\\\\v=10^{12-11}\ m/s\\v=10\ m/s$

Therefore, the atom's maximum speed due to thermal energy provided is 10 m/s.

6 0

4 years ago

A 26.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m th

frozen [14]

Answer:

18 m/s

Explanation:

Force is given by mass × acceleration

F = ma

a = F/m

For the first 11.0 m, there is no friction. Hence, the constant force is applied fully on the crate. The acceleration is

a = 225/26.0 m/s²

By the equation of motion, v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance travelled.

Using the values for the first part of the motion,

v² = 0² + 2 × 8.65 × 11.0

v = 13.8 m/s

This is the initial velocity for the second part of the motion. This part has friction of coefficient 0.20.

The frictional force = 0.20 × weight = 0.20 × 26.0 × 9.8 = 50.96 N

The effective force moving the crate in the second motion = 225 - 50.96 N = 174.04 N

This gives an acceleration of 174.04/26.0 = 6.69 m/s².

Using parameters for the equation of motion, v² = u² + 2as

v² = 13.8² + 2 × 6.69 × 10 = 324.24

v = 18 m/s

8 0

4 years ago

A 60.watt light bulb runs for 5.0 seconds. How much energy does it use?

VikaD [51]

"watt" means "Joule of energy per second"

"60 watts" means "60 Joules per second"

(60 joules per second) x (5 seconds) = 300 Joules of energy

5 0

4 years ago