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salantis [7]
2 years ago
8

A closed, uninsulated system fitted with movable piston, so no matter is exchanged with the surroundings, was assembled. Introdu

ction of 430 J of heat caused the system to expand, doing 238 J of work against a constant pressure of 101 kPa. What is the value of for this process
Physics
1 answer:
xeze [42]2 years ago
8 0

Answer: You do not specify what is being asked for. ∆E? ∆H?

∆E = (430 - 238) J = 192 J

∆H = 430 J

Explanation:

If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.

Therefore ∆H = 430 J

If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w

The question states that 238 J of work are done AND the system expanded

(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)

Therefore, ∆E = (430 - 238) J = 192 J

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A motorist traveling at 17 m/s encounters a deer in the road 46 m ahead. If the maximum acceleration the vehicle’s brakes are ca
AveGali [126]

Answer:

t = 1.29 s

Explanation:

Maximum distance after which the motorist must have to stop is given as

d = 46 m

now the distance after which the vehicle will stop on applying breaks is given as

v_f^2 - v_i^2 = 2 a d

0 - 17^2 = 2(-6) d

d = 24.1 m

now remaining distance is given as

x = 46 - 24.1

x = 21.9 m

now reaction time is given as

t = \frac{x}{v}

t = \frac{21.9}{17}

t = 1.29 s

5 0
3 years ago
A charge q produces an electric field of strength 4E at a distance of d away. Determine the electric field strength at a distanc
gizmo_the_mogwai [7]

Answer:

c.) 36E

Explanation:

The magnitude of the electric field is given by the expression

E=k \frac{q}{d^2} (1)

where k is the Coulomb's constant, q is the charge that generates the field, and d is the distance from the charge.

In this problem, we have that the magnitude of the field at a distance d is 4E, so we can rewrite the previous equation as

4E = k\frac{q}{d^2}

Now we want to determine the electric field at a distance of d'=\frac{1}{3}d away. Substituting into (1), we find

E' = k \frac{q}{d'^2}=k \frac{q}{(\frac{1}{3}d)^2}=9 k \frac{q}{d^2} (2)

We also know that

4E = k\frac{q}{d^2} (3)

So combining (2) with (3), we find a relationship between the original field and the new field:

E' = 9 \cdot (4E) = 36E

7 0
3 years ago
The Heat required to raise the temp. of 20 g water from 25 C to 36 C
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What is the question
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ExtremeBDS [4]
Solid phase. The atoms are tightly packed and vibrate. 
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Answer:

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a) vf = vi + at

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