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Annette [7]
2 years ago
9

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

Physics
1 answer:
ollegr [7]2 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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"The predictions of Einstein’s Theory of General Relativity were tested on a double pulsar system in January of 2004. His equati
Rasek [7]

Answer:

99.95%

Explanation:

A double pulsar system named PSR J0737-3039A/B  in Puppis constellation was discovered in the year 2003. Pulsars are second most densest object in the universe after black holes and they emit radio waves at regular intervals. This pair presented a great and natural setup to test the Theory of General Relativity presented by Einstein in 1915. In this theory Einstein had presented a set of equations on how the space-time fabric will be curved because of the very dense objects such as Neutron stars. It also predicted how the gravitational waves are created because of stars orbiting each other.

A team of astrophysicists led by Michael Kramer, conducted a study on how these gravitational waves will impact the time in which the radio waves emitted by pulsars will reach Earth. The result of the study proved the theory of General Relativity to be accurate up to 99.95%.

8 0
2 years ago
Floor lamps usually have a base with large inertia, while the long body and top have much less inertia. Part A If you want to sh
Novay_Z [31]

Answer:

When a an object is been rotated its resistance capacity to that rotational force is know as rotational inertia  and this mathematically given as

          I = mr^2

   Where m is the mass

                r is the rotation radius

For the spinning of the lamp as a baton to work the location of the center of mass of the floor lamp needs to be located

This is more likely to be located closer to base of the lamp as compared to the top, so success of spinning a floor lamp like a baton is highly likely if the lamp is grabbed closer to the base because that is where the position of its center of mass is likely to be.

Explanation:

5 0
2 years ago
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.55 x 10-4 T) at a distance of 25
antiseptic1488 [7]

we are given in the problem the following dimensions or specifications 
B = 0.000055 T r = 0.25 m constant mu0 = 4*pi*10-7 

The formula that is applicable from physics is 
B = mu0*I/(2*pi*r) I = 2*B*pi*r/mu0 I = 68.75 Amperes 
7 0
2 years ago
Read 2 more answers
Have you ever chewed on a wintergreen mint in front of a mirror in the dark? If you have, you may have noticed some sparks of li
lutik1710 [3]

Answer:

Part a)

E = 3.66 eV

Part b)

\lambda = 508.5 nm

Explanation:

Part a)

change in the energy due to decay of photon is given as

E = h\nu

here we know that

\nu = 8.88 \times 10^{14} Hz

now we have

E = (6.6 \times 10^{-34})(8.88 \times 10^{14})

E = 5.86 \times 10^{-19} J

E = 3.66 eV

Part b)

While electron return to its ground state it will emit a photon of energy 2/3rd of the total energy

so we have

\Delta E = \frac{2}{3}(3.66 eV)

\Delta E = 2.44 eV

now to find the wavelength we have

\Delta E = \frac{hc}{\lambda}

2.44 = \frac{1242}{\lambda}

\lambda = 508.5 nm

3 0
2 years ago
A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it
Degger [83]

Answer:

a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

d)\alpha_2=4.1454\,rad.s^{-2}

e)t_2=1.061\,s

Explanation:

Given that:

  • initial speed of turntable, N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

4 0
2 years ago
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