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Alenkasestr [34]
3 years ago
7

An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads 65 N when the cab is stand

ing still. What is the reading when the cab is moving upward (a) with a constant speed of 7.6 m/s and (b) with a speed of 7.6 m/s while decelerating at a rate of 2.4 m/s2
Physics
1 answer:
lana66690 [7]3 years ago
7 0

Answer:

(A) Reading will be 65 N

(B) Net force on the elevator will be 49.076 N    

Explanation:

We have given the balance force = 65 N

Acceleration due to gravity g=9.8m/sec^2

We know that W=mg

So 65=m\times 9.8

m = 6.632 kg

(a) In first case as the as the speed is constant so the force on the elevator will be 65 N

(B) In second case as the elevator is decelerating at a rate of 2.4m/sec^2

So net acceleration = 9.8-2.4=7.4m/sec^2

So net force on elevator will be = m× net acceleration = 6.632×7.4 = 49.076 N

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Answer:

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6.55 x 10^-2 / 740 = pi * (2.67 x 10^-3)^2 * speed

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6 0
3 years ago
A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left
Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

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  • Distance of support from the left end; x = 3.00m
  • First mass; m1 = 31.3 kg
  • Distance of beam from  the left end( m₁ is attached to ); x_1 = ?
  • Second mass; m_2 = 61.7 kg
  • Distance of beam from  the right of the support( m₂ is attached to ); x_1 = 0.273m

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, m_1g( x-x_1) = m_2gx_2

we divide both sides by g

m_1( x-x_1) = m_2x_2

Next, we make x_1, the subject of the formula

x_1 = x - [ \frac{m_2x_2}{m_1} ]

We substitute in our given values

x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]

x_1 = 3.00m - 0.538m

x_1 = 2.46m

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

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3 years ago
Which part of the drum vibrates when it produces sound?
SVETLANKA909090 [29]

Explanation:

For example, when a drum is struck, the flexible skin (sometimes called a membrane) of the drum vibrates. The compression and expansion of the air on either side of the vibrating membrane produces differences in air pressure. The pressure differences generate a sound wave that propagates outward from the drum surface.

4 0
3 years ago
The displacement vector A and B when added together , give the resultant vector R so that R= A+B use the data in the drawing and
Salsk061 [2.6K]
The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.

Let:

b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°

We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.

Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.

In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
 
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A certain car engine delivers enough force to create 630 N⋅m of torque when the engine is operating at 3200 revolutions per minu
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The appropriate expression for the calculation of power by relating the angular energy in a given time.

In other words the instantaneous power of an angular accelerating body is the torque times the angular velocity

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Our values are given by

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Replacing,

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