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3 years ago

15

Which term defines the energy of motion?

1 answer:

Korvikt [17]3 years ago

4 0

D kinetic energy✨
nnnnnnnnnnnnbbbhhbbbbbbb

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Responsible for reflex, maintain muscles tone and dopamine. 8 letter word

Lady bird [3.3K]

Exercise is the answer hope i helped you

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3 years ago

A typical running track is an oval with 74-mm-diameter half circles at each end. A runner going once around the track covers a d

lisabon 2012 [21]

The centripetal acceleration a is 4.32 $\times$ 10^-4 m/s^2.

<u>Explanation:</u>

The speed is constant and computing the speed from the distance and time for one full lap.

Given, distance = 400 mm = 0.4 m, Time = 100 s.

Computing the v = 0.4 m / 100 s

v = 4 $\times$ 10^-3 m/s.

radius of the circular end r = 37 mm = 0.037 m.

centripetal acceleration a = v^2 / r

= (4 $\times$ 10^-3)^2 / 0.037

a = 4.32 $\times$ 10^-4 m/s^2.

6 0

3 years ago

A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle

liq [111]

Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

$h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft$

Now y = 2500 ft

$cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974$

$\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s$

The angle is changing at 0.0031792338 rad/s

6 0

3 years ago

A box experiencing a gravitational force of 600 N is being pulled to the right with force of 250 N. A 25 N frictional force acts

bearhunter [10]

Answer: 0 NEWTONS

Explanation:

6 0

3 years ago

A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

Let $N_p'\ and\ N_s'$ are the new number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0

3 years ago