Answer:
A. ) K =126. 7 J
B. ) h= 91.1 m.
Explanation:
A)
Assuming no air resistance, once released by the pitcher, the speed must keep constant through all the trajectory, so the kinetic energy of the ball can be expressed as follows:[tex]K = \frac{1}{2}*m*v^{2} = \frac{1}{2}*0.142 kg*(42.24m/s)^{2} = 126.7 J (1)[/tex]
B)
Neglecting air resistance, total mechanical energy must be the same at any point, so, if we choose the ground level as the zero reference level for the gravitational potential energy, and assuming that the ball attains this kinetic energy just before striking ground, this value must be equal to the gravitational potential energy just before be dropped, so we can write the following equality:[tex]U_{o} = K_{f} = 126. 7 J (2)[/tex]
⇒ m*g*h = 126. 7 J
Solving for h, we get:[tex]h = \frac{K_{f}}{m*g} = \frac{126.7J}{0.1420kg*9.8m/s2} = 91.1 m (3)[/tex]
As the distance between an object and the center of the Earth is increased...
a. the object's mass decreases and its weight remains the same
b. both the object's mass and its weight remain the same
c. both the object's mass and its weight decrease
d. the object's mass remains the same and its weight decreases
Explanation:
the answer is D.
have a great day !
As the distance between an object and the center of the Earth is increased, the object's mass remains the same and its weight decreases. Therefore, option D is correct.
What is mass ?A body's mass is an inherent quality. Prior to the discovery of the atom and particle physics, it was widely considered to be tied to the amount of matter in a physical body.
Since the mass of the Earth is evenly distributed throughout its volume, the weight of an object increases with increasing distance from the planet's center until the object lies beneath the surface (in a hole or tunnel). When an object is above the surface, its weight falls as it gets farther from the Earth's center.
when the separation between a point and the Earth's center widens. The weight of the object drops while its mass stays constant.
Thus, option D is correct.
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An element or compound used to enhance a semiconductor is called a(n) ____.
The element named boron can be used to enhance the properties of semiconductors.
What is a semiconductor?A semiconductor is a material that has electronic properties and has the value that falls in between a conductor. It can be a metallic copper or an insulator.
The rise in temperatures leads to a fall in resistivity. The element named boron can improve the electrical properties of the semiconductor as they form the impurities.
Find out more information about the element.
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Need help ASAP please help me please
Answer:
145
Explanation:
A 10kg object rests on a frictionless surface when it is struck by a 300N force. At what rate will it accelerate?
3m/s/s
30m/s/s
0.3m/s/s
300m/s/s
Answer:
I believe the answer is C. 0.3m/s/s
Explanation:
hope this helps :) lemme know if I'm correct
Answer:
Answer option B: 30m/s/s.
Explanation:
Should be the answer. I'm sorry if my answer is wrong not 100% sure. I tried my best.
Statement About
1. Chromosomal and genetic mutations are two types of
mutations.
True or false
PLEASE ANSWER
Answer:
true
Explanation:
A gene mutation is a permanent change in the DNA sequence of a gene. Mutations can occur in a single base pair or in a large segment of a chromosome and even span multiple genes. Mutations can result from endogenous (occurring during DNA replication) or exogenous (environmental) factors. Germline mutations occur in gametes. Somatic mutations occur in other body cells. Chromosomal alterations are mutations that change chromosome structure. Point mutations change a single nucleotide.
Two major categories of mutations are germline mutations and somatic mutations.
Germline mutations occur in gametes. These mutations are especially significant because they can be transmitted to offspring and every cell in the offspring will have the mutation.
Somatic mutations occur in other cells of the body.
A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.58 m. A child applies a force 49.5 N tangentially to the edge of the disk to start it from rest. What is the kinetic energy of the merry-go-round disk (in J) after 2.95 s?
Answer:
The value is [tex]KE = 259.6 \ J[/tex]
Explanation:
From the question we are told that
The weight of the horizontal solid disk is [tex]W = 805 \ N[/tex]
The radius of the horizontal solid disk is [tex]r = 1.58 \ m[/tex]
The force applied by the child is [tex]F = 49.5 \ N[/tex]
The time considered is [tex]t = 2.95 \ s[/tex]
Generally the mass of the horizontal solid disk is mathematically represented as
[tex]m_h = \frac{W}{ g}[/tex]
=> [tex]m_h = \frac{805}{ 9.8 }[/tex]
=> [tex]m_h = 82.14 \ N[/tex]
Generally the moment of inertia of the horizontal solid disk is mathematically represented as
[tex]I = \frac{1}{2} * m * r^ 2[/tex]
=> [tex]I = \frac{1}{2} * 82.14 * 1.58^ 2[/tex]
=> [tex]I = 102.5 \ kg \cdot m^2[/tex]
Generally the net torque experienced by the horizontal solid disk is mathematically represented as
[tex]T = I * \alpha = F * r[/tex]
=> [tex]\alpha = \frac{ F * r }{ I }[/tex]
=> [tex]\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }[/tex]
=> [tex]\alpha = 0.7628[/tex]
Gnerally from kinematic equation we have that
[tex]w = w_o + \alpha t[/tex]
Here [tex]w_o[/tex] is the initial angular velocity velocity of the horizontal solid disk which is [tex]w_o = 0\ rad/s[/tex]
So
[tex]w = 0 + 0.7628 * 2.95[/tex]
=> [tex]w = 2.2503 \ rad/s[/tex]
Generally the kinetic energy is mathematically represented as
[tex]KE = \frac{1}{2} * I * w^2[/tex]
=> [tex]KE = \frac{1}{2} * 102.53 * 2.2503 ^2[/tex]
=> [tex]KE = 259.6 \ J[/tex]
Which of the following is true regarding the speed of earthquake waves?
OA.
S waves travel faster than P waves and surface waves.
ОВ.
Surface waves travel faster than P waves and S waves.
OC.
P waves, S waves, and surface waves all have the same speed.
OD.
P waves travel faster than S waves and surface waves.
Answer:
p waves travel faster than s waves and surface waves
Answer:
p waves travel faster than s waves and surface waves
Explanation:
I took a quiz and got this right.
PROPERTIES OF RADIO WAVES
Answer:
Radio waves are a type of electromagnetic radiation with wavelengths in the electromagnetic spectrum longer than infrared light. Radio waves have frequencies as high as 300 gigahertz (GHz) to as low as 30 hertz (Hz).
Explanation:
What is the magnitude of the centripetal force that must be applied in order for a 2kg ball on a 2.0 m string to spin with uniform circular motion at 5.0 m/s
The ball needs to accelerate with a magnitude a (pointed towards the center of the circle) of
a = (5.0 m/s)² / (2.0 m) = 12.5 m/s²
Then the required tension F in the string would need to be
F = (2 kg) (12.5 m/s²) = 25 N
The free-body diagram below shows the forces acting on a bicycle as the
rider pedals to the right. The vectors are not drawn to scale. The bicycle has a
weight of 800 N and a pedaling force of 250 N. As it moves, it encounters 75
N of air resistance. What is the net force on the bicycle in the x-direction?
Answer:
175 N to the right
Explanation:
I am taking the quiz and this is the correct answer. The pedaling force if 250 N but when it encounters 75 N of air resistance, it reduces to 175 N. This is because the air resistance is going opposite of you.
The weight of the bike and the cyclist are a force that the Earth applies to both of them and that acts vertically and downward, causing an action on the ground. The pedal-pushing force is transmitted from the crank arm to the chainring axis via the transmission forces.
What forces acting on a moving bicycle?When we press the pedals, the force travels to the back wheel, which then applies pressure to the ground. Strength of action. The pavement responds by exerting a force in the opposite direction but in the same direction on the back wheel. Hence, always move forward.
Therefore, Walking and biking are made possible through la friction. The friction created by the tire's pressure on the ground, It prevents the tire from rotating and keeps the lowest part of the wheel on the ground. The wheel is driven by this force, which is transferred to the wheel axle.
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What causes friction between two solids?
Answer:
Friction is when 2 solids move against each other. The cause of friction is adhesion, and surface roughness. Surface roughness is when a surface is rough enough that is causes friction against another surface. Adhesion is when 2 surfaces collide because of thier molecular force.
Hello! How do you answer a question? I will give you 25 points if you answer it. :)
Answer:
press add answer and it will let you answer the question
Explanation:
Answer:
you click on the question, then you click answer, then type what you want, and click the green button on the top right saying dd your answer
Explanation:
1. A rocket is launched straight up into the air. If its entire flight takes 5 seconds....
A) What is the initial velocity of the rocket?
Answer:
The initial velocity of all rocket is zero.A student mixes .075 kg of an unknown substance at 96.5°C with .075 kg of water at 25.0°C. If the final temperature of the system is 31.15°C, what is the specific heat capacity of the substance?
Answer:
The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.
Explanation:
We notice that the student is mixing a substance with a high temperature and another substance with a low temperature, where the first release heat to the latter one until thermal equilibrium is reached. By the First Law of Thermodynamics and assuming that the entire system has no energy interactions with the surroundings, we get the following model:
[tex]\Delta U_{x}+\Delta U_{w} = 0[/tex] (1)
Where [tex]\Delta U_{x}[/tex] and [tex]\Delta U_{w}[/tex] are the changes in internal energy for the unknown substance and water, measured in joules.
By definition of internal energy, we expand the equation above now:
[tex]m_{x}\cdot c_{x}\cdot (T_{o,x}-T_{f,x})+m_{w}\cdot c_{w}\cdot (T_{o,w}-T_{f,w}) = 0[/tex] (2)
Where:
[tex]m_{x}[/tex], [tex]m_{w}[/tex] - Masses of the unknown substance and water, measured in kilograms.
[tex]c_{x}[/tex], [tex]c_{w}[/tex] - Specific heats of the unknown substance and water, measured in joules per kilogram-degree Celsius.
[tex]T_{o,x}[/tex], [tex]T_{f,x}[/tex] - Initial and final temperatures of the unknown substance, measured in degrees Celsius.
[tex]T_{o,w}[/tex], [tex]T_{f,w}[/tex] - Initial and final temperatures of water, measured in degrees Celsius.
Then, we clear the specific heat of the unknown substance:
[tex]c_{x} = \frac{m_{w}\cdot c_{w}\cdot (T_{f,w}-T_{o,w})}{m_{x}\cdot (T_{o,x}-T_{f,x})}[/tex]
If we know that [tex]m_{w} = m_{x} = 0.075\,kg[/tex], [tex]c_{w} = 4186\,\frac{J}{kg\cdot^{\circ}C}[/tex], [tex]T_{f,w} = T_{f,x} = 31.15\,^{\circ}C[/tex], [tex]T_{o,x} = 96.5\,^{\circ}C[/tex] and [tex]T_{o,w} = 25\,^{\circ}C[/tex], then the heat capacity of the unknown substance is:
[tex]c_{x} = \frac{(0.075\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (31.15\,^{\circ}C-25\,^{\circ}C)}{(0.075\,kg)\cdot (96.5\,^{\circ}C-31.15^{\circ}C)}[/tex]
[tex]c_{x} = 393.939\,\frac{J}{kg\cdot ^{\circ}C}[/tex]
The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.