A 2.56×104-kg rocket blasts off vertically from the earth's surface with a constant acceleration. During the motion considered in the problem, assume that g remains constant. Inside the rocket, a 13.6-N instrument hangs from a wire that can support a maximum tension of 27.5 N .
a)Find the minimum time for this rocket to reach the sound barrier (330m/s)
without breaking the inside wire.
b)Find the maximum vertical thrust of the rocket engines under these conditions.
c)How far is the rocket above the earth's surface when it breaks the sound barrier?
a. The minimum time for the rocket to reach the sound barrier is 33.67 seconds.
b. The maximum vertical thrust of the rocket engines under these conditions is 250,893.6 N.
c. The rocket is 5548.1 meters above the Earth's surface when it breaks the sound barrier.
To solve this problem, we'll use Newton's second law of motion (F = ma) and consider the forces acting on the rocket and the instrument inside.
Calculating the minimum time for the rocket to reach the sound barrier without breaking the inside wire.
a) Minimum time to reach the sound barrier:
Given:
Mass of the rocket (m) = 2.56 × [tex]10^4[/tex] kg
Acceleration due to gravity (g) = 9.8 m/[tex]s^2[/tex]
Maximum tension the wire can support (T_max) = 27.5 N
Weight of the instrument (W) = mass of the instrument × acceleration due to gravity = 13.6 N
The forces acting on the instrument inside the rocket are its weight (W) and the tension in the wire (T). At maximum tension, the net force on the instrument is zero.
T - W = 0
T = W
Therefore, the maximum tension in the wire is equal to the weight of the instrument, which is 13.6 N.
Now, let's determine the acceleration of the rocket. The total force acting on the rocket is the sum of the rocket's weight (mg) and the tension in the wire (T).
F_total = mg + T
F_total = (2.56 × [tex]10^4[/tex] kg)(9.8 m/[tex]s^2[/tex]) + 13.6 N
F_total = 250,880 N + 13.6 N
F_total = 250,893.6 N
Since we're assuming the rocket's acceleration is constant.
we can use Newton's second law:
F_total = ma
250,893.6 N = (2.56 × [tex]10^4[/tex] kg)a
Solving for acceleration:
a = 250,893.6 N / (2.56 × [tex]10^4[/tex] kg)
a ≈ 9.8 m/[tex]s^2[/tex]
The acceleration of the rocket is approximately 9.8 m/[tex]s^2[/tex], which is the same as the acceleration due to gravity.
To find the minimum time to reach the sound barrier, we can use the following equation of motion:
v = u + at
where,
v = final velocity (sound barrier velocity = 330 m/s)
u = initial velocity (which is zero since the rocket starts from rest)
a = acceleration
t = time
330 m/s = 0 + (9.8 m/[tex]s^2[/tex])t
Solving for t:
t = 330 m/s / 9.8 m/[tex]s^2[/tex]
t ≈ 33.67 s
Therefore, the minimum time for the rocket to reach the sound barrier without breaking the inside wire is approximately 33.67 seconds.
b) Maximum vertical thrust of the rocket engines:
The maximum vertical thrust of the rocket engines is equal to the total force acting on the rocket, which we calculated earlier:
Maximum vertical thrust = F_total
Maximum vertical thrust ≈ 250,893.6 N
Therefore, the maximum vertical thrust of the rocket engines under these conditions is approximately 250,893.6 N.
c) Distance above the Earth's surface when breaking the sound barrier:
To determine the distance above the Earth's surface when breaking the sound barrier, we can use the following equation of motion:
s = ut + (1/2)at^2
where,
s = distance
u = initial velocity (which is zero)
a = acceleration
t = time it takes to reach the sound barrier (33.67 s).
s = 0 + (1/2)( 9.8 m/[tex]s^2[/tex])[tex](33.67 s)^2[/tex]
s = ([tex]4.9 m/s^2[/tex])(1132.8289 [tex]s^2[/tex])
s ≈ 5548.1 m
Therefore, the rocket is approximately 5548.1 meters (or 5.55 kilometers) above the Earth's surface when it breaks the sound barrier.
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A 5 kW, 230 V motor draws a current of 24 A from the supply. Determine the efficiency of this motor.
The efficiency of motor is 90.58%.To determine the efficiency of the motor, we need to calculate the input power and the output power, and then divide the output power by the input power
The input power can be calculated using the formula:
Input Power = Voltage × Current
Given that the voltage is 230 V and the current is 24 A, we have:
Input Power = 230 V × 24 A
Input Power = 5520 W (or 5.52 kW)
The output power of the motor is given as 5 kW (since it is a 5 kW motor).
Now, we can calculate the efficiency:
Efficiency = (Output Power / Input Power) × 100%
Efficiency = (5 kW / 5.52 kW) × 100%
Efficiency ≈ 90.58%
Therefore, the efficiency of this motor is approximately 90.58%.
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WRITE PLEASE: What distinguishes the 6 kingdoms from each other, according to taxonomic system described in this unit? Be sure to be specific and name each of the kingdoms in your description of their traits. ill mark brainliest
Answer:
:P
Explanation:
the six kingdoms of life, as described in the taxonomic system, are distinct from one another in a variety of ways. each kingdom has its own unique characteristics that set it apart from the others.
the first kingdom is the kingdom animalia, which is composed of multicellular organisms that are heterotrophic and motile. animals are capable of movement and have specialized organs and tissues that allow them to interact with their environment.
the second kingdom is the kingdom plantae, which is composed of multicellular organisms that are autotrophic and sessile. plants are capable of photosynthesis and have specialized organs and tissues that allow them to interact with their environment.
the third kingdom is the kingdom fungi, which is composed of multicellular organisms that are heterotrophic and sessile. fungi are capable of absorbing nutrients from their environment and have specialized organs and tissues that allow them to interact with their environment.
the fourth kingdom is the kingdom protista, which is composed of unicellular organisms that are either autotrophic or heterotrophic and motile. protists are capable of movement and have specialized organelles that allow them to interact with their environment.
the fifth kingdom is the kingdom monera, which is composed of unicellular organisms that are autotrophic and motile. monerans are capable of movement and have specialized organelles that allow them to interact with their environment.
the sixth kingdom is the kingdom archaea, which is composed of unicellular organisms that are autotrophic and motile. archaeans are capable of movement and have specialized organelles that allow them to interact with their environment.
in summary, the six kingdoms of life are distinct from one another in a variety of ways. each kingdom has its own unique characteristics that set it apart from the others. animals are multicellular and heterotrophic, plants are multicellular and autotrophic, fungi are multicellular and heterotrophic, protists are unicellular and either autotrophic or heterotrophic, monerans are unicellular and autotrophic, and archaeans are unicellular and autotrophic.
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2. When an object of mass m slides on a frictionless surface inclined at an angle as shown in the Figure below, the forces acting on it decides the a. acceleration of the object9jo b. speed of the object when it reaches the bottom h L 1 co a
The acceleration of the object in the inclined plane is g sinθ.
The velocity of the object on the inclined plane is √(2gL sinθ).
Given that the inclined surface is a frictionless surface. So, the force of friction is zero. Hence the components of the weight of the object provides the necessary forces to slide the object over the inclined plane.
a) Newton's second law is applied to masses on inclination.
Acceleration due to multiplied by the sine of the angle of inclination provides the acceleration for a frictionless slope of angle in degrees.
The acceleration of the object in the inclined plane is,
a = g sinθ
b) Applying the third equation of motion,
v²- u² = 2as
v² = 2as
v² = 2 x g sinθ x L
Therefore, the velocity of the object on the inclined plane is given by,
v = √(2gL sinθ)
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A kiddie roller coaster car has a mass 100 kilograms. At the top of a hill, it’s moving at a speed of 3 meters/second. After reaching the bottom of the hill, its speed doubles. The car’s kinetic energy at the bottom is what?
(1/2)mv^2 = (1/2) * 100 * (2*3)^2 = 1800 [J]
