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Annette [7]
3 years ago
9

Which changes in an electric motor will make the motor stronger? Check all that apply.

Physics
1 answer:
ollegr [7]3 years ago
8 0
To make the motor turn faster we can:
(a) increase the current
(b) use stronger magnets
(c) push the magnets closer to the coil
(d) put an iron centre piece into the coil
(e) adding more sets of coils
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For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser poin
hoa [83]

Answer:

Slit width is 0.109 mm

Solution:

As per the question:

Wavelength of the light, \lambda = 680\ nm = 680\times 10^{- 9}\ m

Distance from the screen, D = 5.6 m

Bright band in the center, 2x = 8.1 cm

x = 4.05 cm

Now,

To calculate the slit width, w:

From the diffraction:

sin\theta = \frac{\lambda}{w}

For small angle:

sin\theta ≈ tan\theta = \frac{x}{D}

Also

\frac{x}{D} = \frac{\lambda }{w}

Thus

w = \frac{D}{x}\times \lambda

Substituting appropriate values in the above eqn:

w = \frac{6.5}{4.05\times 10^{-2}}\times 680\times 10^{- 9}

w = 0.109 mm

7 0
4 years ago
Physical therapists know as you soak tired muscles in a hot tub, the water will cool down as you heat up. If a 67.9 person at 37
olga55 [171]
Q1=Q2
m1c1(t-t1)=m2c2(t2-t)
67.9kg * c1* (38.7°C-37.1°C)=50.2kg * 4186 J/kg°C * (40.5°C-38.7°C)
67.9kg* c1 * 1.6°C = 50.2kg * 4186 J/kg°C * 1.8°C
108.64 kg°C * c1 = 378246.96 J
c1 = 378246.96J /108.64kg°C
c1=3481.65 J/kg°C
4 0
3 years ago
HELP ASAPP!!!! MARKING BRAINLIEST!
Morgarella [4.7K]

Answer:

Humans inherit 23 chromosomes from each parent

Explanation:

8 0
3 years ago
Read 2 more answers
The rotor blades of a helicopter on the ground with its engine idling rotate at 48rpm. If it takes 12s for the pilot to increase
zhannawk [14.2K]

Answer:

α = 3.59 [rad/s^2]]

Explanation:

First we have to convert the values of the angular speeds of revolutions per minute to radians on second.

wf = final angular velocity = 460 [rpm]

wo= initial velocity = 48 [rpm]

t = time = 12 [s]

460 [\frac{rev}{min}]* [\frac{2\pi rad}{1 rev} ]*[\frac{1min}{60seg} ] = 48.17[\frac{rad}{s} ]\\48 [\frac{rev}{min}]* [\frac{2\pi rad}{1 rev} ]*[\frac{1min}{60seg} ] = 5.02[\frac{rad}{s} ]

now we can replace, in the following equation:

w_{o}=w_{i} +\alpha  *t\\where:\\\alpha = angular acceleration [rad/s^{2}]\\replacing:\\\alpha =\frac{w_{o} -w_{i} }{t} \\\alpha alpha =\frac{48.17-5.02 }{12} \\\alpha =3.59[\frac{rad}{s^{2} } ]

5 0
3 years ago
You have devised an experiment to measure the kinetic coefficient of friction between a ramp and block. You place the block on t
lys-0071 [83]

Answer:

 μ = 0.423

Explanation:

To solve this exercise we must use Newton's second law and kinematics together, let's start using expressions of kinematics to find the acceleration of the body

Let's fix a reference system where the x axis is parallel to the inclined plane, but the acceleration is only on this axis

            x = v₀ t + ½ a t²

The body starts from rest so its initial speed is zero

            a = 2 x / t²

            a = 2 0.5 /0.5²

            a = 4 m / s²

Taking the acceleration of the body, we use Newton's second law, we take the direction up the plane as positive

  X axis

                fr - Wₓ = m a          (1)

  Y Axis  

               N- W_{y} = 0

                N = W_{y}

We use trigonometry to find the components of the weight

            sin 45 = Wₓ / W

           cos 45 = W_{y} / W

           Wₓ = W sin 45

           W_{y} = W cos 45

The out of touch has the expression

             fr = μ N

             fr = μ W_{y}

We substitute in 1

             μ mg cos 45 - mg sin 45 = m a

             W_{y} = (a + g sin 45) / g cos 45

              μ = a / g cos 45 + 1

We calculate

Acceleration goes down the plane, so it is negative

           a = -4 m / s²

            μ = 1- 4 / (9.8 cos 45)

            μ = 0.423

7 0
3 years ago
Read 2 more answers
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