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xxTIMURxx [149]
3 years ago
10

Is the battery sends a current of 10 A through a circuit for one hour, how many coulombs will flow through the circuit?

Physics
1 answer:
garik1379 [7]3 years ago
7 0

1 Ampere = 1 coulomb/second

1 hour = 3,600 seconds

1 coulomb/second = 3,600 coulomb/hour

10 Amperes x 1 hour = 36,000 coulomb/hour
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Two electric charges are moved so that they are twice as far apart as they had originally been. Is the force they experience fro
Natasha_Volkova [10]

Answer:

No.

Explanation:

The force that two particle experience is inversely proportional to the sqare of the distance, this is:

F \  \alpha \  \frac{1}{D^{2}} for a distance D

If we move them so that D is doubled:

\frac{1}{2^{2}.D^{2}  }= \frac{1}{4} \eq  \frac{1}{.D^{2}  } \eq

Then the force they experience is one fourth of the original.

5 0
3 years ago
A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it
Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

L = length of the meter stick = 1 m

m = mass of the meter stick

w = angular speed of the meter stick as it hits the floor

v = speed of the other end of the stick

we know that, linear speed and angular speed are related as

v = r w\\w = \frac{v}{r}

h = height of center of mass of meter stick above the floor = \frac{L}{2} = \frac{1}{2} = 0.5 m

I = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

I = \frac{mL^{2} }{3}

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}

7 0
3 years ago
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit
lapo4ka [179]

Answer:

The  angle is  \theta  =  15.48^o

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  d  =  3.66  \ m

     The time taken is  t =  0.455 \ s

   

The  horizontal component of the speed of the dart is mathematically represented as

      u_x =  ucos \theta

where u is the the velocity at dart is lunched

  so

      distance =  velocity \ in \ the\  x-direction  *  time

substituting values

      3.66 =   ucos  \theta *  (0.455)

 =>   ucos \theta =  8.04  \ m/s

From projectile kinematics the time taken by the dart can be mathematically represented as

         t  =  \frac{2usin \theta }{g}

=>    usin \theta =  \frac{g  * t}{2 }

       usin \theta =  \frac{9.8  * 0.455}{2 }

      usin \theta = 2.23

=>   tan \theta =  \frac{usin\theta }{ucos \theta }  =  \frac{2.23}{8.04}

       \theta  =  tan^{-1} [0.277]

      \theta  =  15.48^o

     

4 0
3 years ago
Belly-flop Bernie dives from atop a tall flagpole into a swimming pool below. His potential energy at the top is 7000 J (relativ
elena55 [62]

Answer:

KE₂ = 6000 J

Explanation:

Given that

Potential energy at top U₁= 7000 J

Potential energy at bottom U₂= 1000 J

The kinetic energy at top ,KE₁= 0 J

Lets take kinetic energy at bottom level =  KE₂

Now from energy conservation

U₁+ KE₁= U₂+ KE₂

Now by putting the values

U₁+ KE₁= U₂+ KE₂

7000+ 0 = 1000+ KE₂

KE₂ = 7000 - 1000 J

KE₂ = 6000 J

Therefore the kinetic energy at bottom is 6000 J.

5 0
3 years ago
Which of the following is a small star that has reached the end of its stellar evolution?
Akimi4 [234]
D.) White Dwarf

It is the smallest star whose mass is approximately equal or greater than 1.4M
Here, M = mass of the Sun.

Hope this helps!
3 0
3 years ago
Read 2 more answers
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