A car travels north along a straight highway at an average speed of 85 km/h. After driving 2.0 km, the car passes a gas station

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 × m

dynamic viscosity = 1.75 × Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ

therefore

= µ ............1

substituting the values

= 1.75 × ×

= 0.175 v

the area between the air and puck is given by

Area =

area =

area = 7.85 × m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 ×

force = 1.374 × v

now applying Newton's second law

force = mass × acceleration

- force =

- 1.374 × v =

solving for t =

the time needed after impact for the puck is 2.18 seconds

Response:

y= 240/901 cos 2t+ 8/901 sin 2t

Clarification:

To determine mass m=weight/g

  m=8/32=0.25

To calculate the spring constant

Kx=mg    (with c=6 inches and mg=8 pounds)

K(0.5)=8               (6 inches converts to 0.5 feet)

K=16 lb/ft

The governing equation for the spring-mass system is

my''+Cy'+Ky=F  

Inserting the known values yields

0.25 y"+0.25 y'+16 y=4 cos 20 t  ----(1) (given C=0.25 lb.s/ft)

Assuming the steady state equation for y is

y=A cos 2t+ B sin 2t

To determine constants A and B, we must equate this with equation 1.

Next, we find y' and y" by differentiating with respect to t.

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now, substitute the values of y", y' and y into equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

By comparing coefficients on both sides

30 A+ B=8

A-30 B=0

From this, we find

A=240/901 and B=8/901

Thus, the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

Answer:

Explanation:

The data indicates that point A is located midway between two charges.

To calculate the electric field at point A, we begin with the field produced by charge -Q ( 6e⁻ ) at A:

= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴

= 13.82 x 10⁻⁶ N/C

This field points towards Q⁻.

A similar field will arise from the charge Q⁺, but it will direct away from Q⁺ toward Q⁻.

To find the resultant field, we add these contributions:

= 2 x 13.82 x 10⁻⁶

= 27.64 x 10⁻⁶ N/C

For the force acting on an electron placed at A:

= charge x field

= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶

= 44.22 x 10⁻²⁵ N

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:

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

Thus,

.......(1)

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

......(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:

......(3)

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

Thus,

......(4)

By dividing equation (4) by (3),

According to equation (2),

This results in a 3.5 fold increase in energy.

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

......(5)

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

.....(6)

Referencing equations (5) and (6)

From equation (2),

Thus, in this particular scenario,

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm