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SVEN [57.7K]
3 years ago
15

A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c

onstant speed of 6.38 m/s is maintained, what is the tension in the string when the object is at the top of the circle.
Physics
1 answer:
Leya [2.2K]3 years ago
8 0

Answer:

T=83.37N

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

\sum F_y: T-mg=F_c

Where F_c is the centripetal force and is given by:

F_c=ma_c=m\frac{v^2}{r}

Replacing and solving for T:

T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N

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A long, thin solenoid has 450 turns per meter and a radius of 1.06 . The current in the solenoid is increasing at a uniform rate
kirill115 [55]

Answer:9.34 A/s

Explanation:

Given

radius of solenoid R=1.06 m

Emf induced E=8.50\times 10^{-6} V/m

no of turns per meter n=450

we know Induced EMF is given by

\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A

Magnetic Field is given by

B=\mu _0ni

thus \frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}

Area of cross-section

A=\pi R^2 where

solving integration we get

E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2

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\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}

\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}

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4 0
3 years ago
A woman is 160160cm tall. What is the minimum vertical length of a mirror in which she can see her entire body while standing up
kipiarov [429]

This question is incomplete, the complete question is;

A woman is 160cm tall. What is the minimum vertical length of a mirror in which she can see her entire body while standing upright.

Hint: Consider the ray diagram below, of the rays that enable her to see her feet and the top of her head.

Use what you know about the law of reflection, together with a bit geometry.

The missing Image is uploaded along this answer below.

Answer:

the minimum vertical length of a mirror in which she can see her entire body while standing upright is 80 cm

Explanation:

Given the data in the question and illustrated in the image below,

From image 2;

The distance from the woman's eyes to the top of her head is represented as b and a represent the distance from her eyes to her feet.

Therefore, since her height is 160 cm

a + b = 160 ------ let this be equation 1

i.e AD + DG = 160 cm

Now, from the same image 2, we will notice that triangle ABC and tringle CBD are similar, so

∠ABC = ∠CBD

AC = CD

since AD = a and AC + CD = A

AC = CD = a/2

Also, triangle DEF and FEG are si,ilar

∠DEF = ∠FEG

so

DF = FG

since DG = b and DF + FG = b

DF = b/2

so the minimum vertical length of a mirror in which the woman can see her entire body while standing upright will be;

⇒ a/2 + b/2  

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from equation 1, a + b = 160

so

⇒ a + b / 2 = 160 / 2 = 80

Therefore, the minimum vertical length of a mirror in which she can see her entire body while standing upright is 80 cm

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