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3 years ago

Newton’s second law of motion can also be described as _____.

Physics

1 answer:

anzhelika [568]3 years ago

3 0

Answer:

F=ma

Explanation:

The Newton's laws of motion tries to explain the how bodies behave and the energy changes when theys are in motion. For the 3 of them to hold, the bodies must be moving in a straight line and with constant velocity.

The second one states that, "the change of momentum of a moving body is directly proportional to the force producing it and it takes place to the direction of force."

From the choices given, the appropriate answer is F=ma

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A spring stretches 1.68cm vertically when a 2.50kg object is suspended from it.Find the distance (in cm) the spring stretches if

Llana [10]

Answer:

2.96 cm

Explanation:

By Hook's law

Force(F) = Spring constant(k) × Extension(d)

F = k × d

Force is the weight of the object, F = W = mg

So we get, mg = kd ⇒ m ∝ d

2.5 ∝ 1.68 --------------(1)

4.4 ∝ d' --------------(2)

From (1) & (2), 4.4/2.5 = d'/1.68

d' = 2.96 cm ⇒ the required extension.

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3 years ago

An ideal parallel-plate capacitor consists of a set of two parallel plates of area Separated by a very small distance 푑. This ca

dolphi86 [110]

Answer:

doubled the initial value

Explanation:

Let the area of plates be A and the separation between them is d.

Let V be the potential difference of the battery.

The energy stored in the capacitor is given by

U = Q^2/2C ...(1)

Now the battery is disconnected, it means the charge is constant.

the separation between the plates is doubled.

The capacitance of the parallel plate capacitor is inversely proportional to the distance between the plates.

C' = C/2

the new energy stored

U' = Q^2 / 2C'

U' = Q^2/C = 2 U

The energy stored in the capacitor is doubled the initial amount.

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3 years ago

A father wanted to explain how the moon shines to his 5-year old child by comparing it to an object that the child uses. Which s

9966 [12]

Answer:

The moon is like a mirror. It reflects light produced by the Sun

Explanation:

5 0

2 years ago

Read 2 more answers

A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i

AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume $v$ and enthalpy $h$ as,

$h_1 = 95.96Btu/lb$ ( $h_1 = h_f$ at $2psia$ )

$v_1 = 0.016238ft^3/lb$ ( $v_1 = v_f$ at $2psia$ )

$w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb$

$w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb$

$h_3 = 1364.0Btu/lb$

$s_3 = 1.5073Btu/lb.R$

( at $P_3 = 1500psia$ & $T_3 = 800^0F$ )

$P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765$

( $S_f$ & $S_{fg}$ when pressure is 2psia)

$h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb$

$n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb$

Therefore,

$q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb$

To calculate the mass flow rate of steam in the cycle, we use the formula

$W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s$

where $1Kj = 0.947817 Btu$

The power output and the rate of heat addition are calculated thus,

$W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW$

$Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s$

The thermal efficiency of the cycle can be found thus;

$n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343$

= 34.3%

5 0

3 years ago

A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.600 rev

natulia [17]

Answer:

The final velocity $\omega_f = 0.4235 \frac{rev}{s}$

Explanation:

Given data

Mass of merry go round $M_m$ = 120 kg

Radius = 1.8 m

Initial angular velocity $\omega_i$ = 0.6 $\frac{rev}{sec}$

Mass of boy $M_{boy}$ = 25 kg

We know that the final velocity is given by

$\omega_f = \frac{\frac{1}{2}M_m \omega_i }{M_{boy} + \frac{1}{2} M_m }$

Put all the values in above formula we get

$\omega_f = \frac{\frac{1}{2}(120) 0.6}{25 + \frac{1}{2} (120) }$

$\omega_f = 0.4235 \frac{rev}{s}$

This is the final velocity.

3 0

3 years ago