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2 years ago

A 20 kg weight is suspended from a string. What is the tension on the string

2 answers:

7 0

The weight is in equilibrium, so the two forces acting on it (tension T pulling upward, and its own weight W pulling it downward) cancel each other. By Newton's second law,

∑ F = T - W = 0

so that

T = W

Calculate the object's weight:

W = (20 kg) (9.8 m/s²) = 196 N

So the tension is T = 196 N as well.

JulsSmile [24]2 years ago

3 0

assume that tension in string a is Ta and in string b is Tb

now taking the component of tensions

Tasin30∘=Tbsin30∘...(1)

it means Ta=Tb

Tacos30∘+Tbcos30∘=mg...(2)

⇒2Tacos30∘=20×10⇒Ta=Tb=3200

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A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$

substitute the values to get the height of image h'

$\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm$

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3 years ago

Which of the following best describe the particles in a liquid.

VashaNatasha [74]

A. fixed volume, changeable shape.

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2 years ago

Help me pls???!!

rosijanka [135]

Answer:

A. I = V / R = 12 / 252 = .048 amps

V = I * R = .048 * 252 = 12 V

V is also the reading the voltage across the battery (12 Volts)

4 0

2 years ago

What is the periodic table of elements? Question 5 options:a chart that organizes all the elements by their atomic number a cha

jok3333 [9.3K]

A chart that organizes all elements by their atomic number.

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3 years ago

An arrow is shot at an angle of 35° and a velocity of 50 m/s. How long does it take to return to its original starting height?

Ostrovityanka [42]

Answer:

4.02 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 35°

Initial velocity (u) = 50 m/s

Acceleration due to gravity (g) = 10 m/s²

Time of flight (T) =?

The time of flight of the arrow can be obtained as follow:

T = 2uSineθ / g

T = 2 × 35 × Sine 35 / 10

T = 70 × 0.5736 / 10

T = 7 × 0.5736

T = 4.02 s

Therefore, the time taken for the arrow to return is 4.02 s

4 0

3 years ago