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3 years ago

Two ways to make peace with your body are to create a positive mental outlook and to

Physics

1 answer:

ozzi3 years ago

4 0

Answer:

The correct answer is option c. "give yourself credit for your attractive body parts".

Explanation:

A common problem that people have is to feel uncomfortable with their body, which results in many serious problems including depression, anxiety and low self-stem. Two ways to make peace with your body are to create a positive mental outlook and to give yourself credit for your attractive body parts. Nobody is perfect, and there is always some things that we like about us and some things that we don't like. Giving yourself credit for your attractive body parts gives motivation, confidence and self appreciation, which helps in making peace with your body.

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Trên bóng đèn có ghi là 12v -6w. Khi đèn sáng bình thường thì dòng điện chạy qua đèn có cường độ là

Annette [7]

Answer:

Xin lỗi, ở đây không có ai nói tiếng Việt, nhưng bạn có thể cuộn hết cỡ xuống dưới để tìm một trang web cho não biết nói tiếng Việt

Explanation:

4 0

3 years ago

Two 1.0 cm * 2.0 cm rectangular electrodes are 1.0 mm apart. What charge must be placed on each electrode to create a uniform el

kvv77 [185]

Answer:

The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

Explanation:

<u>Step 1:</u> calculate the charge on each electrode

Given;

Electric field strength = 2.0 X 10⁶ N/C

The distance between the electrode = 1mm = 1 X 10⁻³ m

Electric field strength (E) = Force (F)/Charge (q)

$E =\frac{Kq}{r^2}$

where;

E is the electric field strength = 2.0 X 10⁶ N/C

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

r is the distance between the electrodes = 1 X 10⁻³ m

q is the charge in each electrode = ?

$q = \frac{Er^2}{K} = \frac{(2X10^6)(1X10^{-3})^2}{8.99 X10 ^9}$ = 0.2225 X 10⁻⁹ C

The charge on each electrode is 0.2225 X 10⁻⁹ C

<u>Step 2:</u> calculate the number of electrons to be moved from one electrode to the other.

1 electron contains 1.602 X 10⁻¹⁹ C

So, 0.2225 X 10⁻⁹ C will contain how many electrons ?

= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)

= 1.4 X 10⁹ electrons

Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

8 0

3 years ago

When an automobile moves with constant velocity, the power developed is used to overcome the frictional forces exerted by the ai

geniusboy [140]

Answer:

The total frictional force is 358.0 newtons

Explanation:

Power is the amount of average work (W) an object does on a period of time (Δt):

$P=\frac{W}{\Delta t}$

Remember average work is average force (F) times displacement (Δs):

$P=\frac{F \Delta s}{\Delta t}$

but displacement over time is average speed $v=\frac{\Delta s}{\Delta t}$ , then:

$P=Fv$ (1)

That is, the power of the car is the force the engine does times the speed of the car. As the question states, if the car is at constant velocity then the power developed is used to overcome the frictional forces exerted by the air and the road, that is by Newton's first law, the force the motor of the car does is equal the force of frictional forces. So, to find the frictional forces we only have to solve (1) for F:

$F=\frac{P}{v}$

Knowing that 1hp is 746W then 30hp=22380W and 1 mile = 1609m then 140 mph = 225308 $\frac{m}{h}$ = $62.5 \frac{m}{s}$ , then:

$F=\frac{22380}{62.5}=358.0 N$

6 0

3 years ago

Read 2 more answers

Which formula can be used to find the tangential speed of an orbiting object?

Luden [163]

The correct formula for calculating the tangential speed of an orbiting object is V(t)=wr.
V(t)= Tangential Speed
w= Angular Velocity
r= Radius of the Path

Hope this helps.

6 0

3 years ago

Read 2 more answers

If an athlete leaps vertically at 4.0m/s, what maximum height does he reach?

joja [24]

Hello
This is a problem of accelerated motion, where the acceleration involved is the gravitational acceleration: $g=-9.81~m/s^2$ , and where the negative sign means it points downwards, against the direction of the motion.

Therefore, we can use the following formula to solve the problem:
$v_f^2 = v_i^2 + 2gS$
where $v_i=4~m/s$ is the initial vertical velocity of the athlete, $v_f=0$ is the vertical velocity of the athlete at the maximum height (and $v_f=0~m/s$ at maximum height of an accelerated motion) and S is the distance covered between the initial and final moment (i.e., it is the maximum height). Re-arranging the equation, we get
$S= \frac{v_f^2-v_i^2}{2g}=0.82~m$

3 0

3 years ago