PLS HELP what force is required to move your 1000 kg car an acceleration of 4 m/s

How can the frequency with which the direction of a current changes be regulated?

sesenic [268]

Answer:

In an inductive circuit, when frequency increases, the circuit current decreases and vice versa.

Explanation:

5 0

3 years ago

on june 21 some earth locations have 24 hours of daylight these locations are all between the latitudes of

andrew-mc [135]

66° N and 90° N

the area of the artic circle in the northern hemisphere

4 0

3 years ago

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.

Explanation:

Depression in freezing point :

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

If molality of the solution in high the depression in freezing point of the solution will be more.
If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

$\frac{p_o-p_s}{p_o}=\chi_{solute}$

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of solution

$\chi_{solute}$ = Mole fraction of solute

$p_s\propto \frac{1}{\chi_{solute}}$

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.

8 0

3 years ago

A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.

Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is 1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

Learn more about energy here: brainly.com/question/13881533

#SPJ1

6 0

2 years ago

A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the

Julli [10]

Answer:

Explanation:

a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV

C = capacitance of the capacitor (in Farads )

V = voltage (in volts) = 100V

C = ∈A/d

∈ = permittivity of free space = 8.85 × 10^-12 F/m

A = cross sectional area = 600 cm²

d= distance between the plates = 0.7cm

C = 8.85 × 10^-12 * 600/0.7

C = 7.59*10^-9Farads

Q = 7.59*10^-9 * 100

Q = 7.59*10^-7Coulombs

Q = 0.759*10^-6C

Q = 0.759µC

b) Energy stored in a capacitor is expressed as E = 1/2CV²

E = 1/2 * 7.59*10^-9 * 100²

E = 0.0000395Joules

E = 39.5*10^-6Joules

E = 39.5µJ

7 0

3 years ago

PLS HELP what force is required to move your 1000 kg car an acceleration of 4 m/s￼

PLS HELP what force is required to move your 1000 kg car an acceleration of 4 m/s