A ball and a thin plate are made from different materials and have the same initial temperature. The ball does not fit through a hole in the plate, because the diameter of the ball is slightly larger than the diameter of the hole. However, the ball will pass through the hole when the ball and the plate are both heated to a common higher temperature. In each of the arrangements in the drawing the diameter of the ball is 1.00 10-5 m larger than the diameter of the hole in the thin plate, which has a diameter of 0.09 m. The initial temperature of each arrangement is 27.3°C. At what temperature will the ball fall through the hole in each arrangement?
Please HELP!!!
hole d=0.09 m
ball D=d+Δd=d+10⁻⁵m=0.09001 m
t=27.3℃
ΔL=αLΔT
D+ΔD = d+Δd
D+ α1•D•ΔT =d + α2•d•ΔT
D-d = α2•d•ΔT - α1•D•ΔT= ΔT(α2•d- α1•D)
ΔT = (D-d)/(α2•d- α1•D)= Δd/(α2•d- α1•D)
Arrangement A:
ball: Gold α1=14•10⁻⁶ (℃⁻)¹
hole: Lead α2 = 29•10⁻⁶ (℃⁻)¹
ΔT = Δd/(α2•d- α1•D) =
=10⁻⁵/(29•10⁻⁶•0.09 - 14•10⁻⁶•0.09001) =
=7.408°.
t=27.3+7.408=34.708℃
Arrangement B:
Steel α1=12•10⁻⁶ (℃⁻)¹
Aluminium α2 = 23•10⁻⁶ (℃⁻)¹
ΔT = Δd/(α2•d- α1•D)=...
Arramgment C:
Quartz α1=0.5•10⁻⁶ (℃⁻)¹
Silver α2 = 19•10⁻⁶ (℃⁻)¹
ΔT = Δd/(α2•d- α1•D)=...
Well, it seems like the ball and the plate are having a bit of a size issue! They just can't seem to squeeze through that hole. But don't worry, with a little heat, they might just find a way to overcome this obstacle.
Now, the key to solving this problem lies in understanding how materials expand when they are heated. When you heat up an object, it tends to expand and increase in size. So, if we heat both the ball and the plate, their diameters will increase.
In this case, the ball has a diameter that is 1.00 * 10^-5 m larger than the hole in the plate, which has a diameter of 0.09 m. So, in order for the ball to fall through the hole, its expanded diameter must be smaller than the hole.
To find out at what temperature the ball will fall through, we need to figure out by how much the ball needs to expand. Since the difference in diameter is 1.00 * 10^-5 m, we can divide this by 2 to get the radius difference: 5.00 * 10^-6 m.
Now, we need to find out how much the ball needs to expand in terms of temperature. For most materials, the coefficient of linear expansion is given. This coefficient tells us how much a material expands or contracts for every degree increase or decrease in temperature.
Let's assume that the coefficient of linear expansion for both the ball and the plate is the same. To find the temperature at which the ball will fall through, we can use the formula:
Change in temperature = (Change in length / Original length) / Coefficient of linear expansion.
Since we are dealing with diameters, we can substitute length with diameter. And since the initial temperature is 27.3°C, we can subtract this from the change in temperature to find the final temperature.
So, the final temperature at which the ball will fall through the hole can be calculated as:
Final temperature = 27.3°C + ( (5.00 * 10^-6 m / 0.09 m) / Coefficient of linear expansion).
Now, without information on the specific materials and their coefficients of linear expansion, it is impossible to give you an exact answer. However, if you have these values, you can plug them into the formula to find the final temperature.
I hope this explanation didn't expand the limits of your patience! Just remember, with a little heat, even the most unlikely things can happen. Good luck!
To determine the temperature at which the ball will fall through the hole in each arrangement, we need to consider the thermal expansion of the ball and the plate.
We can use the equation for thermal expansion:
ΔL = α * L * ΔT
Where:
ΔL is the change in length/size,
α is the coefficient of linear expansion of the material,
L is the initial length/size, and
ΔT is the change in temperature.
In this case, the diameter can be considered as the length, and the coefficient of linear expansion can be assumed to be constant over the temperature range involved.
Let's assume that the hole in the plate has a diameter of D, and the diameter of the ball is D + 1.00 × 10^-5 m. The initial temperature of both the ball and plate is 27.3°C.
To find the temperature at which the ball will fall through the hole, we can set up the following equation:
(2r_ball) - (2r_hole) = α_plate * L_plate * (T_final - T_initial)
Where:
r_ball is the radius of the ball,
r_hole is the radius of the hole,
α_plate is the coefficient of linear expansion of the plate,
L_plate is the initial length of the plate (equal to the diameter),
T_final is the temperature at which the ball falls through the hole,
and T_initial is the initial temperature.
First, let's convert the initial temperature to Kelvin:
T_initial = 27.3°C + 273.15 = 300.45 K
Next, substitute the given values:
(2 * (D + 1.00 × 10^-5)) - (2 * D) = α_plate * D * (T_final - 300.45)
Simplifying the equation:
2 * (1.00 × 10^-5) = α_plate * D * (T_final - 300.45)
Now, we need the coefficient of linear expansion value for the plate material. You can look up this value for the specific material or assume an average value for metals, which is around 1.2 × 10^-5 (1/°C).
For example, if we assume α_plate = 1.2 × 10^-5 (1/°C), and D = 0.09 m:
2 * (1.00 × 10^-5) = (1.2 × 10^-5) * 0.09 * (T_final - 300.45)
Solving for T_final:
T_final - 300.45 = (2 * (1.00 × 10^-5)) / ((1.2 × 10^-5) * 0.09)
T_final = 300.45 + (2 * (1.00 × 10^-5)) / ((1.2 × 10^-5) * 0.09)
Evaluating this expression will give you the temperature at which the ball will fall through the hole in each arrangement.
To determine at what temperature the ball will fall through the hole in each arrangement, we need to consider the principles of thermal expansion.
When objects are heated, they expand due to the increase in thermal energy. In this case, both the ball and the thin plate are heated, causing them to expand. We need to find the temperature at which the expansion of the ball allows it to pass through the hole in the plate.
Let's start by calculating the difference in diameter between the ball and the hole. The diameter difference is given as 1.00 * 10^-5 m.
Next, we need to determine the amount of expansion needed in the diameter of the ball to pass through the hole. This expansion is equal to half of the diameter difference (since both sides of the ball need to expand equally). So, the expansion needed in the diameter of the ball is (1.00 * 10^-5 m)/2 = 5.00 * 10^-6 m.
Now, we can set up the equation to solve for the temperature at which the ball falls through the hole. The equation relates the change in length (or diameter) of an object to its original length (or diameter) and its coefficient of linear expansion (α) multiplied by the change in temperature (ΔT).
ΔL = α * L * ΔT
In this case, we need to find the change in temperature (ΔT) when the expansion (ΔL) of the ball's diameter is equal to the needed expansion of 5.00 * 10^-6 m. The coefficient of linear expansion (α) can differ for different materials, so we will need to look it up for the specific material of the ball.
You can consult a reference book or search online to find the coefficient of linear expansion (α) for the material of the ball. Once you have the value of α, you can rearrange the equation to solve for ΔT:
ΔT = ΔL / (α * L)
Plug in the values you have:
ΔT = (5.00 * 10^-6 m) / (α * 0.09 m)
Solve this equation using the specific α value for the material of the ball to find the temperature change (ΔT) required for the ball to fall through the hole.
Please note that the material of the ball is not specified in the question, so you would need to find that information to complete the calculation.