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Kruka [31]
4 years ago
10

If the mass of both weights is 225 gm, the first mass is located 20∘ north of east, the second mass is located 20∘ south of east

, and the transducer sensitivity is 0.5 volts/Newton, how large a voltage do you expect to measure? Assume the transducer has been properly zeroed so that V = 0 when F3=0. Please express your answers with 1 decimal place.
Physics
1 answer:
Montano1993 [528]4 years ago
7 0

Answer:

The voltage is 2.114 V.

Explanation:

Given that,

Mass of both weights = 225 gm

Transducer sensitivity = 0.5 V/N

The first mass is located 20∘ north of east, the second mass is located 20∘ south of east,

We need to calculate the net equivalent force

Using formula of force

F_{3}=m_{1}g\cos\theta+m_{2}g\cos\theta

F_{3}=2mg\cos\theta

Put the value into the formula

F_{3}=2\times0.225\times10\cos20^{\circ}

F_{3}=4.228\ N

We need to calculate the voltage

Using formula of voltage

Voltage =sensitivity\times F_{3}

Put the value into the formula

Voltage=0.5\times4.228

Voltage =2.114\ V

Hence, The voltage is 2.114 V.

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Answer:

\boxed {\boxed {\sf 2.1 \ kilograms}}

Explanation:

Kinetic energy is the energy an object possesses due to motion. The formula 1/2 the product of mass and the squared velocity.

E_k=\frac{1}{2} mv^2

We know the baseball's kinetic energy is 105 Joules. It is also traveling at a velocity of 10 meters per second. `

First, convert the units of Joules to make unit cancellation easier later in the problem. 1 Joule (J) is equal to 1 kilogram square meter per square second (kg*m²/s²). The baseball's kinetic energy of 105 J is equal to 105 kg*m²/s².

Now we know 2 values:

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Substitute these values into the formula.

105 \ kg*m^2/s^2= \frac{1}{2} m (10 \ m/s)^2

Now we need to solve for m, the mass. Solve the exponent.

  • (10 m/s)²= 10 m/s * 10 m/s = 100 m²/s²

105 \ kg *m^2/s^2 = \frac{1}{2} m (100 \ m^2/s^2)

Multiply on the right side.

105 \ kg *m^2/s^2 =  m (\frac{1}{2} * 100 \ m^2/s^2)

105 \ kg *m^2/s^2 =  m (50 \ m^2/s^2)

The variable, m, is being multiplied by 50 square meters per square second. The opposite of multiplication is division, so we divide both sides by that value.

\frac {105 \ kg *m^2/s^2 }{50 \ m^2/s^2}=  \frac{ m (50 \ m^2/s^2)}{50 \ m^2/s^2}

\frac {105 \ kg *m^2/s^2 }{50 \ m^2/s^2}= m

The units of square meter per square second will cancel out.

\frac {105 }{50} \ kg= m

2.1 \ kg=m

The mass of the baseball is <u>2.1 kilograms. </u>

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Answer:

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The ratio m_2/m_1 of the masses is 9.745

Explanation:

Formula for frequency when mass m_1 is hung on the spring:

f_1=\frac{1}{2\pi}\sqrt{\frac{k}{m_1}}

where:

k is the spring constant

Formula for frequency when mass m_2 is hung on the spring along with m_1:

f_2=\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}

where:

k is the spring constant.

In order to find ratio m_2/m_1, Divide the above equations:

\frac{f_1}{f_2} =\frac{ \frac{1}{2\pi}\sqrt{\frac{k}{m_1}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}}

On Solving the above equation:

\frac{f_1}{f_2} =\frac{\sqrt{\frac{k}{m_1}}}{\sqrt{\frac{k}{m_1+m_2}}}\\(\frac{f_1}{f_2})^{2} =\frac{m_1+m_2}{m_1} \\(\frac{11.8}{3.60})^2= \frac{m_1+m_2}{m_1} \\10.745=\frac{m_1+m_2}{m_1}\\10.745m_1=m_1+m_2\\m_2=10.745m_1-1m_1\\m_2/m_1=9.745

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