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3 years ago

What is the difference between natural rain and acid precipitation

1 answer:

8 0

The key difference between acid rain and normal rain is that the acid rain contains a large amount of sulfur dioxide and nitrogen oxide gases dissolved in it than the normal rain.

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A spacecraft is flying away from the moon toward earth.

Alekssandra [29.7K]

Answer:

it will decrease

Explanation:

According to the law of universal gravitation, the gravitational force exerted by the moon on the spacecraft is equal to the product of their masses and inversely proportional to the square of the distance that separates them. Therefore, as the spacecraft moves away, its distance increases and the force of attraction exerted by the moon decreases.

4 0

3 years ago

A car’s velocity as a function of time is given by Vx (t) = α.t + β.t 2 , where α= 3m/s and β= 0.1m/s 3 . Calculate the average

s344n2d4d5 [400]

The definition of average acceleration allows to find the result for the average acceleration in the given time interval is:

$a_{average}= 1.5 \ \frac{m}{s^2}$

Instantaneous acceleration is defined as the derivative of velocity with respect to time.

a = $\frac{dv}{dt}$

Where a is the acceleration, v the velocity and t the time.

They indicate that the speed of the car is given by the relation.

v = α t + β t²

With α = 3 m / s and β = 0.1 m / s³

Let's make the derivative.

a = α + 2β t

Let's substitute

a = 3 + 2 0.1 t

Average acceleration is the change in velocity in the time interval.

$a_{average} = \frac{\Delta v}{\Delta t }$

Let's find the velocity at the indicated time.

For t = 5 s

v₅ = 3 + 0.1 5²

v₅ = 5.5 m / s

For t = 10 s

v₁₀ = 3 + 0.1 10²

v₁₀ = 13 m / s

Let's calculate the average acceleration.

$a_{average} = \frac{13 - 5.5 }{ 10 - 5 }\\$

$a_{average}= 1.5 \ m/s^2$

In conclusion using the definition of mean acceleration we can find the result for the mean acceleration in the given time interval is:

$a_{average} =$ 1.5 m / s²

Learn more here: brainly.com/question/20057878

7 0

3 years ago

Equal charges, one at rest, the other having a velocity of 104 m/s, are released in a uniform magnetic field. Which charge has t

Yuliya22 [10]

Answer:

case 1 of physics is the answer

4 0

3 years ago

A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a

Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

$Z_{R}$ = R

R = resistance

$Z_{L}$ = jωL

ω = voltage source angular frequency, L = inductance

$Z_{C}$ = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

$Z_{R}$ = 217Ω

$Z_{L}$ = j(220)(0.875) = j192.5Ω

$Z_{C}$ = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$ + $Z_{C}$

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

$v_{R}$ = I× $Z_{R}$

$v_{L}$ = I× $Z_{L}$

$v_{C}$ = I× $Z_{C}$

Given values:

I = (0.0569∠1.147+220t)A

$Z_{R}$ = 217Ω = (217∠0)Ω

$Z_{L}$ = j192.5Ω = (192.5∠π/2)Ω

$Z_{C}$ = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

$v_{R}$ = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

$v_{L}$ = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

$v_{C}$ = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

$v_{R}$ = (12.35∠1.147+220t)V

$v_{L}$ = (10.95∠2.718+220t)V

$v_{C}$ = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

$v_{R}$ = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

$v_{L}$ = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

$v_{C}$ = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0

3 years ago

A train 4.00 3 102 m long is moving on a straight track with a speed of 82.4 km/h. The engineer applies the brakes at a crossing

zysi [14]

Answer:

The value is $t =29.2 \ s$