What are density conversions
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Related Questions
Tobacco use can negatively impact a person's health, family, and finances.a. Trueb. False
Which image shows an example of the electromagnetic force in action?
What is the entropy change of 450 g of water when it changes from liquid to steam at its usual boiling point? For water, = 0.334 MJ/kg, = 2.26 MJ/kg.
The small particles that produce a streak of light upon entering earth’s atmosphere are called
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-T +10g = 10a
-15a =10g
a=-10g/15
a= -1/2 g
Negative a means block is accelerating downwards.
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the spring will be compressed by 0.3072 m
Explanation:
acceleration of elevator=3 m/s²
mass of student= 60 Kg
spring constant=2.5 x 10³ N/m
the force on the student is given by F = m ( g +a)
F=60 (9.8+3)
F=768 N
now the formula for spring force is given by
F= k x
768= 2.5 x 10³ (x)
x=0.3072 m
Final answer:
The spring on which a 60kg student is standing in an elevator accelerating at 3.0 m/s² is compressed by approximately 30.72 cm. This is calculated using Hooke's Law, considering both the weight of the student and the additional force due to the elevator's acceleration.
Explanation:
The situation you've described involves Hooke's Law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. In this case, we can consider the student's weight plus the extra force from the acceleration of the elevator.
To determine the compression of the spring, we use the equation F = kx, where F is the total force exerted on the spring, k is the spring constant, and x is the amount the spring is compressed. Here, the total force (F) includes the weight of the student and the force due to the elevator's acceleration. So, F = mg + ma, where m is the mass of the student, g is the gravity (9.8 m/s²), and a is the acceleration of the elevator.
Substituting the given values, we get F = (60 kg)(9.8 m/s²) + (60 kg)(3.0 m/s²) = 768 N. The compression of the spring (x) is now obtained by rearranging the Hooke's law formula to x = F/k.
This results in x = 768 N ÷ 2.5 x 10³ N/m = 0.3072 m or 30.72 cm. Thus, the spring is compressed by approximately 30.72 cm due to the combined force of the student's weight and the accelerating elevator.
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ultraviolet waves
light waves
sound waves
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Ultrasounds are used in the medical profession for the purpose of diagnosis. Ultrasounds uses high frequency sound waves. Thus, the correct option is D.
What are Ultrasounds?
Ultrasound are the sound waves with frequencies which are higher than the upper audible limit of normal human hearing. Ultrasound is not different from normal audible sound frequencies in its physical properties, except that humans cannot hear these sound.
Ultrasound is used in diagnostics where it is also called as sonography or diagnostic medical sonography. It is an imaging method which uses sound waves to produce images of structures within the body of a person. The images produced can provide valuable information for the purpose of diagnosing and directing treatment for a variety of diseases and conditions.
Therefore, the correct option is D.
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Answer:
D. sound waves
Explanation:
on edg
B) 10 km/hr.
C) 110 km/hr.
D) 1000 km/hr.
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The energy equation, E=12mvx2+12kx2=12kA2, is a useful alternative relationship between velocity and position, especially when energy quantities are also required. If the problem involves a relationship among position, velocity, and acceleration without reference to time, it is usually easier to use the equation for simple harmonic motion, ax=d2xdt2=−kmx (from Newton’s second law) or the energy equation above (from energy conservation) than to use the general expressions for x, vx, and ax as functions of time. Because the energy equation involves x2 and vx2, it cannot tell you the sign of x or of vx; you have to infer the sign from the situation. For instance, if the body is moving from the equilibrium position toward the point of greatest positive displacement, then x is positive and vx is positive.
IDENTIFY the relevant concepts
Energy quantities are required in this problem, therefore it is appropriate to use the energy equation for simple harmonic motion.
SET UP the problem using the following steps
Part A
The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.
Select all that apply.
Select all that apply.
maximum velocity vmax
amplitude A
force constant k
mass m
total energy E
potential energy U at x
kinetic energy K at x
position x from equilibrium
Part B
What is the kinetic energy of the object on the spring when the spring is compressed 5.1 cm from its equilibrium position?
Part C
What is the potential energy U of the toy when the spring is compressed 5.1 cm from its equilibrium position?
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Answer:
Part A
Mass = 50g
Vmax = 3.2m/s
Amplitude= 6cm
Position x from the equilibrium= 5.1cm
Part B
Kinetic energy = 0.185J
Part C
Potential energy = 0.185J
Explanation:
Kinetic energy = 1/2mv×2
Vmax = wa
w = angular velocity= 53.33rad/s
Kinetic energy = 1/2mv^2×r^2 = 0.185J
Part c
Total energy = 1/2m×Vmax^2= 0.256J
1/2KA^2= 0.256J
K= 142.22N/m (force constant)
Potential energy = 1/2kx^2
=1/2×142.22×0.051^2
= 0.185J
Final answer:
To find the kinetic energy of the toy, we need to use the energy equation for simple harmonic motion and the relationship between velocity and position. We can then substitute the known values to calculate the kinetic energy.
Explanation:
In this problem, we are given the amplitude (A) of the oscillation and the maximum velocity (vmax) achieved by the toy. We need to find the kinetic energy (K) of the toy when the spring is compressed 5.1 cm from its equilibrium position.
To solve for the kinetic energy, we can use the energy equation for simple harmonic motion: K = 1/2mvx2, where m is the mass of the object and vx is the velocity of the object at position x. The mass of the object is given as 50 g, which is equal to 0.05 kg.
Since we know the maximum velocity (vmax = 3.2 m/s), we can use the relationship between velocity and position in simple harmonic motion to find the velocity (vx) at a displacement of 5.1 cm from the equilibrium position. The velocity and position in simple harmonic motion are related by vx = ±ω√(A2 - x2), where ω is the angular frequency of the motion.
Substituting the known values into the equations, we can calculate the kinetic energy of the toy.
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