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3 years ago

13

An electric circuit can have no current when a switch is closed. True False

1 answer:

igomit [66]3 years ago

4 0

Answer: False

Explanation: When a switch is closed the circuit is closed and thus the flow of current is on while when the switch is open, the circuit is open, and therefore there will be no flow of current in the circuit.

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A 2430 pound roller coaster starts from rest and is launched such that it crests a 105 ft high hill with a speed of 59 mph. The

Rudik [331]

Answer:

Explanation:

Given

Weight of roller coaster is $W=2430\ pound$

mass of roller coaster $m=\frac{W}{g}=\frac{2430}{32.2}=75.45$

Distance traveled by roller coaster $d=396\ ft$

drag force $f_d=85\ pounds$

velocity at top $v=59 mph\approx 86.53\ ft/s$

Suppose E is the initial energy

Conserving Energy at bottom and top

$E=\frac{1}{2}mv^2+mgh+f_d\cdot d$

$E=0.5\times 75.45\times 86.53^2+2430\times 105+85\times 396$

$E=2.9\times 10^5\ foot-pound$

5 0

3 years ago

During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

$dQ=0$

$dU=-dW$

We need to calculate the number of moles and specific heat

Using formula of heat

$dU=nC_{v}dT$

$nC_{v}=\dfrac{dU}{dT}$

Put the value into the formula

$nC_{v}=\dfrac{-100}{5}$

$nC_{v}=-20\ J/K$

We need to calculate the heat

Using formula of heat

$dQ=nC_{v}(dT_{1})+dW_{1}$

Put the value into the formula

$dQ=-20\times5+25$

$dQ=-75\ J$

We need to calculate the heat capacity for the second process

Using formula of heat

$dQ=nC_{v}(dT_{1})$

Put the value into the formula

$-75=nC_{v}\times(-5)$

$nC_{v}=\dfrac{-75}{-5}$

$nC_{v}=15\ J/K$

Hence, The heat capacity for the second process is 15 J/K.

5 0

3 years ago

A 1,120 kg car is travelling with a speed of 40 m/s. Find its energy.

serious [3.7K]

Answer:

this vehichle has 896,000 jules of energy

Explanation: KE=1/2mv squared, KE is Kinetic energy. m is mass and v is velocity

8 0

2 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

2 years ago

What is an equilibrant?

Stolb23 [73]

I’m pretty sure it is an object with a net force of zero. All forces are balanced and EQUAL

4 0

3 years ago