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Naily [24]
3 years ago
7

An electron experiences a magnetic force with a magnitude of 4.90×10−15 n when moving at an angle of 60.0 ∘ with respect to a ma

gnetic field with a magnitude of 3.70×10−3 t .
Physics
1 answer:
Leya [2.2K]3 years ago
4 0
Missing questions: "find the speed of the electron".

Solution:
the magnetic force experienced by a charged particle in a magnetic field is given by
F=qvB \sin \theta
where
q is the particle charge
v its velocity
B the magnitude of the magnetic field
\theta the angle between the directions of v and B.

Re-arranging the formula, we find:
v= \frac{F}{qB \sin \theta}
and by substituting the data of the problem (the charge of the electron is q=1.6 \cdot 10^{-19} C), we find the velocity of the electron:
v= \frac{F}{qB \sin \theta}= \frac{4.90 \cdot 10^{-15}N}{(1.6 \cdot 10^{-19}C)(3.70 \cdot 10^{-3} T)(\sin 60^{\circ})}=9.56 \cdot 10^6 m/s
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What product is obtained from the aldol condensation of cyclohexanone?
Evgesh-ka [11]

Answer:

First product is FCH-OH chemically known as 2-[2-furyl(hydroxyl)methyl]-Second product is  FCH i.e (2E)-2-[2-furyl-methylene]-cyclohexanone

Explanation:

Please see the attached image for complete chemical reaction of aldol condensation of cyclohexanone

Aldol Condensation is a form of electrophilic substitution reaction in which the alpha carbon in enols or enolate anions is substituted by an electrophile to form carbon-carbon bond. Cyclohexanone also known as the first ketone consists of two alpha-carbons and four potential substitutions i.e alpha-hydrogens but none of the hydrogen on the ring is substituted. Ketones such as cyclohexanone are much more acidic than their parent hydrocarbon.

First product is FCH-OH chemically known as 2-[2-furyl(hydroxyl)methyl]-cyclohexanone  that further undergoes dehydration resulting into FCH i.e (2E)-2-[2-furyl-methylene]-cyclohexanone

Based on the explanations above, the compound formed is shown in the image.  

6 0
3 years ago
A measurement that has both magnitude and direction
Korvikt [17]
<span>A measurement that both magnitude and direction is a vector quantity. An example of this is a moving car. The car exerts force due to its thrust and weight that runs in it. This will give us the magnitude of the car. The resulting motion of the car in terms of displacement, velocity and acceleration that determines its direction makes it a vector quantity. On the other hand, a measurement that has only magnitude is a scalar quantity. The energy exerted by the engine of the car is a scalar quantity.</span>
7 0
3 years ago
Investigators are researching the appearance of a strange movement of objects in what the local community considers a haunted ho
wolverine [178]

Answer:

0.18216 T

Explanation:

N = Number of turns = 219

A = Area = \pi r^2

r = Radius = 1 cm

\omega = Angular speed = 40\times 2\pi

Maximum emf is given by

\epsilon=NBA\omega\\\Rightarrow B=\dfrac{\epsilon}{NA\omega}\\\Rightarrow B=\dfrac{3.15}{219\times \pi 0.01^2\times 40\times 2\pi}\\\Rightarrow B=0.18216\ T

The strength of the magnetic field is 0.18216 T

8 0
3 years ago
A uniform solid disk with a mass of 24.3 kg and a radius of 0.364 m is free to rotate about a frictionless axle. Forces of 90.0
a_sh-v [17]

Answer:

a. -12.7 Nm

b. -7.9 rad/s^2

Explanation:

I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.

Forces applied to the solid disk include:

F_1 = 90.0N\\F_2 = 125N

Other parameters given include:

Mass of solid disk, M = 24.3kg

and radius of solid disk, r = 0.364m

a.) The formula for determining torque (T), is T = r * F

Hence the net torque produced by the two forces is given as a summation of both forces:

T = T_{125} + T_{90}\\= -r(125)sin90 + r(90)sin90\\= 0.364(-125 + 90)\\= -12.7 Nm

b.)  The angular acceleration of the disk can be found thus:

using the formula for the Moment of Inertia of a solid disk;

I_{disk} = {\frac{1}{2}}Mr^2

where M = Mass of solid disk

and r = radius of solid disk

We then relate the torque and angular acceleration (\alpha) with the formula:

T = I\alpha \\-12.7 = ({\frac{1}{2}}Mr^2)\alpha \\\alpha  = -{\frac{12.7}{1.61}} = -7.9 rad/s^2

4 0
3 years ago
A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
7 0
3 years ago
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