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4 years ago

Gravitational potential energy is a function of the _ and the ____ of an object.

Physics

2 answers:

Hunter-Best [27]4 years ago

8 0

Answer:

C. weight, height

bearhunter [10]4 years ago

4 0

The answer is C. Gravitational potential energy is a function of the weight and the height of an object.

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The position function x (6.0 m) cos[(3p rad/s)t p/3 rad]gives the simple harmonic motionof a body. at t 2.0 s, what are the(a) d

Daniel [21]

The correct formula for this simple harmonic motion is:

x = (6.0) cos [ (3 pi) t + pi/3 ]

A. Calculating for displacement x at t = 2:

x =6.0 cos (19/3 pi)
x = 6 (0.5)

x = 3 m

B. Velocity is equivalent to the 1st derivative of the equation. v = dx / dt

Velocity v = dx/dt = -18 pi sin (3pi t + pi/3)

v = - 18 pi sin (6pi + pi/3)

v = 49.0 m/s

C. Acceleration is the 2nd derivative or the 1st derivative of velocity dv / dt
Acceleration a = dv/dt = -54 pi^2 cos(3pi t + pi/3)

a = -54 pi^2 cos(3pi * 2 + pi/3)

a = 386.34 m/s^2

D. Phase = pi/3

E. frequency f= 3pi/2pi = 1.5Hz

F. period = 1/f = 1/1.5 = .6667sec

5 0

4 years ago

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

madreJ [45]

Complete Question

A voltaic cell utilizes the following reaction and operates at 298 K:

3Ce4+(aq)+Cr(s)→3Ce3+(aq)+Cr3+(aq).

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

Answer:

The value is $E^o_{cell} = 2.35 V$

Explanation:

From the question we are told that

The ionic equation is

$3 Ce^{4 +} _{(aq)} + Cr _{(s)} \to 3 Ce^{3+} _{(aq)} + Cr^{3r} _{(aq)}$

Now under standard conditions the reduction half reaction is

$Ce^{4+} + e \to Ce^{3+} ; \ \ E^o_r = 1.61 V$

And the oxidation half reaction is

$Cr^{3+} + 3e^{-} \to Cr ; \ \ \ E^o_o = - 0.74 V$

The emf of this cell under standard conditions is mathematically represented as

$E^o_{cell} = E^o _r - E^o _o$

substituting values

$E^o_{cell} = 1.61 - (- 0.74)$

$E^o_{cell} = 2.35 V$

3 0

3 years ago

Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward force (thr

nordsb [41]

Answer:

$W=3456 N$

Explanation:

Force 1 $F_1=3456$

Force 2 $F_2=2333N$

Acceleration at stage 2 $a_2=0.39$

Generally the weight of the Craft W is given as

W= upward force(thrust)

Therefore

$W=3456 N$

5 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0

3 years ago

The force of the pulley is 652 newtons. It accelerates at 53 m/s. How much does the object weigh. a 1230 grams b 1.230 gram c 12

RoseWind [281]

Answer:

d. 12.30 grams

Explanation:

Given the following data;

Force = 652 N

Acceleration = 53 m/s²

To find how much the object weigh, we would need to determine its mass;

Force = mass * acceleration

652 = mass * 53

Mass = 652/53

Mass = 12.30 grams.

Therefore, the object weighs 12.30 grams.

5 0

3 years ago

Gravitational potential energy is a function of the _____ and the ________ of an object.

Gravitational potential energy is a function of the _ and the ____ of an object.