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3 years ago

On a velocity time graph, what happens when the line crosses the x axis?

Physics

1 answer:

Crank3 years ago

5 0

Answer:

the object is changing direction

Explanation:

During straight line motion, On a velocity vs. time graph, any time the line crosses the x axis, the object is changing direction

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When radio waves try to pass through a city, they encounter thin vertical slits: the separations between the buildings. This cau

Westkost [7]

Answer:

$\theta= sin^{-1}(\frac{\lambda}{a})$

Explanation:

Assuming we have to find

Find the angle theta to the first minimum from the center of the central maximum (Expressing answer in terms of λ and a.):

a= avg separation between the building.

λ = wavelength

for single slit diffraction we have

a sinθ =nλ

for the first minimum n=1

a sinθ =λ

therefore,

$\theta= sin^{-1}(\frac{\lambda}{a})$

4 0

4 years ago

The unit of length most suitable for measuring the thickness of a cell phone is a ( megameter, meter, millimeter, nanometer ) Th

puteri [66]

Answer: 1.millimeter 2.meter

Explanation:

a phones thickness can be measured with such a small measurement and a tree is not big enough for a megameter but not small enough for a millimeter so it'd be a meter

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

A spring is 4 cm long. A student hangs a weight of 2N. It is now 5.5 cm long.

jeka94

(i)1.5cm
(ii)4.5cm
(iii)10cm

4 0

2 years ago

You prepare tea with 0.250 kg of water at 85.0 ºC and let it cool down to room temperature (20.0 ºC) before drinking it. Essenti

vladimir2022 [97]

Answer:

232 J/K

Explanation:

The amount of heat gained by the air = the amount of heat lost by the tea.

q_air = -q_tea

q = -mCΔT

q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)

q = 68,000 J

The change in entropy is:

dS = dQ/T

Since the room temperature is constant (isothermal):

ΔS = ΔQ/T

Plug in values (remember to use absolute temperature):

ΔS = (68,000 J) / (293 K)

ΔS = 232 J/K

5 0

3 years ago