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mojhsa [17]
3 years ago
11

Tina has been dieting for a total of 13 weeks she lost 3 lb on the first week of her diet but gain the back of pound on the seco

nd week on each remaining week of her diet she lost 2 lb how many pounds has Tina lost in all a. 26 lb B. 22 lb C. 20 lb D. 24 lb
Physics
1 answer:
PolarNik [594]3 years ago
6 0

Answer:

D. 24 lb

Explanation:

Tina has been dieting for 13 weeks

First week she lost 3 pounds

Next week she gained 1 pound and did not lose any. This will be subtracted as she has gained a pound

The remaining 11 weeks she lost 2 pounds per week

Weight lost in the 11 weeks = 11×2 =22 pounds

Total weight lost

3-1+22 = 24 lb

Tina has lost 24 pounds in total during the 13 weeks

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A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet
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Answer:

Acceleration = 2.35 m/s^{2}

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = \frac{mg}{2} = 0.5 mg

Friction along incline = umg cos30° = mg 0.3\times \frac{\sqrt{3}}{2}

Friction along incline  = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = \frac{net force}{mass} = 0.24 g = 2.35 m/s^{2}

The height of incline = 8 m

Length of the inclined edge = 16 m

v^{2}=u^{2}+2as

v^{2}= 2\times 0.24 \times 9.8\times 16

v= 8.67 m/s

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QUESTION 1<br> 67.2 kilometers = how many decameters
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Answer: the answer should be 6,720 decameters.

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List five situations in which temporary wiring would be used
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an emergency

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temp joining components

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At which point does the planet have the least gravitational force acting on it?
Elza [17]

Answer:

At which point does the planet have the least gravitational force acting on it?

Explanation:

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A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (
SIZIF [17.4K]

The solution to the questions are given as

  • t=40.39 \mathrm{sec}
  • \varepsilon &=(0.12v)e^{0.057t}
  • the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}

\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$

\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

\varepsilon &=(0.12v)e^{0.057t}

(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}

c)

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

brainly.com/question/13076734

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2 years ago
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