Find the resultant of two components, 3km west and 4 km south. How do I solve this using the Pythagorean theorem?

A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery

GuDViN [60]

Answer:

at the beginning: $2.3\cdot 10^{-10} F$

when the plates are pulled apart: $1.1\cdot 10^{-10} F$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=k \epsilon_0 \frac{A}{d}$

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

$A=0.210 m^2$ is the area of the plates

$d=0.815 cm=8.15\cdot 10^{-3} m$ is the separation between the plates at the beginning

Substituting into the formula, we find

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F$

Later, the plates are pulled apart to $d=1.63 cm=0.0163 m$ , so the capacitance becomes

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F$

4 0

3 years ago

The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont

ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

$T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}$

But heat capacity of object B is twice that of object A

$T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K$

Therefore, the final temperature of both objects is 400 K.

4 0

2 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)

Cerrena [4.2K]

Answer:

B false

Explanation:

dont have any explanation

8 0

2 years ago

Read 2 more answers

A 94.7 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.75 r

Marta_Voda [28]

Answer:

The angular velocity of the platform is 1.114 rad/s.

Explanation:

Step 1: Given data

Mass of the horizontal circular platform = 94.7 kg

Mass of the monkey = 21.1 kg

Initial angular velocity = 1.75 rad/s

A monkey drops a 9.25 kg bunch of bananas

They hit the platform at 4/5 of its radius from the center

Model the platform as a disk of radius 1.63 m

Step 2: Calculate the moment of inertia of the disk

I = ½ * m * r² = ½ * 94.7 * 1.63² = 125.80

Step 3: Calculate the initial angular momentum

I = 125.80 * 1.75 = 220.15

Step 4: Calculate the moment of inertia for the bananas

For the bananas, r = 4/5 * 1.63 = 1.304 m

I = 9.25 * 1.304² = 15.73

Step 5: Calculate Moment of inertia for the monkey

I = 21.1 * 1.63² = 56.06

Step 6: Total moment of inertia = 125.80 + 15.73 + 56.06 = 197.59

Step 7: Calculate final angular momentum = 197.59 * ω

197.59 * ω = 220.15

ω = 220.15 / 197.59

This is approximately 1.114 rad/s.

3 0

3 years ago

Thomas found an organism in his backyard and is trying to classify it into a kingdom. The organism has small roots, and Thomas f

Anvisha [2.4K]

Answer:

Kingdom Plantae

Explanation:

Plants are multi-celled, have roots, and are found in the backyard. I also did research but I can't put it in here or else my answer gets deleted. :)

4 0

3 years ago