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3 years ago

Simple Circuit and Ohm's Law Check-for-Understanding

Physics

2 answers:

kobusy [5.1K]3 years ago

7 0

Current only.

(But in the real world, also power supplied by the battery, power dissipated by the resistor, temperature of the resistor, and temperature of the air around the resistor.)

Vanyuwa [196]3 years ago

4 0

Answer:

current, only

Explanation:

current:I

voltage:U

resistance:R

formula: I=U/R

Increasing the battery cause the increasing in the voltage. Resistance does not normally change. And the current would increase.

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A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1

pav-90 [236]

I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
$F_a=F_cf-F_g$
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
$F_a=m\frac{v^2}{r}-mg=179 $N$
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
$F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N$
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
$F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}$

6 0

3 years ago

A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c

viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = $71\times 10^{- 9} C$

Q' = + 42 nC = $42\times 10^{- 9} C$

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

$\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E} = 708.03 N/C$

Now,

Electric field at the mid-point due to charge Q' is given by:

$\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E'} = 418.84 N/C$

Now,

The net Electric field is given by:

$\vec{E_{net}} = \vec{E} - \vec{E'}$

$\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C$

5 0

3 years ago

You measure the velocity of a drag racer that accelerates with constant acceleration. You want to plot the data and determine th

Ivenika [448]

Answer:

a) Linear equation

Explanation:

Definition of acceleration

$a=\frac{dv}{dt}\\$

if a=constant and we integrate the last equation

$v(t)=v_{o}+a*t$

So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

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2 years ago

A SI unit used to measure force, equal to less than one-quarter of a pound, is the what

bulgar [2K]

The SI unit of force is the Newton.

1 newton is the force that accelerates a 1 kilogram mass
at the rate of 1 meter per second².

1 pound of force is equivalent to roughly 4.448 newtons.

(1 newton is equivalent to roughly 0.225 pounds of force.)

4 0

2 years ago

what are three activities in your everyday life that involve electromagnetic energy? and explain what type of electromagnetic en

mihalych1998 [28]

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in some way with MICROWAVE energy, since I work as a microwave

system design engineer. Most of the time it's just paperwork, but there

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8 0

3 years ago