What distance will a car cover while uniformly accelerating from 12m/s to 26m/in 14 seconds

In a solution,a Is the substance being dissolved

Elena-2011 [213]

In a solution, a SOLUTE is the substance being dissolved.

In a solution, a SOLVENT is the substance in which something is dissolved.

For example in salt water, the salt is the solute. And the water is the solvent. Usually, which substance has a larger amount in the solution can be called a solvent. It doesn't matter if it's a gas, solid, or liquid. They can all be solute or solvents.

4 0

4 years ago

Let 10 RE be the distance of the satellite A from the center of the Earth, where RE is the radius of the Earth. What is the grav

stepan [7]

Answer:

a = 0.098 m/s²

Explanation:

The satellite, at any distance from the center of the Earth, is subject to the attractive force due to the Earth, according to the Newton´s Universal Law of gravitation, as follows:

Fg = G*ms*mE / (rse)²

According to Newton´s 2nd Law, neglecting any other force acted upon the satellite, we can write the following equation:

Fg = ms*a = G*ms*mE / (rse)²

⇒ a = G*mE / (rse)² (1)

As the distance between the satellite and the center of the Earth is 10 times the radius of the Earth, replacing this value in (1), we have:

a = G*mE / (10*RE)² = G*mE/(RE)² * (1/100)

but G*mE/(RE)², is just g, the acceleration due to gravity on the surface of the earth, so the gravitational acceleration due to Earth at satellite A, is as follows:

a = g*(1/100) = 0.01*g = 0.098 m/s²

7 0

3 years ago

The difference in heights of the liquid in the two sides of the manometer is 43.4 cm when the atmospheric pressure is 755 mm hg.

Schach [20]

The atmospheric P is greater than the P in the flask, since the Hg level is lacking down lower on the side open to the atmosphere.

43.4 cm x (10 mm / 1 cm) = 435 mm

the density of Hg is 13.6 / 0.791 = 17.2 times better than the liquid in the manometer. This means that 1 mmHg = 17.2 mm of manometer liquid.

435 mm manometer liquid x (1 mm Hg / 17.2 mm manometer liquid) = 25.3 mm Hg

The pressure in the flask is 755 - 25.3 = 729.7 mmHg.

729.7 mmHg x (1 atm / 760 mmHg ) = 0.960 atm.

4 0

3 years ago

A 0.200 H inductor is connected in series with a 88.0 Ω resistor and an ac source. The voltage across the inductor is vL=−(12.0V

Step2247 [10]

Answer:

a. (VL)R/ωL[1 - cos[ωt]] = (10.84 V)[1 - cos[(487rad/s)t]]

b. 1.084 mV

Explanation:

a. Since it is a series circuit, the current in the inductor is the same as the current in the resistor.

Now, the voltage across the inductor vL = -Ldi/dt.

So, the current, i = -1/L∫vLdt.

Now, vL = −(12.0V)sin[(487rad/s)t] and L = 0.200 H

Substituting these into i, we have

i = -1/L∫vLdt

= -1/0.200H∫[−(12.0V)sin[(487rad/s)t]]dt.

= -[−(12.0V)]/0.200H∫[sin[(487rad/s)t]]dt.

= 60V/H∫[sin[(487rad/s)t]]dt

Integrating i, we have

i = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)t]] + C

at t = 0, i(0) = 0

0 = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)× 0]] + C

0 = 60V/H ÷ [(487rad/s)[-cos[0]] + C

0 = 60V/H ÷ [(487rad/s)[-1]+ C

C = 60V/H ÷ [(487rad/s)

So, i = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)t]] + 60V/H ÷ [(487rad/s)

i = 60V/H ÷ [(487rad/s)[1 - cos[(487rad/s)t]]

i = (0.123A)[1 - cos[(487rad/s)t]] = VL/ωL[1 - cos[ωt]] where ω = 487rad/s and VL = 12.0 V and L = 0.200 H

So, the voltage across the resistor vR = iR where R = resistance of resistor = 88.0 Ω

So, vR = iR = VL/ωL[1 - cos[ωt]] × R = (VL)R/ωL[1 - cos[ωt]]

= (0.123A)[1 - cos[(487rad/s)t]] × 88.0 Ω

= (10.84 V)[1 - cos[(487rad/s)t]]

b. vR at t = 2.00 ms = 0.002 s

So, vR = (10.84 V)[1 - cos[(487rad/s)(0.002)]]

= (10.84 V)[1 - cos[0.974]]

= (10.84 V)[1 - 0.9999]

= (10.84 V)(0.0001)

= 0.001084

= 1.084 mV

3 0

3 years ago

Two small identical speakers are connected (in phase) to the same source. The speakers are 3 m apart and at ear level. An observ

zhenek [66]

Answer:

b. 2 m

Explanation:

Given that:

the identical speakers are connected in phases ;

Let assume ; we have speaker A and speaker B which are = 3 meter apart

An observer stands at X = 4m in front of one speaker.

If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is: <u> </u>

From above; the distance between speaker A and speaker B can be expressed as:

$\sqrt{3^2 + 4^2 } \\ \\ = \sqrt{9+16 } \\ \\ = \sqrt{25} \\ \\ = 5 \ m$

The path length difference will now be:

= 5 m - 4 m

= 1 m

Since , we are to determine the least intense sound; the destructive interference for that path length will be half the wavelength; which is

= $\dfrac{1}{2}*4 \ m$

= 2 m

8 0

4 years ago

What distance will a car cover while uniformly accelerating from 12m/s to 26m/in 14 seconds​

What distance will a car cover while uniformly accelerating from 12m/s to 26m/in 14 seconds