_________ = voltage / resistance a. power b. energy c. charge d. current

Lewis structure for BeCI2

ludmilkaskok [199]

Answer:

you can see it in the picture

5 0

2 years ago

A stone is thrown, run a velocity of 15m/s is projected of an elevation of 30° to the horizontal calculate the time rate of flig

frozen [14]

Answer:

1.53 seconds

Explanation:

Applying,

T = 2usin∅/g................ Equation 1

Where, T = time of flight, u = initial velocity, ∅ = angle of projectile to the horizontal, g = acceleration due to gravity

From the question,

Given: u = 15 m/s, ∅ = 30°

Constant: g = 9.8 m/s²

Substitute these values in equation 1

T = 2(15)(sin30°)/9.8

T = 15/9.8

T = 1.53 seconds

Hence the time rate of flight is 1.53 seconds

3 0

2 years ago

A plane electromagnetic wave, with wavelength 5 m, travels in vacuum in the positive x direction with its electric vector E, of

Dafna1 [17]

Answer:

Explanation:

E₀ = 229.1 V/m

E = E₀ / √2 = 229.1 / 1.414 = 162 V/m

B = E / c ( c is velocity of em waves )

= 162 / (3 x 10⁸) = 54 x 10⁻⁸ T

rate of energy flow = ( E x B ) / μ₀

= 162 x 54 x 10⁻⁸ / 4π x 10⁻⁷

= 69.65 W per m².

5 0

3 years ago

A 20-kg child running at 1.4 m/s jumps onto a playground merry-go-round that has mass 180 kg and radius 1.6m. She is moving tang

Dominik [7]

Answer:

ωf = 0.16 rad/s

Explanation:

Moment of inertia of the child = mr² = 20(1.6²) = 51.2 kg•m²

Moment of Inertia of the MGR = ½mr² = ½(180)1.6² = 230.4 kg•m²

(ASSUMING it is a uniform disk)

Initial angular momentum of the child = Iω = I(v/r) = 51.2(1.4/1.6) = 44.8 kg•m²/s

Conservation of angular momentum

44.8 = (51.2 + 230.4)ωf

ωf = 0.15909090...

4 0

2 years ago

A diagram of a closed circuit with a power source on the left labeled 6 V. There are 3 resistors in parallel, separate paths, co

AfilCa [17]

Answer:

Current: 1.0 Amperes

The minimum current is flowing through path D

Explanation:

We first find the equivalent resistance to the three resistors in parallel ( which is the total resistance of the circuit) via the equation:

$\frac{1}{R_e} =\frac{1}{R_B}+\frac{1}{R_C}+\frac{1}{R_D}\\\frac{1}{R_e} =\frac{1}{10}+\frac{1}{20}+\frac{1}{50}=0.17\\R_e=(1/0.17)\Omega\\R_e=5.88 \Omega$

with this info, we can estimate the current going through branch A using Ohm's Law, and the information that the power source is 6 V:

$V=I*R\\6V=I*5.88\Omega\\I=\frac{6}{5.88} Amp\\I=1.02A$

where the current comes in units of Amperes since all other the quantities are given in the SI system, and we can round this answer to 1.0 Amp following the request to round it to the tenth.

The current will be the lowest through the branch with the largest resistor due to the fact that less current will flow through the path of more resistance.

Than means that the lowest current will be registered through branch D where the 50 $\Omega$ resistor is.

8 0

3 years ago

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