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2 years ago

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Apply the impulse-momentum relation and the work-energy theorem to calculate the maximum value of t if the cake is not to end up

on the floor. assume that the cake moves a distance d while still on the tablecloth and therefore a distance r?d while sliding on the table top. assume that the friction forces are independent of the relative speed of the sliding surfaces. you can easily try this trick yourself by pulling a sheet of paper out from under a glass of water, but have a mop handy just in case!

1 answer:

loris [4]2 years ago

8 0

Puto chupame el semen ok? right?

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Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1

7nadin3 [17]

Answer:

A). σ = 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B). $\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C). $U=10.322$ J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

$\sigma =k.\varepsilon _{0}.E$

where, E is resultant electric field = 1.2 x $10^{6}$ V/m

$\varepsilon _{0}$ is permittivity of free space = 8.85 x $10^{-12}$ $C^{2}$ /N- $m^{2}$

k is dielectric constant = 3.6

∴ $\sigma =k.\varepsilon _{0}.E$

= 3.6 x 8.85 x $10^{-12}$ x 1.2 x $10^{6}$

= 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

$\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )$

$\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )$

$\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C).

Area of the plate, A = 2.5 $cm^{2}$

= 2.5 x $10^{-4}$ $m^{2}$

diameter of the plate, d = 1.8 mm

= 1800 m

∴ Total energy stored in the capacitor

$U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad$

$U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800$

$U=10.322$ J

4 0

2 years ago

Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for

Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0

2 years ago

An electric clock is hanging on a wall. As you are watching the second hand rotate, the clock's battery stops functioning, and t

Setler [38]

Answer:

B. W is positive and a is negative

Explanation:

As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have

initial angular velocity is termed as positive angular velocity

$\omega = positive$

now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed

so we will have

$\alpha = negative$

so here correct answer is

B. W is positive and a is negative

8 0

2 years ago

A bucket of water experiencing a gravitational force of 525 N is pulled up from a water well. The net force in the y-direction i

lukranit [14]

Answer:

6n!!!!!!!!!!!!!!!!!!

Explanation:

nnnn

8 0

2 years ago

An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular accele

leva [86]

To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.

Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.

The average angular acceleration

$\alpha = \frac{\omega_f - \omega_0}{t}$

Here

$\alpha$ = Angular acceleration

$\omega_{f,i} =$ Initial and final angular velocity

There is not initial angular velocity,then

$\alpha = \frac{\omega_f}{t}$

We know that the relation between the tangential velocity with the angular velocity is given by,

$v = r\omega$

Here,

r = Radius

$\omega$ = Angular velocity,

Rearranging to find the angular velocity

$\omega = \frac{v}{r}}$

$\omega = \frac{30}{0.20} \rightarrow$ Remember that the radius is half te diameter.

Now replacing this expression at the first equation we have,

$\alpha = \frac{30}{0.20*6}$

$\alpha = 25 rad /s^2$

Therefore teh average angular acceleration of each wheel is $25rad/s^2$

3 0

2 years ago