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3 years ago

11

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular

acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?

1 answer:

RUDIKE [14]3 years ago

6 0

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, $\omega _i$ = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

$\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad$

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

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How is momentum conserved is a Newton's cradle when one steel ball hits the other

Vikentia [17]

Answer:

Newton's Cradle experiment perfectly demonstrates the law of conservation of momentum which states that in a closed system, momentum before the collision is equal to momentum after the collision of the system.

As the first ball swings in the air, it gains momentum. When it strikes the second ball, it loses momentum and second ball gains equal amount of momentum. The second ball transfers the momentum to third, then fourth and till the last. The last ball when gains the same momentum swings up in the air. This continues. This experiment is done in drag free condition. This means there is no loss of momentum or opposing forces present.

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3 years ago

Which one of the following statements concerning emf is true?

Andrej [43]

Answer:

C) Emf is the force exerted on a single charge to move it from one terminal to the other of an emf device.

Explanation:

Electromotive force, abbreviation E or emf, electrical energy per unit supplied from an energy source, such as an electrical generator or a battery. In the engine or battery, electricity is transformed from one source to another as the device works on the electrical charge that is passed within itself.

In physics it is defined as Work done by a cell or battery in moving a unit charge from its one terminal to the other, against the potential.

Hence C is correct

3 0

3 years ago

1. Calcular la masa de mercurio que pasó de 30 °C hasta 120 °C y absorbió 4400 cal. Calor específico del

timofeeve [1]

Answer:

Masa, m = 0.088 kg

Explanation:

Given the following data;

Temperatura inicial = 30°C

Temperatura final = 120°C

Capacidad calorífica específica = 138J/kg.K

Calor absorbido, Q = 4400 cal.

Para encontrar la masa;

La capacidad calorífica viene dada por la fórmula;

$Q = mct$

Dónde;

Q representa la capacidad calorífica o la cantidad de calor.

m representa la masa de un objeto.

c representa la capacidad calorífica específica del agua.

dt representa el cambio de temperatura.

dt = T2 - T1

dt = 120 - 30

dt = 90°C to kelvin = 273 + 90 = 363K

Sustituyendo en la fórmula, tenemos;

$4400 = m*138*363$

$4400 = 50094m$

$m = \frac {4400}{50094}$

Masa, m = 0.088 kg

7 0

3 years ago

Question 1(Multiple Choice Worth 4 points)

ArbitrLikvidat [17]

Question 1: Amount of energy
Question 2: Freezing ice
Question 3: Chemical change
Question 4: More energy is given off by the new bonds formed than is needed to break the original bonds.
Question 5: The resulting mass would be 35 grams.

4 0

3 years ago

. A block of mass 4.0 kg on a spring has displacement as a function of time given by, x (t) = (1.0 cm) cos [(2.0 rad/s) t + 0.25

svlad2 [7]

Answer and Explanation:

1. Evaluate the function x(t) at t=0.5

$x(t)=1cos(2t+0.25)\\x(0.5)=cos(2*0.5+0.25)=cos(1+0.25)=cos(1.25)=0.3153223624\approx0.31cm$

2. The period of motion T can be calculated as:

$T=\frac{2\pi}{\omega}$

Where:

$\omega=2rad/s$

So:

$T=\frac{2\pi}{2}=\pi\approx3.14s$

3. The angular frequency can be expressed as:

$\omega=\sqrt{\frac{k}{m} }$

Solving for k:

$k=(\omega)^2*m=(2)^2*4=4*4=16\frac{N}{m}$

4. Find the derivate of x(t):

$\frac{dx}{dt} =v(t)=-2sin(2t+0.25)$

Now, the sine function reach its maximum value at π/2 so:

$2t+025=\frac{\pi}{2}$

Solving for t:

$t=\frac{\frac{\pi}{2} -0.25}{2} =0.6603981634s$

Evaluating v(t) for 0.6603981634:

$v(0.6603981634)=-2sin(2*0.6603981634+0.25)=-2sin(\frac{\pi}{2} )=-2*1=2$

So the maximum speed of the block is:

$v(0.6603981634)=2cm/s=0.02m/s$

In the negative direction of x-axis

5. The force is given by:

$F=kx$

The cosine function reach its maximum value at 2π so:

$2t+0.25=2\pi$

Solving for t:

$t=\frac{2\pi-0.25}{2} =3.016592654s$

Evaluating x(t) for 3.016592654:

$x(3.016592654)=cos(2*3.016592654+0.25)=cos(2\pi)=1cm=0.01m$

Therefore the the maximum force on the block is:

$F=16*0.01=0.16N$

6 0

3 years ago