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4 years ago

If for every action force an equal and opposite reaction force exists, how can anything ever be accelerated?

1 answer:

8 0

If there are two equal and opposite forces on the SAME THING, then the thing doesn't accelerate. You're right about that. But the action and reaction forces act on two different things. The bullet and the rifle. The ball and the bat. The airplane and the air. etc. So BOTH can accelerate.

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What is energy? it's from my physics book

s344n2d4d5 [400]

Energy is the capacity of doing work

4 0

3 years ago

Read 2 more answers

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

$\Sigma F=ma_{c}=m\frac{v^{2}}{R}$

Where:

m is the mass of the car
v is the speed
R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

$F_{f}=m\frac{v^{2}}{R}$

$F_{f}=2100\frac{18^{2}}{83}$

$F_{f}=8197.60 \: N$

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0

3 years ago

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ν0. find the minimum en

shusha [124]

The energy carried by the incident light is
$E=hf$
where h is the Planck constant and f is the frequency of the light. The threshold frequency is the frequency that corresponds to the minimum energy needed to eject the electrons from the metal, so if we substitute the threshold frequency in the formula, we get the minimum energy the light must have to eject the electrons:
$E=hf=(6.63 \cdot 10^{-34}Js)(5.64 \cdot 10^{14}s^{-1})=3.74 \cdot 10^{-19}J$

4 0

4 years ago

We need to find the launch velocity of our new marble launcher. we know that it will launch a 25g marble to a distance of 73 cm,

White raven [17]

The launch velocity of the marble launcher is 34.65 m/s

Given that the launch velocity of marble launcher, launches a 25g marble to a distance of 73 cm (0.73 m) and the marble roll up to 6.2 meters before stopping. The launch height is 20 cm (0.2 m).

The time for landing can be calculated by the second equation of motion formula:

h = ut + $\frac{1}{2}$ g $t^{2}$

Let u = 0

0.2 = 0×t + $\frac{1}{2}$ × 9.8 × $t^{2}$

$t^{2}$ = $\frac{0.2}{4.9}$

$t^{2}$ = 0.04

t = 0.2s

Now, the launch velocity of the marble launcher can be calculated by:

Speed = Distance / Time

Speed = $\frac{0.73 + 6.3}{0.2}$

Speed = $\frac{6.93}{0.2}$

Speed = 34.65 m/s

Therefore, the launch velocity of the marble launcher is 34.65 m/s

Know more about Launch velocity: -brainly.com/question/18883779

#SPJ9

3 0

1 year ago

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