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Kruka [31]
3 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]3 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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A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force
djyliett [7]

Answer:

a) U = 0 J    

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

4 0
3 years ago
A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is
Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

           P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

         P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

            P_atm V₁ = (P_atm + ρ g H) V₂

             V₂ = V₁    P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

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Answer: The motions shifts.

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What is the momentum of a 5 kg object that has a velocity of 1. 2 m/s? 3. 8 kg • m/s 4. 2 kg • m/s 6. 0 kg • m/s 6. 2 kg â
nevsk [136]

The momentum of a 5kg object that has a velocity of 1.2m/s is 6.0kgm/s.

<h3> MOMENTUM:</h3>

Momentum of a substance is the product of its mass and velocity. That is;

Momentum (p) = mass (m) × velocity (v)

According to this question, an object has a mass of 5kg and velocity of 1.2m/s. The momentum is calculated thus:

Momentum = 5kg × 1.2m/s

Momentum = 6kgm/s.

Therefore, the momentum of a 5kg object that has a velocity of 1.2m/s is 6.0kgm/s.

Learn more about momentum at: brainly.com/question/250648?referrer=searchResults

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3 years ago
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