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Fed [463]
3 years ago
15

In moving out of a dormitory at the end of the semester, a student does 1.20 x 104 J of work. In the process, his internal energ

y decreases by 5.46 x 104 J. Determine each of the following quantities (including the algebraic sign): (a)W, (b)U, and (c)Q.
Physics
1 answer:
Zielflug [23.3K]3 years ago
8 0

Answer:

(a) W=1.20×10⁴J

(b) U= -5.46×10⁴J

(c) Q= -4.26×10⁴J

Explanation:

Given that student does 1.20×10⁴J work

(a) W=1.20×10⁴J

Work done by student,so positive sign

During the process, his internal energy decreases by 5.46×10⁴J.

(b) U= -5.46×10⁴J.

As the Energy decreases therefore negative sign

For (c) Q

We know the formula

Q=W+U\\Q=1.2*10^{4}+(-5.46*10^{4} )\\ Q=-4.26*10^{4}J

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An airplane flies at airspeed (relative to the air) of 280 km/h . The pilot wishes to fly due North (relative to the ground) but
MrRa [10]

Answer:

the pilot should head the plane 7.547^{\circ} towarrds south- west

Solution:

The airspeed of the airplane, v = 280 km/h

The velocity of the wind, v' = 52 km/h South-west

Angle, \theta = 225^{\circ}

Now, measured angle in the clockwise direction from North:

sin225 = sin(\pi + 45) =  - sin 45^{\circ}

Now,

vsinx - v'sin45 = 0

280sinx = 52sin45

x = sin^{- 1}(\frac{52}{280}\times \frac{1}{\sqrt{2}})

x = 7.547^{\circ} south- west

4 0
3 years ago
PLEASE HELP ME!!!!!!!!!
dexar [7]
7.5 I think because it can not be 9 because it’s not close to 50
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3 years ago
En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

7 0
3 years ago
3) A tolley of mass 4kg, moving with a velocity
Leokris [45]

Answer:

3.57 m/s

Explanation:

The sum of the 2 momentums Is equal the finale momentums. so if momentums Is q, v Is velocity and m Is Mass, q3=m1*v1+m2**v2=16+9=25 m*kg/s

q3=m3*v3

v3=q3/m3=25/(4+3)=3.57m/s

8 0
3 years ago
If a book has a mass of 3 kg, what is the book's weight in N?
sesenic [268]

Answer:

29.4 N

Explanation:

F = ma

F = (3 kg) (9.8 m/s²)

F = 29.4 N

5 0
3 years ago
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