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2 years ago

Two objects made of the same material are heated to 60 oC and 90 oC. According to

Physics

2 answers:

earnstyle [38]2 years ago

6 0

Answer:

The newton law of cooling is:

dQ/dt = -h*a(To - Ta)

Where h and a are constants, To is the temperature of the object and Ta is the temperature of the ambient.

this means that initially if in both cases Ta is the same, the object with higher temperature cools faster.

Now, obviously, if we want to know which object cools faster to a given temperature, for example, 30°C, the object with 60°C will reach that temperature in less time than the other one. (this is because the hot object cools faster at the beginning, but as it starts to lose temperature, the rate at which it colls starts to decrease)

nlexa [21]2 years ago

4 0

Assuming that there is in a vacuum, the two object will cool at the same rate, because the objects are made of the same material they will have the same cooling rate, assuming the surrounding temperature is the same.

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A research submarine has a 30-cm-diameter window that is 8.1 cm thick. The manufacturer says the window can withstand forces up

malfutka [58]

The pressure at a certain depth underwater is:

P = ρgh

P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth

The pressure exerted on the submarine window is:

P = F/A

P = pressure, F = force, A = area

The area of the circular submarine window is:

A = π(d/2)²

A = area, d = diameter

Set the expressions for the pressure equal to each other:

F/A = ρgh

Substitute A:

F/(π(d/2)²) = ρgh

Isolate h:

h = F/(ρgπ(d/2)²)

Given values:

F = 1.1×10⁶N

ρ = 1030kg/m³ (pulled from a Google search)

g = 9.81m/s²

d = 30×10⁻²m

Plug in and solve for h:

h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)

h = 1540m

5 0

2 years ago

The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v

velikii [3]

Answer:

$\theta = n\pi/2, {\rm where~n~is~an~integer.}$

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

$\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k$

Similarly,

$\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k$

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

$\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)$

So,

$\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)$

At time t = 0, this equation becomes

$0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}$

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In order to solve this problem, we must first find out the value of each line on the number line. However, we can make this problem more simple by ignoring every interval except for the ones between 0 and 6. There are three total intervals in between 0 and 6 (including 6 and excluding 0). Therefore, we can do 6/2, and get an interval value of 2. This means that each line adds a value of 2. Since the car is only one line past zero, we only have to add one value of 2. Since 0 + 2 = 2, our final answer is C. 2.

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2 years ago

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