Vf² = V₁² + 2ad
(12 m/s) ² = (0 m/s) ² + 2 (a) (50m)
a = 1.44 m/s ²
Answer:
[tex]\boxed {\boxed {\sf 1.44 \ m/s^2}}[/tex]
Explanation:
We are asked to find the block's rate of acceleration.
We are given the final velocity, initial velocity, and distance. Therefore, we will use the following kinematic equation:
[tex]{v_f}^2={v_i}^2+2ad[/tex]
The block starts at rest or 0 meters per second. The block reaches a final velocity of 12 meters per second. The distance traveled is 50 meters.
[tex]v_i[/tex]= 0 m/s [tex]v_f[/tex]= 12 m/s [tex]d[/tex]= 50 mSubstitute the values into the formula.
[tex](12 \ m/s)^2 = (0 m/s)^2 + 2a (50 \ m)[/tex]
Solve the exponents.
(12 m/s)² = 12 m/s * 12 m/s = 144 m²/s²(0 m/s)² = 0 m/s * 0 m/s = 0 m²/s²[tex]144 \ m^2/s^2= 0 m^2/s^2 + 2a (50 \ m)[/tex]
The 0 can be subtracted from both sides, or simply canceled.
[tex]144 \ m^2/s^2 = 2a (50 \ m)[/tex]
Multiply 2 and 50 meters.
[tex]144 \ m^2/s^2 = (2 *50 \ m)* a[/tex]
[tex]144 \ m^2/s^2 = (100 \ m)* a[/tex]
We are solving for a, so we must isolate the variable. It is being multiplied by 100 meters. The inverse of multiplication is division, so we divide both sides by 100 meters.
[tex]\frac {144 \ m^2/s^2}{100 \ m}= \frac{(100 \ m)*a}{100 \ m}[/tex]
[tex]\frac {144 \ m^2/s^2}{100 \ m}=a[/tex]
[tex]1.44 \ m/s^2 =a[/tex]
The block's rate of acceleration is 1.44 meters per second squared.
When converted to a household measurement, 9 kilograms is approximately equal to a
Answer:
D) 19.8 lbs
Explanation:
1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.
1kg is also equal to 2.205 lbs. 9*2.205=19.8416
9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system
a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.
This leaves us with 19.8 lbs
2. In these diagrams, the sunlight is coming from the left, as shown by the arrows. Which
diagram is accurate and why? (1)
a) A is accurate because the half of the Moon that is facing the sun is it by the sun, and the
other half is dark.
b) B is accurate because the whole Moon is bright as long as there are no shadows
covering parts of it.
c) C is accurate because the Moon is almost as close as it can be to the sun, and so the
Moon is almost fully lit by the sun.
d) All three diagrams are sometimes accurate because the Moon has some lighter-colored
rock and some darker-colored rock, and the half of the Moon with the lighter-colored
rock rotates to face different directions.
Answer:
Explanation:
a) A is accurate because the half of the Moon that is facing the sun is it by the sun, and the other half is dark.
The most accurate diagram of the Sun and Moon is: a. A is accurate because the half of the Moon that is facing the sun is it by the sun, and the other half is dark.
What is a Moon?A Moon can be defined as the natural satellite of planet Earth that shines due to the Sun's reflected light (sunlight), as it revolves around the Earth from west to east in 29½ days.
The phases of the Moon.Based on astronomical information, there are four (4) primary Moon phases and these are:
New MoonFirst QuarterFull MoonThird Quarter (Last Quarter)In conclusion, diagram A is the most accurate because half of the Moon that faced the Sun is lit by the Sun while the other half is dark.
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Does bouncing or sticking produce more impulse?
Answer:
yes
Explanation:
Bouncing is a way to increase impulse. Because an object that bounces changes directions the force of impulse must be absorbed then generated by the target object. (Impulse is nearly doubled.)
I really need your help I will mark you brainliest!
I have been stuck for 15 minutes!!
Answer the following questions about Hurricane Harvey:
a. Why did hurricane Harvey travel from the east (Africia) twards the west (Florida) when our weather starts in the west and travels east?
B. why did hurricane harvey rotate counterclockwise? Do all hurricanes rotate counterclockwise?
The term that is used to quantity the amount of heat in the evaporator is
Answer:
Enthalpy
The quantity of heat contained in a material as measured from a particular base or point.
- The rocket now has a thruster that malfunctions and is now pushing the rocket in the wrong direction. What is the new net force on the rocket if it is now accelerating at 12mls²
From your previous question
Mass=30kg
Acceleration=12m/s^2
[tex]\\ \sf\longmapsto F=ma[/tex]
[tex]\\ \sf\longmapsto F=30(12)[/tex]
[tex]\\ \sf\longmapsto F=360N[/tex]
Given,
Acceleration = 12 m/s²
According to your previous question you posted,
Mass = 30 KG
Solution:
We know that the formula of force according to the Newton's second law is:
[tex]\longmapsto \sf \red{f = m \times a}[/tex]
[tex]\sf \longmapsto \ 30 \times 12[/tex]
[tex]\sf \longmapsto \: \boxed {\sf \blue3 \red60}[/tex]
Answer:
[tex]\sf \boxed{ \bf \: force = 360\: N}[/tex]
4. A gas's solubility is best in a liquid solvent when the solution is under high or low pressure
Answer:
low pressure
Explanation:
Please show your workmkmmkekdmnesnndndndndn
Answer:
Explanation:
Remark
The question really is how long does the projectile stay in the air if it was just dropped from a height of 46 meters.
Givens
vi = 0
a = 9.81 m/s^2
h = 46 m
t = ?
Solution
d = vi*t + 1/2 a t^2
46 = 0 + 1/2 * 9.81 * t^2 multiply both sides by 2
92 = 9.81 * t^2 divide by 9.81
t^2 = 9.38 Take the square root of both sides
sqrt(t^2) = sqrt(9.38)
t = 3.06 seconds.
In 3.06 seconds the object will hit the ground, thus stopping it. How far horizontally does it go in that time.
d = r * t
d=?
r = 102 m/s
t = 3.06 seconds
d = 102 * 3.06
d = 312.4 meters.
HELP HELP!
Which of the following surfaces would have the lowest amount of friction?
a. Glass
b. Sand
c. Wood
d. Ice
Answer:
d. Ice
Explanation:
Please HELP!!!!!!!!!!!!!!!!!!!!!
Using Newton's First Law, explain why, in a frictionless environment, a car that is under motion will not stop moving?
Answer:
According to Newton's first law of motion, an object maintains its state unless a force acts on it. Therefore, a moving car does not change its direction and keeps its speed unless a force acts on it.
dsfhfdskjhfdkshjkfdhskjdfhkjshjkfhsdkjf
Answer:
Abcdfeghijklmnopqrstuvwxyz
Explanation:
They are in order
What forces act on an airplane to keep it in the air? If the airplane is traveling at a constant velocity, what could be said about the magnitudes of the forces?
Let's see
[tex]\\ \sf\longmapsto \overrightarrow{drag}[/tex]
[tex]\\ \sf\longmapsto \overleftarrow{Thrust}[/tex]
[tex]\\ \sf\longmapsto Lift\:Force\uparrow[/tex]
[tex]\\ \sf\longmapsto Acceleration\:due\:to\: gravity\downarrow[/tex]
Net force is 0
QUESTION 5 When 235U is bombarded with one neutron, fission occurs and the products are three neutrons, 94Kr and?
a. 139Ba
b. 141Ba
c. 139Ce
d. 139Xe
If the law allows abortion based on sex, should we also allow abortion based on race, ethnicity, or skin color? Briefly explain why you believe this is right or wrong.
Answer:
No because you don't have to be black or white you don't have to be a pacific race to just get it like lets say for example a asian person come in and they want to get a abortion but the doctor say no then that person can end up killing her self or when the baby is born she could end you know doing something bad. But no you do not have to be a pacific person to just get it
Explanation:
A rock falls to the ground with a force of 300N and air resistance pushes back up on the rock with a
force of 45N. Which is true of the rock?
Answer:
what are the choices
Explanation:
don't worry i'll edit it
Starting from rest, a student pulls a (25.0 A) kg box a distance of (4.50 B) m across the floor using a (134 C) N force applied at 35.0 degrees above horizontal. The coefficient of kinetic friction is 0.220 between the floor and the box. Find the work done by the pulling force. Give your answer in joules (J) and with 2 significant figures.
A 25.0 kg box is moved 4.50 m by a force of 134 N applied at 35.0° above horizontal. The work done by the pulling force is 4.9 × 10² J.
Starting from rest, a student pulls a 25.0 kg box a distance of 4.50 m across the floor using a 134 N force applied at 35.0 degrees above horizontal.
We can calculate the work (W) done by the pulling force using the following expression.
[tex]W = F \times d \times cos\theta = 134 N \times 4.50 m \times cos35.0\° = 4.9 \times 10^{2} J[/tex]
where,
F: puling forced: displacementθ: angle between F and dThe friction between the floor and the box. does not influence the work done by the pulling force.
A 25.0 kg box is moved 4.50 m by a force of 134 N applied at 35.0° above horizontal. The work done by the pulling force is 4.9 × 10² J.
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Given below are the names of some objects and materials: Water, basket ball, orange, sugar, globe, apple and earthen pitcher Group them as Eatables and Non eatables.
hope it helps
Pls help thx need asap!!
Answer:
10N right
Explanation:
First, we need to find the base of the triangle to find the force acting on the right of the object.
We have been given the hypotenuse and we need to find the adjacent, so we are using cosSOH, CAH, TOAcos(θ) = adjacent ÷ hypotenuseNext, sub in the values (we don't know the adjacent so i've called it x):
cos(60) = x ÷ 60To find x, we need to multiply both sides by 60
60 × cos(60) = xx = 30 NNow we know all the forces, we can work out the net force:
Both of the 40N cancel out because they are opposite forcesThe force on the right of the object is 10N stronger than the force on the left (30N - 20N), so the object would move to the rightwhen air gets hot, it rises and forms…
Answer:
Gas
Explanation:
Hot air rises because gases expand as they heat up. When air heats up and expands, its density also decreases.
If a rock is thrown vertically upward from the surface of Mars with velocity of 20 m/s, its height (in meters) after t seconds is h = 20t − 1.86t2. (a) What is the velocity (in m/s) of the rock after 2 s? 12.56 Correct: Your answer is correct. m/s (b) What is the velocity (in m/s) of the rock when its height is 25 m on its way up? On its way down? (Round your answers to two decimal places.) up 8.07 Incorrect: Your answer is incorrect. m/s down -8.07 Incorrect: Your answer is incorrect. m/s
The velocity of the rock after 2s is 12.56 m/s.
The velocity of the rock at 25 m upwards is 14.68 m/s.
The velocity of the rock at 25 m downwards is -15.97 m/s.
The given parameters;
velocity of the rock, u = 20 m/sheight of the rock, h = 20t - 1.8t²The velocity of the rock after 2s is calculated as follows;
[tex]v = \frac{dh}{dt}= 20 -2(1.86)t \\\\v = 20 - 3.72t\\\\at \ t= 2 \ s \\\\v = 20 - 3.72(2) = 12.56 \ m/s[/tex]
The velocity of the rock when the height is 25 m:
h = 20t - 1.8t²
25 = 20t - 1.8t²
1.8t² - 20t + 25 = 0
solve for the time of motion "t" using quadratic formula
a = 1.8, b = -20, c = 25
[tex]t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(-20) \ \ +/- \ \ \sqrt{(-20)^2 - 4(1.8\times 25)} }{2(1.8)}\\\\t = 9.67s, \ \ or \ 1.43 \ s[/tex]
The value of t that will give 25 m upwards using the motion model is 1.43 s;
h = 20(1.43) - 1.86(1.43)² = 25 m
The velocity of the rock at 25 m upwards (t = 1.43 s) is calculated as follows;
[tex]v = \frac{dh}{dt} = 20 -3.72t\\\\v = 20 - 3.72(1.43) \\\\v = 14.68 \ m/s[/tex]
The velocity of the rock at 25 m downwards (t = 9.67 s) is calculated as;
[tex]v = \frac{dh}{dt} = 20 - 3.72t\\\\v = 20 - 3.72(9.67)\\\\v = -15.97 \ m/s[/tex]
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four particles connected by rigid rods of negligible mass where y1 = 5.70 m. the origin is at the center of the rectangle. The system rotates in the xy plane about the z axis with and angular speed of 5.60 rad/s. a) calculate the moment of inertia of the system about the z axis. b) Calculate the rotational kinetic energy
The definition of moment of inertia and kinetic energy we can find the results for the questions are:
a) The moment of inertia is: I = 32.49 kg m²
b) The kinetic energy is: K = 509.4 J
a) The moment of inertia is a scalar quantity that has the same function as mass, it is given by
I = m r²
They indicate that the length of the bar is y₁ = 5.70 m, if the two rigid bars form the square that supports the masses, the distance from the center to the end where each mass is:
r = y₁ / 2
r = 5.70 / 2 = 2.85 m
The moment of inertia of the system is.
I = 4 m r²
Let's calculate.
I =4 m 2.85²
I = 32.49m kg m²
If we assume that the mass of each point particle is:
m = 1 kg
I = 32.49 kg m²
b) They ask for the rotational kinetic energy, if the angular velocity is.
w = 5.60 rad / s
The kinetic energy for rotational motion is.
K = ½ I w²
Let's calculate.
K = ½ 32.49 5.60²
K = 509.4 J
In conclusion with the definition of moment of inertia and kinetic energy we can find the results for the questions are:
a) The moment of inertia is: I = 32.49 kg m²
b) The kinetic energy is: K = 509.4 J
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Question 2 of 10
You are riding a bicycle. If you apply a forward force of 125 N, and you and
the bicycle have a combined mass of 82 kg, what will be the forward
acceleration of the bicycle? (Assume there is no friction.)
A. 1.67 m/s2
B. 3.37 m/s2
C. 0.66 m/s2
D. 1.52 m/s2
O
Answer:
D - 1.52 m/s2
Explanation:
force: 125 N
Mass: 82 kg
you just divide the force by the mass which would be approximately 1.52
the hanging
mass would need to be
Answer:
what do you need help with I can try to help you
1. According to the picture, which is the least dense?
O A. the chess piece
O B. the toy soldier
O C. the liquid in the container
O D. There is no way to tell,
According to the picture, the chess piece is the least dense. The correct option is A.
The amount of mass that a substance has in relation to its volume is referred to as its density. It is a measurement of how closely together the molecules or particles are arranged within a given object or material.
Typically, floating objects are less dense than the liquid they are in. The fact that the chess piece is floating on the water suggests that it is less dense than the liquid.
It might be an alternative, though, if the toy soldier is also floating and not submerged. More visual context is necessary for precise determination.
Thus, the correct option is A.
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--[50 POINTS]--
1)A block of mass 25 kg is placed on flat ground. The coefficient of static friction and kinetic friction are 0.73 and 0.16
a.If a person pushes the block and the block is moving, what will be the acceleration of the block?
2) A block has a mass of 79 kg. The coefficient of static and kinetic friction between the sled and the ground is 0.87 and 0.37. Person A tries to pull the block with 210N, but fails.
a) Person B successfully pulls the sled with 909N. What is the acceleration of the sled?
Newton's second law allows us to find the results for the acceleration of the blocks are:
1) The acceleration is a = 559 m / s²
2) The acceleration is a = 7.88 m / s²
Newton's second law states that the net force is equal to the product of the mass and the acceleration of the body.
∑ F = m a
Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body.
The reference system is a coordinate system with respect to which the decomposition of the forces is carried out, in the attached we have a free body diagram of the system.
1) They indicate that the body mass is 25 kg.
y-axis
N - W = 0
N = W = m g
x-axis
F -fr = ma
The friction force is a macroscopic force that results from the sum of all the microscopic interactions between the two surfaces, it has the formula
fr = μ N
Where fr is the friction force, N the normal and very the friction coefficient.
This friction coefficient has two values:
Static. For when with there is not relative motion between the two surfaces. Dynamic. When there is relative motion between the two surfaces.
We substitute.
F - μ m g = m a
a) The system moves which is the acceleration.
Suppose that the force that star to move the system keeps constant, just before the system begins to move the coefficient of friction is static, let's find the applied force.
F = μ m g
F = 0.73 25 9.8
F = 178.85 N
The block begins to move and the friction coefficient decreases to the dynamic value, we look for the acceleration.
a = [tex]\frac{F - \mu \ m g}{m}[/tex]
a = [tex]\frac{178.85 - 0.16 \ 25 \ 9.8 }{25}[/tex]
a = 5.59 m / s²
2) In this case the mass of the block is 79 kg and the applied force is
F = 909 N
We look for acceleration.
a = [tex]\frac{909 - 0.37 \ 79 \ 9.8 }{79}[/tex]
a = 7.88 m / S²
In conclusion using Newton's second law we can find the results for the acceleration of the blocks are:
1) The acceleration is a = 559 m / s²
2) The acceleration is a = 7.88 m / s²
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I need HELP! IMMEDIATELY!!!
Answer:
Option A
Explanation:
N is Chloroplast, P is a large centralized vacuole and R is the cell wall which are found only in plant cells. Hence, the given cell is a plant cell.
S = Rough Endoplasmic Reticulum
P = Vacuole
O = Nucleus
T = Smooth Endoplasmic Reticulum
U = Golgi Apparatus
N = Chloroplast
U = Cell Membrane
M = Mitochondria
R = Cell wall
[tex]\rule[225]{225}{2}[/tex]
Hope this helped!
~AH1807A 20 kg box rests on the ground. Round all answers to the hundredths, if necessary. What is the weight of the box?
Weight=Applied force
[tex]\\ \sf\longmapsto F=mg[/tex]
[tex]\\ \sf\longmapsto F=20(10)[/tex]
[tex]\\ \sf\longmapsto F=200N[/tex]
Answer:
Weight=200
Explanation:
Given,
[tex]\sf\longmapsto \: mass = 20 \: kilograms[/tex]
We know that–
[tex]\sf\longmapsto \: acceleration \: \: because \: of \: gravity = 10m/s^2[/tex]
[tex]\sf\longmapsto \: Weight= \: is \: F = m•g[/tex]
Let's start solving!
Weight = 20×10
Weight=200
helphelphelp due in 5 minutes answer the question below
if you double the force an object what will hapen to the accelration
Explanation:
the acceleration will double because force is directly proportional to the acceleration
The Canadian bobsled team hit the brakes on the sled they are traveling East in so that it decelerates at a rate of 0.43
m/s^2.
a) How long does it take to stop if it travels 85 m before coming to rest?
b) Draw the velocity vector.
c) Draw the acceleration vector.
Answer:
Explanation:
85 = ½(0.43)t²
t = √(2(85)/0.43)
t = 19.883380...
t = 20 s
v→ 8.55 m/s initial, 0 m/s final
a← 0.43 m/s²