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4 years ago

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A machine part rotates at an angular speed of 0.76 rad/s; its speed is then increased to 2.96 rad/s using an angular acceleratio

n of 1 rad/s 2 . Find the angle through which the part rotates before reaching this final speed. Answer in units of rad.

1 answer:

son4ous [18]4 years ago

5 0

Answer:

θ = 4.092 °

Explanation:

Given that

Initial Angular speed ,ωi = 0.76 rad/s

Final angular speed ,ωf= 2.96 rad/s

The angular acceleration ,α = 1 rad/s²

We know that

ωf² = ωi ²+ 2 α θ

θ=Angle rotates

2.96² = 0.76²+ 2 x 1 x θ

$\theta=\dfrac{2.96^2-0.76^2}{2}\ degrees\\\theta=4.092\ degrees\\$

Therefore the angle rotates by machine part will be 4.092 degrees.

θ = 4.092 °

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A concave mirror with a radius of curvature of 20 cm has a focal length of

xxTIMURxx [149]

Answer:

A concave mirror has a radius of curvature of 20 cm. What is it's focal length? If an object is placed 15 cm in front of it, where would the image be formed? What is it's magnification?

The focal length is of 10 cm, object distance is 30 cm and magnification is -2.

Explanation:

Given:

A concave mirror:

Radius of curvature of the mirror, as C = 20 cm

Object distance in-front of the mirror = 15 cm

a.

Focal length:

Focal length is half of the radius of curvature.

Focal length of the mirror = $\frac{C}{2}$ = 10 cm

According to the sign convention we will put the mirror on (0,0) point, of the Cartesian coordinate open towards the negative x-axis.

Object and the focal length are also on the negative x-axis where focal length and image distance will be negative numerically.

b.

We have to find the object distance:

Formula to be use:

⇒ $\frac{1}{focal\ length}= \frac{1}{image\ distance} + \frac{1}{object\ distance}$

⇒ Plugging the values.

⇒ $\frac{1}{-10} =\frac{1}{image\ distance}+\frac{1}{-15}$

⇒ $\frac{1}{-10} -\frac{1}{-15}=\frac{1}{image\ distance}$

⇒ $\frac{1}{-10} + \frac{1}{15}=\frac{1}{image\ distance}$

⇒ $\frac{-3+2}{30} =\frac{1}{image\ distance}$

⇒ $\frac{-1}{30} =\frac{1}{image\ distance}$

⇒ $-30\ cm=image\ distance$

Image will be formed towards negative x-axis 30 cm away from the pole.

c.

Magnification (m) is the negative ratio of mage distance and object distance:

⇒ $m=-\frac{image\ distance}{object\ distance}$

⇒ $m=-\frac{(-30)}{(-15)}$

⇒ $m=-2$

The focal length of the concave mirror, is of 10 cm, object distance is 30 cm and magnification is -2.

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3 years ago

Q1:A large tank is filled with water. The pressure on the base of the fish tank is 4000N/m². The base of the tank is a rectangle

Ymorist [56]

Answer:

F = 36 kN

Explanation:

It is given that,

The pressure on the base of the fish tank is 4000N/m².

The base of the tank is a rectangle measuring 2.0m by 4.5m.

Area of the base of the tank is 9 m²

We need to find the force on the base caused by the base of the water. Pressure on the base of the tank is given by the force acting per unit area such that,

$P=\dfrac{F}{A}\\\\F=P\times A\\\\F=4000\times 9\\\\F=36000\ N\\\\F=36\ kN$

So, the force of 36 kN is acting on the base by the base of water.

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4 years ago

A box with sides of length 40 cm is made up of six pieces of concrete of thickness 5 mm. The inside of the box is filled with 10

Damm [24]

Answer:

d)2.13 C s⁻¹

Explanation:

Rate of flow of heat through walls

= $\frac{KA(T_2-T_1)}{d}$

K = .33

A = 6 X .4 X .4 =0.96

T₂-T₁ = 30+40 = 70

d = 5 x 10⁻³

Put these data in the relation above

Rate of flow of heat

= $\frac{.33\times.96\times70}{5\times10^{-3}}$

= 4435.2 Js⁻¹

Specific heat of gas = 2.5 R = 20.785 J

Rise in temp = $\frac{4435.2}{100\times20.785}$

= 2.13 degree celsius.

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3 years ago

Organize the three states of matter from least kinetic energy to greatest kinetic energy

Yuri [45]

Particle arrangement and movement

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<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em>

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3 years ago

Two thin rectangular sheets (0.16 m x 0.44 m) are identical. In the first sheet the axis of rotation lies along the 0.16-m side,

malfutka [58]

Answer:

$t'=1.124s$

Explanation:

$I_b*a_b=I_a*a_a$

$a_a=\frac{w_f}{t'}$

$a_b=\frac{w_f}{t}$

$I_b*\frac{w_f}{t}=I_a*\frac{w_f}{t'}$

$I_b*\frac{1}{t}=I_a*\frac{1}{t'}$

$I=\frac{1}{3}*M*r^2$

$I_a=\frac{1}{3}*M*r_a^2$

$I_b=\frac{1}{3}*M*r_b^2$

$\frac{1}{3}*M*r_b^2*\frac{1}{t}=\frac{1}{3}*M*r_a^2*\frac{1}{t'}$

$t'=(\frac{r_a}{r_b})^2 *t$

$t'=(\frac{0.16m}{0.44m})^2*t$

$t'=\frac{16}{121}*8.5s$

$t'=1.124s$

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3 years ago