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16 days ago

8

Suppose that an owner of the same dog breed has also taken some measurements. They notice that the surface area of the dog has i

ncreased by a factor of 3 over a period of four years. Determine the change in the dog's relative surface area in this time period.

1 answer:

inna [2.2K]16 days ago

7 0

Answer:

The surface area of the dog changes from A to 3A

Explanation:

It is stated that the dog's surface area has increased by a factor of 3 over four years.

We need to calculate the change in the relative surface area of the dog over this timeframe.

Let’s assume the initial surface area is A.

Since the surface area has been multiplied by 3,

it follows that the surface area after four years is equal to 3×A = 3A.

Thus, the dog's surface area transitions from A to 3A.

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The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

inna [2210]

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The velocity of the center of mass for the two-block system just prior to their collision is 2.9489 m/s

Explanation:

Provided information:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = collision time = 0.5 s

Question: What is the velocity of the center of mass for the two-block system right before the blocks collide, vf =?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It’s essential to calculate the required force:

$F=(m+m)*g*sin\theta$

Here, g = acceleration due to gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

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20 days ago

Read each scenario below. Then select the answer that best completes each sentence.

Ostrovityanka [2208]

Answer:

The power used by raul's microwave must match the power consumed by katrina's because both microwaves took different durations to accomplish the same heating task.

Explanation:

The power output from a car engine is equivalent to that of a bicycle since both perform the same amount of work over time. Both raul and katrina shared a frozen meal, heating each portion in different microwaves. Katrina's portion was warm in one minute, whereas raul's portion required two minutes. Therefore, the power utilized by raul's microwave aligns with that of katrina's, given that it took longer to achieve the same result.

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26 days ago

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A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above t

inna [2210]

Answer:

The answer to the specified question will be " $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$ ".

Explanation:

Referring to the question,

$\sum F_{x}$

⇒ $TCos \theta-F_{s}=0$

⇒ $T_{m}Cos \theta =F_{s}$ ...(equation 1)

$\sum F_{y}$

⇒ $TSin \theta+F_{N}=m_{g}$

⇒ $M_{g}-TSin \theta=F_{N}$ ...(equation 2)

Now,

From equation 1 and equation 2, we conclude

⇒ $T_{m} Cos \theta = \mu_{s}F_{N}$

By substituting the value of $F_{N}$ , we derive

⇒ $T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)$

⇒ $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$

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16 days ago

A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a

inna [2210]

If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

$C=\frac{\epsilon_0A}{d}$

In this equation, $\epsilon_0$ refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

$C \alpha \frac{1}{d}$ .......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

$\frac{C_1}{C_2} =\frac{d_2}{d_1}$ ......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

$E_1=\frac{Q^2}{2C_1}$ ......(3)

Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.

Thus,

$E_2=\frac{Q^2}{2C_2}$ ......(4)

By dividing equation (4) by (3),

$\frac{E_2}{E_1} =\frac{C_1}{C_2}$

According to equation (2),

$\frac{E_2}{E_1} =\frac{C_1}{C_2}=\frac{d_2}{d_1}$

This results in a 3.5 fold increase in energy.

$\frac{E_2}{E_1} =\frac{d_2}{d_1}=3.5\\ d_2=3.5*2.63 mm\\ =9.205 mm=9.21 mm$

Scenario (2)

If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.

The initial energy is described as

$E_1=\frac{1}{2} C_1V^2$ ......(5)

The final energy when the plate separation transitions to d₂ can be written as:

$E_2=\frac{1}{2} C_2V^2$ .....(6)

Referencing equations (5) and (6)

$\frac{E_2}{E_1} =\frac{C_2}{C_1}$

From equation (2),

$\frac{E_2}{E_1} =\frac{C_2}{C_1}=\frac{d_1}{d_2}$

Thus, in this particular scenario,

$\frac{E_2}{E_1} =\frac{d_1}{d_2}\\d_2=\frac{d_1}{3.5} \\ =\frac{2.63 mm}{3.5} \\ =0.109 mm=0.11 mm$

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm

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12 days ago

The weight of your car will also affect its _____.

inna [2210]

Stopping distance.

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4 days ago

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