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3 years ago

Help plssssss asap:(((

Physics

2 answers:

horrorfan [7]3 years ago

7 0

Answer:

5 blocks south

Explanation:

polet [3.4K]3 years ago

5 0

Answer:

5 blocks south

Explanation:

She ran 8 block south, then 3 blocks north.

8 - 3 = 5.

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Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation d

Schach [20]

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

$h_{fg}$ = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

$T_{L}$ = -18.77°C = 254.23 K

At saturation pressure 800 kPa, the saturation temperature is

$T_{H}$ = 31.31°C = 304.31 K

Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.

Thus, $q_{reject}$ = $h_{fg}$ = 171.82 kJ/kg

We know COP of heat pump

COP = $\frac{T_{H}}{T_{H}-T_{L}}$

= $\frac{304.31}{304.31-254.23}$

= 6.076

Therefore, Work out put, W = $\frac{q_{reject}}{COP}$

= 171.82 / 6.076

= 28.27 kJ/kg

8 0

3 years ago

Do solar panels create enough energy to power the car throughout the day?

JulijaS [17]

Answer:

Yes, but only if it's sunny.

Explanation:

As you know, solar panels generate energy through the sun's rays of light (better known as sunlight). Therefore, as long as the sun is shining high in the sky, the car will generate electricity and be able to function. If this vehicle was only powered by solar panels, it would not function during the night, in cloudy areas, and/or in dark places (such as parking garages or home garages).

Hope this helps!

5 0

2 years ago

If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo

lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

temperature at t = 0

$T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}$

T(0) = 95°C

temperature after half hour of cooling

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

t = 30 minutes

$T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}$

$T( 30 ) = 20 + 75 \times 0.5488$

T(30) = 61.16° C

average of first half hour will be equal to

$T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt$

$T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30$

$T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30$

$T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]$

$T = \dfrac{1}{30}[600 - 2058.04 + 3750]$

T = 76.39°C

4 0

3 years ago

A clay vase on a potter's wheel experiences an angular acceleration of 5.69 rad/s2 due to the application of a 16.0-n m net torq

Digiron [165]

The equivalent of the Newton's second law for rotational motions is:
$\tau = I \alpha$
where
$\tau$ is the net torque acting on the object
$I$ is its moment of inertia
$\alpha$ is the angular acceleration of the object.

Re-arranging the formula, we get
$I= \frac{\tau}{\alpha}$
and since we know the net torque acting on the (vase+potter's wheel) system, $\tau=16.0 Nm$ , and its angular acceleration, $\alpha = 5.69 rad/s^2$ , we can calculate the moment of inertia of the system:
$I= \frac{\tau}{\alpha}= \frac{16.0 Nm}{5.69 rad/s^2} =2.81 kg m^2$

8 0

3 years ago

A truck is traveling at 27 m/s down the interstate highway where you are changing a flat tire. frequency of 185 Hz.

Reil [10]

Answer:

(a) the observed frequency is 200 Hz

(b) the observed frequency is 188 Hz.

Explanation:

speed of the truck, Vs = 27 m/s

frequency of the truck as it approaches, Fs = 185 Hz

(a) Apply Doppler effect to determine the frequency you will hear.

As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.

$F_s = F_o [\frac{V}{V_S + V} ]$

Where;

Fo is the observed frequency which is the frequency you will hear.

V is speed of sound in air

$F_s = F_o [\frac{V}{V_S + V} ]\\\\185 = F_o [\frac{340}{27 + 340} ]\\\\185 = F_o (0.926)\\\\F_o = \frac{185}{0.926}\\\\F_o = 199.78 \ Hz$

$F_o = 200 \ Hz$

(b) Apply the following formula for a moving observer and a moving source;

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )$

The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz$

5 0

3 years ago