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djyliett [7]
1 year ago
12

You have been abducted by aliens and find yourself on a strange planet. Fortunately, you have a meter stick with you. You observ

e that, if the stick oscillates around one end as a physical pendulum, it has a period of 0.85 s. What is the acceleration of gravity on this planet?
Physics
2 answers:
defon1 year ago
0 0

Answer:

36.4276\ m/s^2

Explanation:

m = Mass of stick

L = Length of stick = 1 m

h = Center of mass of stick = \dfrac{1}{2}=0.5\ m

g = Acceleration due to gravity

T = Time period = 0.85 s

Time period is given by

T=2\pi\sqrt{\dfrac{I}{mgh}}

Moment of inertia is given by

I=m\dfrac{L^2}{3}

T=2\pi\sqrt{\dfrac{m\dfrac{L^2}{3}}{mgh}}\\\Rightarrow T=2\pi\sqrt{\dfrac{L^2}{3gh}}\\\Rightarrow g=\dfrac{4\pi^2L^2}{3hT^2}\\\Rightarrow g=\dfrac{4\pi^2\times 1^2}{3\times 0.5\times 0.85^2}\\\Rightarrow g=36.4276\ m/s^2

The acceleration of gravity on this planet is 36.4276\ m/s^2

Bond [772]1 year ago
0 0

Answer:

54.6 m/s^2

Explanation:

length of the pendulum, l = 1 m

time period of the pendulum, T = 0.85 s

According to the formula of time period of pendulum

T = 2\pi \sqrt{\frac{l}{g}}

g = \frac{4\pi ^{2}l}{T^{2}}

g = \frac{4\times 3.14\times 3.14\times 1}{0.85\times 0.85}

g = 54.6 m/s^2

Thus, the acceleration due to gravity at this planet is 54.6 m/s^2.

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A 1500 W radiant heater is constructed to operate at 115 V. (a) What will be the current in the heater? (b) What is the resistan
OlgaM077 [116]

Answer:

a) I = 13.04 A

b)  R = 8.82 ohms

c) 1291.87 kilocalories are generated an hour.

Explanation:

let P be the power of the heater, V be the voltage of the heater, I be the current of the heater, R be the resistance.

a) we know that:

P = I×V

I = P/V

  = (1500)/(115)

  = 13.04 A

Therefore, the current of the heater is 13.04 A

b) we now have voltage and current, according to Ohm's law:

R = V/I

  = (115)/(13.04)

  = 8.82 ohms

Therefore, the resistance of the heating coil is 8.82 ohms.

c) the number of kilocalories generated in one hour by the heater is just the energy the heater produces in one hour which is given by:

E = P×t

  = (1500)(1×60×60)

  = 5400000 J

since 1 calorie = 4.81 J

1 kilocalorie = 0.001 calories

E = 5400000/4.18 ≈ 1291866.029 calories ≈1291.87 kilocalories

Therefore, 1291.87 kilocalories are produced/generated in one hour.

8 0
2 years ago
Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -2.00 nC and is at x = 5.00 c
tatuchka [14]

Answer:

q₁= +0.5nC

Explanation:

Theory of electrical forces

Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

To solve this problem we apply Coulomb's law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

o solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F=K*q₁*q₂/d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)  

d: distance between the charges in meters

Data:

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Data

q₃=+5.00 nC =+5* 10⁻⁹ C

q₂= -2.00 nC =-2* 10⁻⁹ C

d₂= 5.00 cm= 5*10⁻² m

d₁= 2.50 cm=  2.5*10⁻² m

k = 8.99*10⁹ N*m²/C²

Calculation of magnitude and sign of q1

Fn₃=0 : net force on q3 equals zero

F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.

F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.

We propose the algebraic sum of the forces on q₃

F₂₃ - F₁₃=0

\frac{k*q_{2} *q_{3} }{d_{2}^{2}  } -\frac{k*q_{1} *q_{3} }{d_{1}^{2}  }=0

We eliminate k*q₃ of the equation

\frac{q_{1} }{d_{1}^{2}  } = \frac{q_{2} }{d_{2}^{2}  }

q_{1} =\frac{q_{2} *d_{1} ^{2} }{d_{2}^{2}  }

q_{1} =\frac{2*10^{-9}*2.5^{2}*10^{-4}   }{5^{2}*10^{-4}  }

q₁= +0.5*10⁻⁹ C

q₁= +0.5nC

4 0
1 year ago
A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "
blsea [12.9K]

Answer:

a) Q = C*emf

b)  Reduction in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are connected in series with the battery

Using Kirchoff's voltage law, sum of all voltages in the circuit is zero

Let V_{R} = Voltage dropped across the Resistor

V_{c} = Voltage dropped across the capacitor

Applying KVL;

emf - V_{R}  - V_{c} = 0\\.........................(1)

Since the connection is in series, the same current flow through the circuit

V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C

Putting V_{c} and V_{R} into equation (1)

emf - IR - Q/C = 0

At the final charge, the capacitor in fully charged, and current drops to zero due to equilibrium

I = 0A\\emf = Q/C\\Q = C* emf

b) Current starts running through the plate because as the sheet of plastic is inserted between the plates both the electric field intensity and the electric potential reduces. The charge also reduces, then current flows

c) The current through the resistor is the current through the entire circuit ( series connection)

I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\

Putting the values of t and I₀ into the formula for I written above

I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}

d) NB: The initial charge on the capacitor = C * emf

The final charge will be:

Q = K* Q_{initial} \\Q_{initial}  = C *emf\\Q_{final}  = KCemf

4 0
2 years ago
A migrating salmon heads in the direction n 45° e, swimming at 2 mi/h relative to the water. the prevailing ocean currents flow
Sophie [7]
Calculate for the x and y-components of the velocities involved in this item.

 2 mi/h (45° east)
  x-component = (2 mi/h)(sin 45°)
                     = 1.41 mi/h
  y-component = (2 mil/h)(cos 45°)
                     = 1.41 mi/h

4 mi/h (east)
  x-component = 4 mi/h
  y-component = 0 mi/h

Adding up the corresponding components:
     x-component = 5.4142 mi/h

     y-component = 1.4142 mi/h

Calculating for the resultant,
        R = sqrt ((x²) + (y²))
   
        R = sqrt ((5.4142 mi/h)² + (1.4142 mi/h)²)
             R = 5.60 mi/h

Answer: 5.6 mi/h
5 0
2 years ago
A sharpening wheel is traveling at 5 rad/s, it slows down to rest in 30 seconds while sharpening an axe. What is its angular acc
Ratling [72]

Answer:

Angular acceleration = 0.167 rad/s^2

Explanation:

Given

Initial Angular velocity (w1) = 5 rad/s

Final Angular velocity (w2) = 0 rad/s

Time taken to change velocity from w1 to w2 = 30 seconds

Angular acceleration is equal to the change in angular velocity to the time taken for making thing change

Hence, Angular acceleration

\frac{w_2 -w_1}{t} \\\frac{5-0}{30}\\0.167rad/s^2

3 0
1 year ago
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