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2 years ago

What must be true about two objects if heat is flowing between them

Physics

1 answer:

Vedmedyk [2.9K]2 years ago

4 0

Answer:

The objects must be different temperatures.

Explanation:The objects must be different temperatures. eat will flow from Object 1 to Object 2 in examples 1, 2, and 4, and heat will flow from Object 2 to Object 1 in Example 3. Which factors affect heat transfer between a warm and a cool substance?

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a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t

lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0

3 years ago

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling

Natalka [10]

Answer:

H = 5 m

Explanation:

As the person leaves the slide horizontally so the time taken by the person to hit the water is given as

$t = 0.672 s$

so we can find the vertical velocity by which person will hit the water using kinematics

$v_y = u_y + at$

$v_y = 0 + (9.81)(0.672)$

$v_y = 6.6 m/s$

now the speed of the person at the end of the slide is given as

$v = \frac{L}{t}$

$v = \frac{5}{0.672}$

$v_x = 7.44 m/s$

now by energy conservation we can find the initial height

$mgH = \frac{1}{2}m(v_x^2 + v_y^2)$

$H = \frac{1}{2g}(v_x^2 + v_y^2)$

$H = \frac{1}{2(9.81)}(7.44^2 + 6.6^2)$

$H = 5 m$

6 0

3 years ago

Read 2 more answers

Metal sphere 1 has a positive charge of 7.00 nc . metal sphere 2, which is twice the diameter of sphere 1, is initially uncharge

MariettaO [177]

Answer:

2.33 nC, 4.67 nC

Explanation:

when the two spheres are connected through the wire, the total charge (Q=7.00 nC) re-distribute to the two sphere in such a way that the two spheres are at same potential:

$V_1 = V_2$ (1)

Keeping in mind the relationship between charge, voltage and capacitance:

$C=\frac{Q}{V}$

we can re-write (1) as

$\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$ (2)

where:

Q1, Q2 are the charges on the two spheres

C1, C2 are the capacitances of the two spheres

The capacitance of a sphere is given by

$C=4 \pi \epsilon_0 R$

where R is the radius of the sphere. Substituting this into (2), we find

$\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 R_2}$ (3)

we also know that sphere 2 has twice the diameter of sphere 1, so the radius of sphere 2 is twice the radius of sphere 1:

$R_2 = 2R_1$

So the eq.(3) becomes

$\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 2R_1}$

And re-arranging it we find:

$Q_2 = 2Q_1$

And since we know that the total charge is

$Q_1 + Q_2 = 7.00 nC$

we find

$Q_1 = 2.33 nC\\Q_2 = 4.67 nC$

3 0

4 years ago

What is the electrical force between q1 and q2? Recall that k = 8.99 × 109 N•meters squared over Coulombs squared.. 4.3 × 10 N 3

Rudiy27

Answer:

Explanation:

Incomplete question but for understanding.

We want to find the electrical force between two charges, then you can use the coulombs law which states that the force of attraction or repulsion between two charges is directly proportional to the product of the two charges and inversely proportional to the square of their distance apart,

So,

F = kq1•q2 / r²

Where k is a constant and it is given as

K = 8.99 × 10^9 Nm²/C²

q1 and q2 are the charges and in this question it is not given, so the question is incomplete. Let assume that,

q1 = - 1.609 × 10^-19 C electron

q2 = 1.609 × 10^-19 C proton

Since unlike charges attract, then it is force of attraction

Also, r is the distance apart and it is not given, let assume the distance between the two charges is 2 × 10^-5m

Then,

F = kq1•q2 / r²

F = 8.99 × 10^9 × 1.609 × 10^-19 × 1.609 × 10^-19 / (2 × 10^-5)²