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3 years ago

In a circuit, the supplied voltage is 1.5 volts, and the resistor value is 3 ohms. What is the current in the circuit?

1 answer:

6 0

Using Ohm's Law:

V = IR

Where V = Voltage in Volts, I = Current in Ampere, R = Resistance in Ohms

V = IR

1.5 = I * 3

1.5 = 3I

3*I = 1.5

I = 1.5/3

I = 0.50 A

Current in the Circuit is 0.50 Ampere.

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Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.

andreev551 [17]

Answer:

the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = $m_A$

then, the mass of object B = $m_B = \frac{m_A}{4}$

work done on object A = -18 J

work done on object B = -18 J

let $v_i$ be the initial speed

let $v_f$ be the final speed

For object A;

$K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\$

$v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\$

Thus, the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

To obtain the change in the final speed of object B, apply the following equations.

$K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B = \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\$

$\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i$

Thus, the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

3 0

2 years ago

A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is

pychu [463]

Answer:

The final velocity of the runner at the end of the given time is 2.7 m/s.

Explanation:

Given;

initial velocity of the runner, u = 1.1 m/s

constant acceleration, a = 0.8 m/s²

time of motion, t = 2.0 s

The velocity of the runner at the end of the given time is calculate as;

$v = u + at$

where;

v is the final velocity of the runner at the end of the given time;

v = 1.1 + (0.8)(2)

v = 2.7 m/s

Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.

7 0

2 years ago

A ball starts from rest and undergoes uniform acceleration of 2.50m/s^2. What is the velocity of the ball 4s later?

erica [24]

Explanation:

Given:

v₀ = 0 m/s

a = 2.50 m/s²

t = 4 s

Find: v

v = at + v₀

v = (2.50 m/s²) (4 s) + 0 m/s

v = 10 m/s

4 0

2 years ago

Read 2 more answers

Is an iron rod a good conductor of heat and electricity

Afina-wow [57]

Yes It is all metal is a good heat conductor

4 0

3 years ago

Starting from rest, a particle that is confined to move along a straight line is accelerated at a rate of 5.0 m/s2. Which one of

Elanso [62]

Answer:

c) The slope is not constant and increases with increasing time.

Explanation:

The equation for the position of this particle (starting from rest is)

$s = at^2/2 = 5t^2/2 = 2.5t^2$

We can take derivative of this with respect to time t to get the equation of slope:

$s' = (2.5t^2)' = 2*2.5t = 5t$

As time t increase, the slope would increases with time as well.

6 0

3 years ago