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2 years ago

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an object having a core temperature of 1700 is removed from a furnace and placed in an environment having a constant temperature

of 80. one hour after being removed, the core temperature is 1300. how many hours will it take for the core temperature of the object to drop to 850

1 answer:

IRINA_888 [86]2 years ago

4 0

Answer: It will be take 2.6 hours

Explanation: Please see the attachments below

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A system delivers 1275 j of heat while the surroundings perform 855 j of work on it. calculate ∆esys in j.

kakasveta [241]

The first law of thermodynamics says that the variation of internal energy of a system is given by:
$\Delta U = Q + W$
where Q is the heat delivered by the system, while W is the work done on the system.

We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system

So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J

So, the variation of internal energy of the system is
$\Delta U = -1275 J+855 J=-420 J$

6 0

2 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

2 years ago

You are driving downhill on a rural road with a 3% grade at a speed of 45 mph. While playing on the side of the road, a child ac

Dennis_Churaev [7]

Answer: a) 95.07m b) 81.88 m

Explanation:

a)

For finding the distance when vehicle is going downhill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31-0.03)}$

Stop sight distance= 95.07 m

b)

For finding the distance when vehicle is going uphill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31+0.03)}$

Stop sight distance= 81.88 m

5 0

2 years ago

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅ = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0

2 years ago

The velocity of a 10.0 kg object that has 720 J of kinetic energy is m/s. (Report the answer to two significant figures.)

nikdorinn [45]

Kinetic energy<span> is the </span>energy<span> of motion. An object that has motion - whether it is vertical or horizontal motion - has </span>kinetic energy<span>. It is expressed as:

KE = mv^2 /2

720 = 10.0v^2 /2
v = 12 m/s

Hope this answers the question. Have a nice day.</span>

4 0

2 years ago

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