A kind of foliated metamorphic rock a. coal b. limestone c. marble d. slate

Explain the law of conservation of energy. Give a specific example using kinetic and potential energy that shows how energy is c

Ede4ka [16]

Answer:

As you said you already know, energy cannot be created or destroyed.

Explanation:

You cannot gain energy or lose energy, it can only be converted. So if you start on a 3m high hill and go down it, your potential energy is equal to mgh, and if you get to the bottom of the hill, your KE would be equal to your PE at the top, and when you start going up another hill again, the maximum height you can reach is 3m, because energy cannot be created or destroyed, and your mass and gravitational acceleration are the same, so therefore you can only reach the same height you started from due to the conservation of energy.

7 0

3 years ago

Three resistors of 4.0 ohms, 6.0 ohms, and 12 ohms are connected in parallel to an applied potential difference of 12 volts. (1)

lesantik [10]

Answer:

(1) 2 ohms

(2) 12 Volts Across each resistor

(3) I₁ = 3 A, I₂ = 2 A, I₃ = 1 A

Explanation:

From the question,

(1) Equilvalent Resistance (Rt) for parallel connection is

1/Rt =(1/R₁)+ (1/R₂) + (1/R₃)

Where R₁ = 4 ohms, R₂ = 6 ohms, R₃ = 12 ohms

1/Rt = 1/4 +1/6 +1/12

1/Rt = (3+2+1)/12

1/Rt = 6/12

1/Rt = 1/2

Rt = 2 ohms.

(2) Since the resistors are connected in parallel, They will have the same potential difference across them,

Hence the P.d across each resistor = 12 Volts.

(3) For R₁,

I₁ = V/R₁ = 12/4

I₁ = 3 A.

For R₂,

I₂ = V/R₂

I₂ = 12/6

I₂ = 2 A

For R₃,

I₃ = V/R₃

I₃ = 12/12

I₃ = 1 A

4 0

3 years ago

The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu

BartSMP [9]

Answer:

hello your question is incomplete attached below is the missing part

answer : short period oscillations frequency = 0.063 rad / sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= $( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns} ) = 0$

where :

$w_{np} = Natural frequency of plugiod oscillation$

$\alpha _{p} = damping ratio of plugiod oscilations$

comparing the general form with the given equation

$w^{2} _{np}$ = 18.2329

$w^{2} _{ns} = 0.003969$

hence the short period oscillation frequency ( $w_{ns}$ ) = 0.063 rad/sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

8 0

3 years ago

A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it

Savatey [412]

The charge of the object must be $1.11 \times e^{-5} \text { coulomb }$

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

$Electric field, E=\frac{\text { Force }(F)}{q}$