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3 years ago

Which of the following describes the variable that responds to a change?

Physics

2 answers:

timama [110]3 years ago

8 0

Dependent variable is your answer.

Hitman42 [59]3 years ago

8 0

Answer:

b. Dependent variable

Explanation:

Dependent variable is also referred to as the responding variable.It is what is measured or studied in the experiment. The dependent variable responds to the independent variable. It's what changes as a result of the changes to the independent variable.

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The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pre

IrinaVladis [17]

Answer:

A) T1 = 269.63 K

T2 = 192.59 K

B) W = -320 KJ

Explanation:

We are given;

Initial volume: V1 = 7 m³

Final Volume; V2 = 5 m³

Constant Pressure; P = 160 KPa

Mass; m = 2 kg

To find the initial and final temperatures, we will use the ideal gas formula;

T = PV/mR

Where R is gas constant of helium = R = 2.0769 kPa.m/kg

Thus;

Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K

Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K

B) world one is given by the formula;

W = P(V2 - V1)

W = 160(5 - 7)

W = -320 KJ

6 0

3 years ago

(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 48.0 µC. What is the voltage of the

frosja888 [35]

Answer:

a.6 V

b.24 V

Explanation:

We are given that

a. $C=8\mu F=8\times 10^{-6} F$

$1\mu =10^{-6}$

Q= $48\mu C=48\times 10^{-6} C$

We know that

$V=\frac{Q}{C}$

Using the formula

$V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V$

b. $Q=192\mu C=192\times 10^{-6} C$

$V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V$

8 0

3 years ago

Which statement is true regarding the vectors shown.

kap26 [50]

Answer:

Y-> + F-> +G ->=E->

Explanation:

7 0

3 years ago

PLEASE HELP!!! i’ll mark brainliest if you’re correct

dedylja [7]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

The net force acting on the block is ~

$\qquad \sf \dashrightarrow \: 10 + 5$

$\qquad \sf \dashrightarrow \: 15 \:N$

So, the Answer in the boxes will be ~

$\boxed{ \sf15} \: \boxed{ \sf N}$

7 0

3 years ago

A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =

natali 33 [55]

With acceleration

$\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j$

and initial velocity

$\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i$

the velocity at time t (b) is given by

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

We can get the position at time t (a) by integrating the velocity:

$\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du$

The particle starts at the origin, so $\mathbf x(0)=\mathbf0$ .

$\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du$

$\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}$

$\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j$

Get the coordinates at t = 8.00 s by evaluating $\mathbf x(t)$ at this time:

$\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j$

$\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j$

so the particle is located at (x, y) = (64.0, 64.0).

Get the speed at t = 8.00 s by evaluating $\mathbf v(t)$ at the same time:

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j$

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

This is the velocity at t = 8.00 s. Get the speed by computing the magnitude of this vector:

$\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}$

5 0

3 years ago