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Firdavs [7]
2 years ago
6

HELP ASAP PLEASE!!!

Physics
2 answers:
Montano1993 [528]2 years ago
7 0

Answer:

The answer is A, B, C, D

Explanation:

This is because gravity is the weakest force of the four fundamental forces, so it automatically cancels letter E

Vesnalui [34]2 years ago
6 0
It’s A,B,C,D
hope this helps
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Everyday high and low tides on earth will happen daily about ______________. *
Murrr4er [49]
I think every 6 hours would be correct
7 0
2 years ago
A metal sphere of radius 2.0 cm carries an excess charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sph
Nonamiya [84]

Answer:

The electric field is  E = 7.5 *10^{6} \ N/C

Explanation:

From the question we are told that

    The radius of the metal sphere is  R = 2.0 \ cm  =  0.02 \ m

     The excess charge which the metal sphere carries is  q =  3.0 \mu C  =  3.0*10^{-6} \ C

      The distance of the position being to the center is D = 6.0 \ cm  = 0.06 \ m

       The coulomb constant is   k =9*10^{9} \  N \cdot m^2 /C^2

Generally the electric field is mathematically represented as  

        E = \frac{k *  q}{D^2}

substituting values

        E = \frac{9*10^{9} *  30.*10^{-6}}{(0.06)^2}

      E = 7.5 *10^{6} \ N/C

5 0
2 years ago
A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in
Troyanec [42]

Answer:

The value is  \epsilon =  3.84 *10^{-5} \  V

Explanation:

From the question we are told that

  The diameter of the ring is  d =  18 \ mm  =  0.018 \  m

   The length of the solenoid is l = 25 \ cm  =  0.25 \ m

   The diameter of the solenoid is  D = 5.0 \ cm  = 0.05 \ m

    The number of turns is  N = 1500

   The change in  current in the solenoid is   \Delta  I   = 20 \ A

   The time taken is  \Delta  t  = 1 \ s

Generally the radius of the ring is  

     r = \frac{d}{2}

=>  r = \frac{0.018 }{2}

=>  r = 0.009 \ m

Generally the area of the ring is mathematically represented as  

      A = \pi r^2

=>   A = 3.142 *  0.009^2    

=>   A = 2.545 *10^{-4}\ m^2

Generally the induced emf is mathematically represented as

       \epsilon  =  A * \frac{dB}{dt}

Here    

         \frac{dB }{dt} =  \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}

Here  \mu_o is the permeability of free space with value  

         \mu_o =  4\pi *10^{-7} \ N/A^2

So  

     \frac{dB }{dt} =   4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}

=>  \frac{dB }{dt} =   0.150816\  T/s

So

     \epsilon =   0.150816 *  2.545 *10^{-4}

=>   \epsilon =  3.84 *10^{-5} \  V

3 0
3 years ago
A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi
Mamont248 [21]

Answer:

a=368.97\ m/s^2

Explanation:

Given that,

Initial angular velocity, \omega=0

Acceleration of the wheel, \alpha =7\ rad/s^2

Rotation, \theta=14\ rotation=14\times 2\pi =87.96\ rad

Let t is the time. Using second equation of kinematics can be calculated using time.

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s

Let \omega_f is the final angular velocity and a is the radial component of acceleration.

\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s

Radial component of acceleration,

a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2

So, the required acceleration on the edge of the wheel is 368.97\ m/s^2.

6 0
3 years ago
Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains
OLEGan [10]

Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

C = 252000 J/326.7 kg°C

<u>C = 771.35 J/kg°C</u>

5 0
3 years ago
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