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3 years ago

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An object is represented by the dot on a motion map. What is the best description for the motion of this object? The object is m

oving with a constant speed. The object is moving with a constant acceleration. The object is standing still. The object is falling.

2 answers:

ankoles [38]3 years ago

5 0

The object is standing still.

alexira [117]3 years ago

5 0

Answer:

C. The object is standing still

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When the input voltages of a difference amplifier are 5.1 v and 6.4 v, the output voltage is 64.7 v. The inputs are changed to 4

daser333 [38]

Answer:

a) Differential mode gain = 48

b) Common mode gain = 0.4

c) CMRR = 120

Explanation:

The output of a difference amplifier is related to the input by the equation:

$V_{0} = A_{1} V_{1} + A_{2} V_{2} \\$

When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes

6.4 A₁ + 5.1 A₂ = 64.7.....................(1)

When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes

5.6 A₁ + 4.9 A₂ = 35.7.....................(2)

Multiply equation (1) by 5.6 and (2) by 6.4

35.84 A₁ + 28.56A₂ = 362.32.....................(3)

35.84 A₁ + 31.36 A₂ = 228.48....................................(4)

Subtract equation (3) from (4)

2.8 A₂ = -133.84

A₂ = -133.84/2.8

A₂ = -47.8

Put the value of A₂ into equation (1)

6.4 A₁ + 5.1 (-47.8) = 64.7

6.4 A₁ = 64.7 + 243.78

A₁ = 308.48/6.4

A₁ = 48.2

a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)

Common mode gain = 0.4

b) Differential mode gain = (A₁ -A₂)/2

Differential mode gain = (48.2 - (-47.8))/2

Differential mode gain = 96/2

Differential mode gain = 48

c) Common Mode Rejection Ratio (CMRR)

$CMRR = |\frac{Differential Mode Gain}{Common Mode Gain} |$

$CMRR = |\frac{48}{0.4} |\\CMRR = 120$

4 0

2 years ago

A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor

Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force $F_{1}=6\ N$

Second force $F_{2}=10\ N$

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

$\tau=Force\times lever\ arm$

So, Here,

$\tau_{2}=5 \tau_{1}$

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

$F_{2}\times d_{2}=5\times F_{1}\times r_{1}$

$r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}$

Where, $F_{1}$ =First force

$F_{1}$ =First force

$F_{2}$ =Second force

$r_{1}$ = distance

Put the value into the formula

$r_{2}=\dfrac{5\times6\times0.2}{10}$

$r_{2}=0.6\ m$

Hence, The moment arm is 0.6 m

6 0

3 years ago

Combine these three velocity vectors into a resultant: 3.0 m/s north, 4.0 m/s east 1.0 m/s west. Identify the resultant vector

Slav-nsk [51]

Answer:

The resultant vector is 1 m/s

Explanation:

The resultant vector is 1 m/s west based on triangle law of vector addition, when two sides of a triangle is represented by two vectors, the resultant vector is the third side of the triangle.

5 0

3 years ago

What is the relationship between the electric force and each one of the charges?

Bess [88]

Explanation: Electrostatic force is directly related to the charge of each object. So if the charge of one object is doubled, then the force will become two times greater.

3 0

2 years ago

Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4

zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is $c_p=1.0\ Btu/lbm.F$ , and the steam properties as, A-4E:

$h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R$

Using the energy balance for the system:

$\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s$

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

$\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s$

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

$\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R$

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0

3 years ago