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1 month ago

You're riding a unicorn at 25 m/s and come to a uniform stop at a red light 20m away. What's your acceleration?

Physics

2 answers:

Ostrovityanka [3.2K]1 month ago

8 0

The result is -15.625 m/s².

Acceleration signifies the alteration of velocity over a specified duration. It can be calculated with this formula:

$a = \dfrac{vf-vi}{t}$

Where:

vf = final velocity

vi = initial velocity

t = time

Let’s examine the information provided in your query:

Initially, the vehicle was traveling at 25 m/s before coming to a halt. Thus, it was in motion and subsequently ceased moving, indicating that the final velocity is 0 m/s.

However, we notice that the problem does not provide a time value. We need to determine the time taken from when it was in motion to when it reached the traffic light located 20 m away.

The time can be calculated using the kinematics equation:

$d = \dfrac{vi+vf}{2} *t$

We derive the equation by substituting the known values first.

$20m = \dfrac{25m/s+0m/s}{2}(t)$

$20m = 12.5m/s{2}(t)$

$\dfrac{20m}{12.5m/s}=t$
$1.6s=t$

The duration from when it was in motion until it stopped is 1.6s. Now we can utilize this in our acceleration calculation.

$a = \dfrac{0m/s-25m/s}{1.6s}$

$a = \dfrac{-25m/s}{1.6s}$

$a = -15.625m/s^{2}$

It is important to note that the acceleration is negative, indicating the vehicle slowed down.

Maru [3.3K]1 month ago

5 0

My acceleration is 15.63 m/s² (decelerating)

To explain:

The formula for acceleration can be expressed as:

Where

vi = initial velocity = 25 m/s $v^{2} _{f} -v^{2} _{i} =2ax$

vf = final velocity = 0

x = distance = 20 m

Substituting values and solving for acceleration "a" results in:

The negative value of acceleration signifies that it is decelerating.

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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 [3103]

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

6 0

1 month ago

A sculptor has asked you to help electroplate gold onto a brass statue. You know that the charge carriers in the ionic solution

Maru [3345]

Answer:

Explanation:

Amount of gold deposited = 0.5 g

Gold's molar mass = 197 g/mol

Time duration, t = 6 hours

= 6 × 3600

= 12600 s

Calculation of moles: mass/molar mass

= 0.5/197

= 0.00254 mole

Assuming

Au --> Au+ + e-

Faraday's constant = 9.65 x 10^4 C mol-1

Charge, Q = 96500 × 0.00254

= 244.924 C

Relation: Q = I × t

Thus, I = 244.924/12600

= 0.011 A

= 11.34 mA.

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Yuliya22 [3333]

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A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s

ValentinkaMS [3465]

Answer:

0.6

Explanation:

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