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2 years ago

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Given a triangle with sides X = 6.35 cm and Y = 12.25 cm with an angle of 90 degrees between them, find the length of the hypote

nuse and the size of the other two angles.

1 answer:

never [62]2 years ago

4 0

Answer:

the hypotenuse = 13.78 cm

Ф = 27.44°

θ = 62.56°

explanation:

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Two blocks, with masses m and 3m, are attached to the ends of a string with negligible mass that passes over a pulley, as shown

olasank [31]

Answer:

a) v = √ g x , b) W = 2 m g d , c) a = ½ g

Explanation:

a) For this exercise we use Newton's second law, suppose that the block of mass m moves up

T-W₁ = m a

W₃ - T = M a

w₃ - w₁ = (m + M) a

a = (3m - m) / (m + 3m) g

a = 2/4 g

a = ½ g

the speed of the blocks is

v² = v₀² + 2 ½ g x

v = √ g x

b) Work is a scalar, therefore an additive quantity

light block s

W₁ = -W d = - mg d

3m heavy block

W₂ = W d = 3m g d

the total work is

W = W₁ + W₂

W = 2 m g d

c) in the center of mass all external forces are applied, they relate it is

a = ½ g

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2 years ago

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Suppose a certain scale is not calibrated correctly, and as a result, the mass of any object is displayed as 0.75 kilogram less

Papessa [141]

Answer

A certain scale is not calibrated correctly,

Mass displayed as 0.75 kilogram less than its actual mass.

The scale of display will be equal to '1'

The correlation between actual mass and displayed mass is 1 because correlation is independent of change of origin and scale and correlation between any value and value minus 0.75 will be one.

So, the correct answer will be "1"

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2 years ago

Air pressure is 1.0 · 105 N/m2, air density is 1.3 kg/m3, and the density of soft drinks is 1.0 · 103 kg/m3. If one blows carefu

natita [175]

Answer:

v = 27.456 m/s

Explanation:

The support pressure needed of the water in the straw can be calculated by the formula

Given that,

P = r*g*h

= 1000*9.8*0.05 Pa.= 490 Pa

This pressure is compensated by 0.5*r*v^2 of the air,

Hence,

0.5*1.3*v^2 = 490

velocity of air blown into the straw =

v = 27.456 m/s

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2 years ago

Which type of wave can transmit energy through a vacuum?

m_a_m_a [10]

The wave that can transmit energy through a vacuum is a electromagnetic wave

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2 years ago

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Two objects (45.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h

Oksanka [162]

a) The acceleration of the objects is $3.56 m/s^2$

b) The tension in the string is 280.8 N

Explanation:

a)

We start by writing the equations of motion for the two masses attached to the pulley.

For the heavier mass, we have:

$m_1 g - T = m_1 a$ (1)

where

$m_1 = 45.0 kg$ is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

T is the tension in the string

a is the acceleration of the system (here we assumed that the heavier mass accelerates downward)

For the lighter mass, we have

$T-m_2 g = m_2 a$ (2)

where

T is the tension in the string

$m_2 = 21.0 kg$ is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

a is the acceleration of the system (here we assumed that the lighter mass accelerates upward)

From (1) we get

$T=m_1g - m_1 a$

And substituting into (2),

$(m_1 g - m_1 a)-m_2 g = m_2 a\\(m_1 -m_2)g = (m_1+m_2)a\\a=\frac{m_1 - m_2}{m_1+m_2}g=\frac{45-21}{45+21}(9.8)=3.56 m/s^2$

b)

From the previous part of the problem we got an expression for the tension in the string:

$T=m_1g - m_1 a$

Where we have

$m_1 = 45.0 kg$

$g=9.8 m/s^2$

$a=3.56 m/s^2$ is the acceleration, found in part a)

Susbtituting, we find

$T=(45.0)(9.8)-(45.0)(3.56)=280.8 N$

Learn more about forces and acceleration:

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2 years ago