Answer:
5.88 m/s
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 7 m/s
Acceleration (a) = –0.6 m/s²
Distance (s) = 12 m
Final velocity (v) =.?
We, can determine how fast the sprinter is running when he crosses the goal line as follow:
v² = u² + 2as
v² = 7² + (2 × –0.6 × 12)
v² = 49 + (–14.4)
v² = 49 – 14.4
v² = 34.6
Take the square root of both side
v = √34.6
v = 5.88 m/s
Therefore, the sprinter is running at a speed of 5.88 m/s when he crosses the goal line.
Michael reaches the end of his route and slowly comes to a stop, reaching 1 poin
a final velocity of O m/s and ending his 2 m/s run. If it took him 4 seconds to
come to a stop, what was his acceleration? Don't forget units and
directions if appropriate?
Your answer
A test rocket is launched by accelerating it along a 200.0 m, 35 degree incline at 1.25 m/s/s starting
from rest. The instant the rocket leaves the incline, its engines turn off, it is subject only to gravity,
and air resistance can be ignored. Find
a) the maximum height above the ground the rocket reaches
Answer:
The maximum height above the ground the rocket reaches is 123.1 m.
Explanation:
Let's find the final velocity at a distance of 200 m:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
Where:
[tex]v_{f}[/tex] is the final speed =?
v₀ is the initial speed =0
a is the acceleration = 1.25 m/s²
d is the distance = 200 m
[tex]v_{f} = \sqrt{2ad} = \sqrt{2*1.25 m/s{2}*200 m/s} = 22.4 m/s[/tex]
Now, when the engines of the rocket turn off and it is subject only to gravity, the height reached is:
[tex] v_{fy}^{2} = v_{0y}^{2} - 2gh [/tex]
Where:
[tex]v_{f}[/tex] = 0
[tex]h = -\frac{v_{fy}^{2} - v_{0y}^{2}*sin(\theta)}{2g} = \frac{(22.4*sin(35))^{2}}{2*9.81 m/s^{2}} = 8.4 m[/tex]
Finally, the maximum height above the ground is:
[tex] h_{max} = h + H [/tex]
Where H is the vertical component of the 200.0 meters.
[tex]h_{max} = h + H = 8.4 m + 200.0 m*sin(35) = 123.1 m[/tex]
Therefore, the maximum height above the ground the rocket reaches is 123.1 m.
I hope it helps you!
A box rests on the (frictionless) bed of a truck. The truck
driver starts the truck and accelerates forward. The box!
immediately starts to slide toward the rear of the truck
bed. Discuss the motion of the box, in terms of Newton's
laws, as seen (a) by Mary standing on the ground beside
the truck, and (b) by Chris who is riding on the truck
(Fig. 4–35).
Answer:
follows are the solution to the given point:
Explanation:
In point A:
The meet is next to the truck on the floor, and for Mary, its box is fixed wrt. Since no power occurs, no speed is going to be happening. So, Mary's box seems to be at rest.
In point B:
Chris and as well as the box ride mostly on the truck because the truck was accelerating, it has been said that Chris will be in a non-inert position mostly on the truck. For Chris, its box is thrown backward as well as the box is moved because of another force.
Que fuerza se obtendrá en el embolo mayor de una prensa hidráulica cita area es de 100cm² cuando en el embolo menor área igual a 15cm² se aplica una fuerza de 200N?
Responder:
1.333,33 N
Explicación:
Dado que :
Área del pistón más grande (A1) = 100 cm²
Área del pistón más pequeño (A2) = 15 cm²
Fuerza del pistón más pequeño (F1) = 200N
Fuerza del pistón más grande = x
Usando la relación:
Presión del pistón más grande = presión del pistón más pequeño
Presión = Fuerza / Área
Por lo tanto,
F1 * A1 = F2 * A2
200 * 100 = F2 * 15
20000 = 15F2
F2 = 20000/15
F2 = 1.333,33 N
PLEASE HELP... can you help with either one of them or both. And explain how
Answer:
#6 is 14 miles in 7 mins
Explanation:
you need to get the unit rate (how many miles per min) then multiply from there so you take 10/5 (because there 5 mins) and get 2 miles per min, then you take 2 an multiply it by 7 (because you want to know how far she went in 7 min) and you get 14
Answer:: 14miles in 7 mins
If you threw a rock somewhere with no gravity or friction the rock would..
A.) Continue at the same speed in the same direction.
B.) Stop.
C.) Move forward quickly then slowly.
D.) Move forward slowly then fall.
Answer:
A
Explanation:
Once the stone is thrown, with direction vectors and initial velocity will keep on traveling for infinite period unless there is another force to stop it.
For example, Newton's 1st law of Motion: ''Every inanimate body will continue its state of rest/motion unless acted upon by an external unbalanced force''
Since there is no friction, drag or any viscous force, our stone will travel for infinite period of time.
I hope this helps :)
Electromagnetic induction is the process of producing an electric current by moving a loop of wire ____.
Answer:
through a magnetic field.
Explanation:
a 24 N force is being applied to the right, while a 345 N force is being applied to the left of a 18 kg crate.
What is the acceleration of the crate?
Answer:
a = 17.83 m/s²
Explanation:
Given:
Fnet = 345N - 24N
m = 18kg
Unknown:
a or Acceleration of the crate
Formula and solution:
Fnet = F1 + F2
Fnet = 345N - 24N
Fnet = 321N
a = Fnet/m
a = 321 kg m/s² ÷ 18 kg
Final Answer:
a = 17.83 m/s²
A diagram of the forces being applied to a box is provided. If the net force acting on the box is 10N toward the right, what is the magnitude of force applied by the boy pulling to the left? Record your answer in the grid below. PLEASE HELP!!!!!!!!!!!!
Answer:
14 N.
Explanation:
Let the force applied by the boy pulling to the left be x.
From the question given above, the following data were obtained:
Net force (Fₙ) = 10 N toward the right
Force applied by the boy pulling to the left = x
Next, we shall determine the total force in the left direction. This can be obtained as follow:
Force in the left direction (Fₗ) = x + 8
Next, we shall determine the total force in the right direction. This can be obtained as follow:
Force in the right direction (Fᵣ) = 11 + 21 = 32 N
Finally, we shall determine the force applied by the boy pulling to the left direction (i.e the value of x) as follow:
Net force (Fₙ) = 10 N toward the right
Force in the left direction (Fₗ) = x + 8
orce in the right direction (Fᵣ) = 32 N
Fₙ = Fᵣ – Fₗ (since the net force is toward the right direction)
10 = 32 – (x + 8)
Clear bracket
10 = 32 – x – 8
10 = 32 – 8 – x
10 = 24 – x
Collect like terms
10 – 24 = – x
–14 = –x
Divide both side by –1
x = –14/–1
x = 14 N
Thus, the force applied by the boy pulling to the left direction is 14 N.
How do the molecules of cold water differ from the molecules of hot water?
-Answer choices
The molecules of cold water have fewer heat molecules mixed with them.
The molecules of cold water have more mass than the molecules of hot water.
The molecules of cold water are larger than the molecules of hot water.
The molecules of cold water are closer together than the molecules of hot water.
:)
Answer:
The molecules of cold water move slower than the molecules of hot water. The molecules of cold water move at the same speed as the molecules of hot water.
The molecules of cold water are closer together than the molecules of hot water. Heat energy make the molecules more mobile and make them far apart.
What is intermolecular forces?Intermolecular force is the force which binds the molecules of a compound or atoms in a molecules. There are different types of intermolecular forces such as ionic bonding, covalent bonding, hydrogen bond etc.
In solid state the intermolecular force between molecules is high and they are closely packed. In liquid phase, molecules have small space to move and ln gas phase, molecules are far apart with very less intermolecular force.
When water is heated, the heat energy weaken the intermolecular force between the molecules. Hence, in cold water, molecules are closer compared to hot water.
Find more on intermolecular forces:
https://brainly.com/question/9007693
#SPJ2
8. The human body is a what? conductor, insulator, electrolyte, or low reaistive
Answer:
The human body is a Conductor
What is the first step needed to solve ? (1 point)
(03.04 LC)
What is the first step needed to solve (3/4)x - 3 = 18
Select one:
a. Add 3 to both sides
b. Subtract 18 from both sides
c. Multiply both sides by 3
d. Divide both sides by 4
Answer:
answer is add 3 to both sides
a 1200 kg car travels at a constant velocity of 15 m/s. what is the net force on the car?
Answer:
The answer is 18,000 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × accelerationFrom the question we have
force = 1200 × 15
We have the final answer as
18,000 NHope this helps you
Distance between the centres of charged body is inevrsley proportional to the magnitude of electric force.
TRUE or FALSE
Answer:
TRUE
Explanation:
what do you mean by quantization of energy?
Answer:
The quantization of energy refers to the fact that at subatomic levels, energy is best thought of as occuring in discreet "packets" called photons. The red and blue photons are therefore "quantized" just as dollar bill denominations are "quantized". Each photon contains a unique amount of discreet energy.
Why are copies of stored digital data better than copies of stored analog
data?
A. The quality of the copies is the same, but digital data can be
copied more quickly.
B. Digital data contain only two different values, so it is easy to
distinguish between them.
c. Digital data contain many continuous values, so it is easy to
distinguish among them.
D. When many copies are made, the quality of digital copies
decreases, but not as much as that of analog copies.
Answer:
letter D.
Explanation:
When many copies are made, the quality of digital copies
decreases, but not as much as that of analog copies.
When many copies are made, the quality of digital copies decreases, but not as much as that of analog copies. Option D is correct.
What is digital data?Digital data is information represented as a string of discrete symbols, each of which can take only one of a finite number of values from some alphabet
The primary aim of digital memory is to enable the storage and retrieval of binary data.
The benefits of digital storage of information over analog procedures are similar to the advantages of digital communication of information over analog communication.
Copies of stored digital data are better than copies of stored analog data. When many copies are made, the quality of digital copies decreases, but not as much as that of analog copies.
Hence option D is correct.
To learn more about the digital data refer to the link;
https://brainly.com/question/16412578
A charge of 50-μC is placed on the y axis at y = 3.0 cm and a 77-μC charge is placed on the x axis at x = 4.0 cm. If both charges are held fixed, what is the magnitude of the initial acceleration of an electron (me = 9.1 x 10-31 kg) released from rest at the origin?
Answer:
Explanation:
We shall first calculate the electric field at origin .
Field due to first charge
= KQ / R²
= 9 x 10⁹ x 50 x 10⁻⁶ / .03²
E₁ = 5 x 10⁸ N/C
Field due to second charge
= KQ / R²
= 9 x 10⁹ x 77 x 10⁻⁶ / .04²
E₂ = 4.33 x 10⁸ N/C
Resultant field E = √ ( 5² + 4.33² ) x 10⁸
= 6.6 x 10⁸ N/C
Force on electron = E x q
= 6.6 x 10⁸ x 1.6 x 10⁻¹⁹ N
= 10.56 x 10⁻¹¹ N
Acceleration = force / mass
= 10.56 x 10⁻¹¹ / 9.1 x 10⁻³¹
= 1.16 x 10²⁰ m /s²
Answer:
[tex]a=1.16\times 10^{20}\ m/s^2[/tex]
Explanation:
Given that,
First charge, [tex]q_1=50\ \mu C = 50\times 10^{-6}\ C[/tex]. It is placed at y = 3 cm
Second charge, [tex]q_2=77\ \mu C = 77\times 10^{-6}\ C[/tex] . It is placed at x = 4 cm.
Force on electron due to q₁.
[tex]F_1=\dfrac{kq_1e}{r^2}\\\\F_1=\dfrac{9\times 10^9\times 1.6\times 10^{-19}\times 50\times 10^{-6}}{(0.03)^2}\\\\F_1=8\times 10^{-11}\ N[/tex]
Force on electron due to q₂.
[tex]F_2=\dfrac{kq_1e}{r^2}\\\\F_2=\dfrac{9\times 10^9\times 77\times 10^{-6}\times 1.6\times 10^{-19}}{(0.04)^2}\\\\F_2=6.93\times 10^{-11}\ N[/tex]
The net force on the electron is given by :
[tex]F=\sqrt{F_1^2+F_2^2} \\\\F=\sqrt{(8\times 10^{11})^2+(6.93\times 10^{-11})^2} \\\\F=1.058\times 10^{-11}\ N\\\\\text{or}\\\\F=10.58\times 10^{-11}\ N[/tex]
Let a is the acceleration of the electron. Net force is given by :
F = ma
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{10.58\times 10^{-11}}{9.1\times 10^{-31}}\\\\a=1.16\times 10^{20}\ m/s^2[/tex]
So, the acceleration of the electron is [tex]1.16\times 10^{20}\ m/s^2[/tex].
Which of the following careers is most likely to require business skills?
Systems Analyst
Hardware Engineer
Software Engineer
Database Administrator
Answer:
C. Software Engineer
PLEASE HELP ME BOO!! A rocket, initially at rest, steadily gains speed at a rate of 17.0m/s2 for 5.50s during take-off.
Answer:
[tex]v\: is : 93.5 \: m {(s)}^{ - 1}[/tex]
Explanation:
[tex]given : \\ acceleration \: \boxed{a} = 17.0 \: m {(s)}^{ - 2} \\ time\: \boxed{t} = 5.50 \: s \\ but \:\: \boxed{a} = \frac{v}{t} \\ v = at = 17 \times 5.5 = 93.5 \\ the \: velocity\: \boxed{v} \: is : 93.5 \: m {(s)}^{ - 1}[/tex]
if the velocity is what you require.....cos there was no clear statement of the required value.
♨Rage♨
Answer:
Vjhc log vcvjkvjkgvjcjfkfj
If F = G(m1, x xm2)/r2
m2 = ?
I need help!
Question 2
1
Without gravity, the stars, planets and all celestial objects in the universe would
Answer:
fly apart
Explanation:
gravity holds everything in place by constantly attracting both of the objects
without gravity, the objects would not be attracted anymore. Hence, they fly apart.
Find the acceleration of an object when the force is 850 N and the mass is 85 kg.
Answer:
ok sooooo,
F=MA
you are looking for a
you have F= 850 N and M= 85 kg
you divide the force with the mass
you get A= 10 m/s²
Which atom is most likely to accept electrons to form an ionic bond?
A. a mercury ion with a negative 2 charge
B. a potassium ion with a negative 1 charge
C. radon, a noble gas with 8 valence electrons
D. sulfur, a nonmetal with 6 valence electrons
Answer:
D.
Explanation:
Why does lighting usually
lighting usually strike build
buildings and electric poles ?
GREETINGS!
lightening usually strikes the metal pole and cables above the building because they provide a safe low resistance pathway to the lightening to move from building from the concrete steel to the ground below, thats how building doesnt get any damage and high electric charge passes through it
HOPE IT HELPS YOU!
Can someone help me with this please
Answer:
b
a
Explanation:
A cat leaps to try to catch a bird. If the cat's jump was at 60° off the ground and its initial velocity was 60 m/s what is the highest point of its trajectory, neglecting air resistance?
Answer:
137.75 m
Explanation:
The following data were obtained from the question:
Angle of projection (θ) = 60°
Initial velocity (u) = 60 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (H) =.?
The maximum height reached by the car can be obtained as follow:
H = u² Sine²θ /2g
H = 60² × (Sine 60)² / 2 × 9.8
H = 3600 × 0.866² / 19.6
H = 137.75 m
Therefore, the maximum height reached by the cat is 137.75 m
2. What is the net force exerted on a 80.0 kg race-car driver while the race
car is accelerating from 0 to 50.0 m/s in 5.50 s?*
Answer:
The net force exerted on the race-car driver is 727.3 N
Explanation:
Acceleration
According to Newton's second law of motion, the acceleration of an object a is directly proportional to the net force F and inversely proportional to the object's mass m:
[tex]F=m.a[/tex]
Since we don't have the value of the acceleration, we must use the kinematic equations to calculate it.
The main formula for accelerated motion is:
[tex]v_f=v_o+a.t[/tex]
Where vf is the final speed of the object, vo is its initial speed, a is the acceleration, and t is the time.
The race-car driver accelerates from vo=0 to 50 m/s in t=5.5 s. We can calculate the acceleration, solving the above equation for a:
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
[tex]\displaystyle a=\frac{50-0}{5.5}[/tex]
[tex]a=9.09\ m/s^2[/tex]
Now we calculate the net force:
[tex]F=80\ Kg\cdot 9.09\ m/s^2[/tex]
[tex]\boxed{F=727.3\ N}[/tex]
The net force exerted on the race-car driver is 727.3 N
You are pushing a box North in the hallway, at 20n, and a friend gets in front of the box and goes in the opposite direction, at 30n. What direction is the box going at? How much force does the box have going in that direction? (PLEASE HELP)
Answer:
The box is going south (opposite of north) at 10 N.
plz mark brainliest!!:DD
Date
Class
Name
Chapter 4 Directed Reading
Overview
Work and
Energy
Directions: Complete the concept map using the terms in the list below.
distance
mechanical energy potential energy
machines
motion
work
force
object
multiply
motion
9
tim
equa
Is
can be
made
RARIAr with
th
at
6
fore
0.
tropafer
00-00
00 0
O
Energ
duo
to
stored in
position
and
Copyright Glencow a division of The McGraw-Hill Companie te
motion
kinetic
energy
Work and Energy 15
Answer:
work and energy , copyright
Maddie is the fastest runner on her soccer team and loves to kick the ball. Using Maddie as an example, explain how each of Newton's laws of motion could apply to Maddie and the soccer ball as she lines up, kicks, and scores a goal for her team.
Answer:
Explanation:
When she lines up------1st law: the ball will be at rest until she acts on the ball
The first law state that "In the first law, an object will not change its motion unless a force acts on it."
When she kicks the ball----2nd law: the ball constantly changes direction and velocity depending on the applied force by her
The second law states that "In the second law, the force on an object is equal to its mass times its acceleration."
When she scores a goal-----3rd law: the force with which she kicks the ball to score a goal is equal to the reaction of the goal net
The third laws states that "In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction."