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Dmitrij [34]
3 years ago
5

They want a mixture of 150 grams of water at 60 degrees Celsius, using hot water at 90 degrees Celsius and cold water at 14 degr

ees Celsius, how many grams of water should be used per face?​
Physics
1 answer:
Gnesinka [82]3 years ago
6 0

Answer:

Explanation:

Here,

Final mass(m)=150g

Final Temperature(t)=60°C

Temperature of hot water(t1)=90C

Temperature of cold water(t2)=14C

Let the grams of water to be used per face be m1 grams and (150-m1) grams for hot and cold water respectively.

We know that,

Heat lost by hot body=Heat gained by cold body

or, m1 s1 Δt1=m2 s2 Δt2

or, m1 (t1-t)=(150-m1) (t-t2)  [∵s1=s2, SHC of water is taken to be same]

or, m1 (90-60)=(150-m1) (60-14)

or, 30m1=6900-46m1

or, 76m1=6900

∴m1=90.79g

(150-m1)=59.21g

Hence, 90.79g and 59.21 g hot and cold water must be used respectively.

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In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its sp
Slav-nsk [51]

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

so

m = 4000 / 2.20462 =  1814.37 kg

Initial velocity V_{i} = 60 mph = 26.8224 m/s

Final velocity V_{f} = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = \frac{1}{2}m(  V_{i}² - V_{f}² )

we substitute

Δk = \frac{1}{2}×1814.37( (26.8224)² - (13.4112)² )

Δk = \frac{1}{2} × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

so

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

4 0
3 years ago
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0
3 years ago
An object travels at 15 meters in 10 seconds east. What is it’s velocity?
Juli2301 [7.4K]
A new jxo rib is excellent is.
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If there was a small sphere freely suspended in plane poiseuille flow, how would you go about using linearity to prove that the
Blizzard [7]

Answer Explanation :

Poiseuille equation: this equation is used for non ideal flow this is used for the calculation of pressure in laminar flow it is physical law we know that fluid in laminar flow, flows across the pipe whose diameter is larger than the length of pipe

in mathematical form the equation can be expressed as

Q = \frac{(P_2-P_1) r^{4}\times \pi}{8\times\eta \times l }

where η is the cofficient of viscosity

now if we assume a small sphere of radius a is suspended freely in the plane of the laminar flow then for assuring that the sphere does not migrate with the flow we have to calculate the rate of flow of the liquid

8 0
3 years ago
If a 42kg rolling object slows from 11.5m/s to 3.33m/s how much work did friction do
geniusboy [140]

Answer: 1608.39 J

Explanation: Given that the

mass M = 42kg

U = 11.5m/s

V = 3.33m/s

how much work did friction do

Work done = Force × distance

Work done = Ma × distance

But acceleration a = V/t

Work done = M × V/t × d

Work done = M × V × d/t

Where d/t = velocity

Therefore,

Work done = M × U × V

Work done = 42 × 11.5 × 3.33

Work done = 1608.39 J

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