Organized elements into four groups based on properties

A space craft is traveling in interplanetary space at a constant velocity. what is the estimated distance traveled by the space

mina [271]

Answer:

A. 2.30 x 10^2 kil

Explanation:

3/2 = 1.5 and 0.72/2 = 0.36

1.93 + 0.36 = 2.29

2.29 = about 2.30

7 0

3 years ago

A thin uniform rod (mass = 0.53 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plan

gladu [14]

Answer:

(a) L = 0·73 m

(b) 4·39 × $10^{-3}$ J

Explanation:

(a) From the figure, consider the torque about the point where the rod is attached because if we consider another point then there will be hinge forces acting on the rod at the point of attachment

Let L m be the length of the rod and β be the angle between the rod and the vertical

Let α be the angular acceleration of the rod

As the force of gravity acts at the centre so from the figure, the torque about the point of attachment will be 0·53 × g ×(L ÷ 2) ×sinβ

Assuming that the value of amplitude of this oscillation to be small

As torque = moment of inertia × angular acceleration

0·53 × g ×(L ÷ 2) ×sinβ = ((0·53 × L²) ÷ 3) × α (∵ moment of inertia of the rod from the point of attachment)

<h3>For small oscillations, α = ω² × β</h3>

After substituting the value of α and solving we get

ω = √((3 × g) ÷ (2 × L))

Time period = (2 × π) ÷ ω = (2 × π) ÷ √((3 × g) ÷ (2 × L))

∴ (2 × π) ÷ √((3 × g) ÷ (2 × L)) = 1·4

Substituting the value of g as 9·8 m/s² and solving we get

L = 0·73 m

(b) At the maximum amplitude condition the velocity will be 0 and potential energy will be maximum and maximum kinetic energy will be attained at the lowest point and hinge forces will not do work as the point of attachment is not moving

∴ Taking the reference for finding the potential energy as the lowest point

<h3>Maximum potential energy = Maximum kinetic energy </h3><h3>As total energy is constant, since there is no dissipative force</h3>

Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 (∵ increment in height is (L × (1 - cosβ)) ÷ 2

∴ Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 After substituting the value we get

Maximum potential energy = 4·39 × $10^{-3}$ J

∴ Maximum kinetic energy = 4·39 × $10^{-3}$ J

4 0

3 years ago

12. Unmanned Space Probe A 2500 kg unmanned space probe is moving in a straight line at a constant speed of 300 m/s. Control roc

Lena [83]

answer of both parts are attached below

4 0

4 years ago

Read 2 more answers

A mass suspended from a spring is oscillating up and down as indicated. Consider the following possibilities. A At some point du

dangina [55]

A mass suspended from a spring is oscillating up and down, (as stated but not indicated).

A). At some point during the oscillation the mass has zero velocity but its acceleration is non-zero (can be either positive or negative). <em>Yes. </em> This statement is true at the top and bottom ends of the motion.

B). At some point during the oscillation the mass has zero velocity and zero acceleration. No. If the mass is bouncing, this is never true. It only happens if the mass is hanging motionless on the spring.

C). At some point during the oscillation the mass has non-zero velocity (can be either positive or negative) but has zero acceleration. <em>Yes.</em> This is true as the bouncing mass passes through the "zero point" ... the point where the upward force of the stretched spring is equal to the weight of the mass. At that instant, the vertical forces on the mass are balanced, and the net vertical force is zero ... so there's no acceleration at that instant, because (as Newton informed us), A = F/m .

D). At all points during the oscillation the mass has non-zero velocity and has nonzero acceleration (either can be positive or negative). No. This can only happen if the mass is hanging lifeless from the spring. If it's bouncing, then It has zero velocity at the top and bottom extremes ... where acceleration is maximum ... and maximum velocity at the center of the swing ... where acceleration is zero.

7 0

3 years ago

A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is used to heat a house. The mass flow rate

Savatey [412]

(a)

Use â€œsaturated refrigerant-134a pressure tableâ€, to find $ , h , and of refrigerant-134a, at 320kPa (P) .

$h_{1} h_{g}$ = 251.93kJ/kg

$s_{1} (s_g}$ ) =0.93026 kJ/kg-K

$v_{1}(v_{g})$ =0.063681 m/kg

we solve $h_{6}$ ,

$h_{6}$ = 282.62 kj/kg

$h_{3} (h_{f)}$ = 127.25 kJ/kg

We use this formula for finding $Q_{n}$

$Q_n} =( h_{2} - h_{3} )$

= 38.8 kw

(b)

Find the COP of the heat pump ( $COP_{R}$ ) .

( $COP_{R}$ ) = $q_{L} /w_{in}$

$COP_{R}$ = (282.62 -127.25) / (282.62-251.93)

= 5.06

What is evaporator pressures ?

Valves for regulating evaporator pressure

Although the compressor suction pressure may be lower, evaporator pressure regulation (EPR) valves can be employed in the suction line to prevent the evaporator pressure from dropping below a set or controlled value.

An EPR valve is used for the following tasks:

1. Prevent potential damage to a liquid chilling evaporator from the liquid freezing.

2. Prevent frost from accumulating on an air-cooling evaporator when it is near freezing point or when operation cannot be interrupted by a brief failure.

3. Permit the operation of two or more evaporators with the same compressor at various load temperatures.

4. Vary the evaporator pressure in accordance with a fluctuating load that is managed by the load temperature.

Learn more about evaporator pressures visit this link:

brainly.com/app/ask?q=What+is+evaporator+pressures+

#SPJ4

7 0

1 year ago