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.
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]
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.
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.
brainly.com/question/12389810
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.
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.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:
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.