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3 years ago

Can someone help me on #18 - #26? At least one ☝️ I really need help!!!

Physics

1 answer:

Novosadov [1.4K]3 years ago

3 0

#18). (I think. It's the one that starts with "Compare..."
Gravity ALWAYS attracts.
The force between electric charges can attract or repel ...
it depends whether the charges are the same kind or opposite kinds.

#19).
With both gravity and electric charges, the force between them quickly
becomes weaker when the distance between them increases.

#20).
I don't think it changes.
If the doorknob gets charged by something that TOUCHES it, so that
charges can flow into it from the other object or out of it, then the total
amount of electric charge on it might change. But the question says
that the doorknob is charged by an "electric field", so nothing touched it,
and charges couldn't flow into it or out of it. The only way it got charged
was by the charges it already had in it getting moved around ... electrons
in one part of the knob moving over to the other side. Then it would act
as if it was charged ... if you touched it, you might get zapped.

#21)., #22)., #23).
You're supposed to draw a graph to answer these.
It's a very easy graph to draw, and you should do it.
Label the x-axis 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Label the y-axis 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2 .
Put the four points on the graph ... A, B, C, and D.
For each point, the 'battery voltage' is the number on the x-axis
and the 'Current' is the number on the y-axis.
Then draw a line through the points.
When you have the graph to look at, you can easily answer 21, 22, and 23.

24). I'm not sure, and I don't want to guess.

25). Did you ever move a coil of wire near a magnet in class ?
This is the same situation, only the magnet is moving and the wire is still.
The result will be the same.

26). Magnets have two poles that attract
the opposite kind and repel.
Now you copy the map and fill in the other side.

This is a lot of work for 5 points, so I left some parts for you to do.
Another reason I did that is: You'll learn a lot more that way.

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3 years ago

What is the equivalent resistance of the circuit?

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3 years ago

Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s

nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

$x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}$

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

$v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}$

with this value of vo you calculate the max time:

$t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s$

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

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TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0

3 years ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ $\frac{du}{dy}$

= µ $\frac{v}{h}$ ............1

put here value

= 1.75× $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

and

area between air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

and now apply newton second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0

3 years ago