3. Which resistor experiences a larger voltage drop across it?

If a ball is launched horizontally at 40 m/s

vaieri [72.5K]

Answer:

40m/s

Explanation:

The horizontal component of velocity remains constant because there are no external forces in that direction

By applying motion equations, V= U+ at

where ,

v - final velocity
u - initial velocity
a-acceleration
t - time

v = u +at

As no force act on the ball ( we neglect air resistance here) no acceleration is seen,

So v = u = 40m/s

4 0

3 years ago

What is the speed of an 800 kg automobile if it has a kinetic energy of 9.00 x 10^J?

MA_775_DIABLO [31]

Ek = 1/2 mv^2

9 × 10^4 = 1/2 × 800 × v^2

9 × 10^4/400 = 400 v^2 / 400

9 × 10^4/400 = v^2

√225 = v

15 ms⁻¹ = v

That's the only way I know how to work it out

I think in this case velocity and speed would be considered the same because me

s = d/t and v=d/t

one is distance travelled and the other is displacement of a body

7 0

2 years ago

Imagine you have just witnessed a small avalanche on a mountain while skiing, and two slushy snowballs just crashed together in

inn [45]

We have that the momentum p is given by the formula p=mv where m is the mass and v is the velocity. Since for A p=-14kgm/s and m=7, we have that the velocity is -14/7=-2m/s. Hence its speed is 2 m/s.
For b we have that p=15kgm/s and v=3m/s. Because m=p/v, we have m=3kg.
We also have that the momentum is conserved in this system. Hence, the net sum of the momentum of the 2 snowballs equals the momentum of the single giant ball. Hence, p(total)=p(combined)=-14+15=1kgm/s (momentum is a vector; the positive sign means that it tends to the positive direction).

6 0

4 years ago

Read 2 more answers

This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broa

Montano1993 [528]

Answer:

$2.77287\times 10^{15}\ m$

Explanation:

P = Power = 50 kW

n = Number of photons per second

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

$\nu$ = Frequency = 781 kHz

r = Distance at which the photon intensity is i = 1 photon/m²

Power is given by

$P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s$

Photon intensity is given by

$i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m$

The distance is $2.77287\times 10^{15}\ m$

3 0

3 years ago

200-grams of computer chips with a specific heat of 0.3 kJ/kg·K are initially at 25°C. These chips are cooled by placement in 0.

balu736 [363]

Answer:

a. -0.01324 kJ/K, b. = 0.03233 kJ/K , c. = 0.01909, Yes the process is possible

Explanation:

Heat transfer will occur between the chip and the surrounding fluid. Then, finally they will attain a common equilibrium temperature and heat transfer will stop. Now, if we assume that, after heat transfer, chip will attain the temperature of fluid, that is, -34 C,, So , to check whether this is possible

Amount of energy lost by the chip = m . c . (T(i) - T(f))

= 0.2 x 0.3 (25 + 34) = 3.54 KJ

Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

Note that heat transfer will change the internal energy of the fluid. Do not consider enthalpy change, as this is not a problem involving fluid flow in and out of the system

M[(μf+ x . μfg) - μf] = m(xμfg)

Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)

3.54 = 0.1 . X x 203.29

x = 0.1741, which is the dryness fraction of fluid at the final state.

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

0.1 x μfg= 20.329 kJ, which is much greater than 3.45 kJ, therefore, it is certain that the state of fluid will be at -34 C only and at the saturation pressure of 69.56 KPa. So the chip will come to attain the temperature of -34 C.

a. Write the equation for the change of entropy in the chips

ΔSchips = mchips . c . ln(T2/T1), where mc is the mass of chips, c is the specific heat of chips, T2 is the temperature at state 2 and T1 is the temperature at state 1

Substitute mc = 0.2 kg, c = 0.3kJ/kg.K, T1 = 25 + 273, T2 = -34 + 273

delSchips = 0.2 x 0.3 x ln [(-34+273)/ (25+273)]

= -0.01324 kJ/K

There fore the change in entropy of the chips is -0.01324 kJ/K

b. Entropy change of fluid R- 134a

ΔS2 = m[Sfinal – S initial]

= m[Sf + x . Sfg - Sf]

= 0.2 x (0.1741 x 0.92859)

= 0.03233 kJ/K

c. Calculate the total change in the entropy of the entire system

delS = delSchips + delSR -134a

= -0.01324 + 0.03233

= 0.01909

Since the total change in entropy of the entire system is positive that exactly explains that the actual processes are happening in the direction of increase of entropy therefore, the process is possible.

6 0

4 years ago