The acceleration when a 44.7 N net force is applied to 7.7 kg object is 5.8 m / s²
According to Newton's second law of motion,
F = m a
F = Force
m = Mass
a = Acceleration
m = 7.7 kg
F = 44.7 N
a = F / m
a = 44.7 / 7.7
a = 5.8 m / s²
Newton's second law of motion states that the force is equal to the rate of change of momentum. It can also be said as acceleration of an object is directly proportional to the net force acting on the object.
Therefore, the acceleration of the object is 5.8 m / s²
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Clare de Iles is shopping. She walks 16m North to the end of an aisle. She then makes a right hand turn and walks 21m Eastward to the end of the aisle. Determine th emagnitude of Claire's resultant displacement. And the angle of her resultant.
The magnitude and direction of Clare's resultant displacement is 26.4 m and 37.3⁰ respectively.
What is the resultant displacement?The magnitude of Clare's resultant displacement is calculated as follows;
her initial position, a = 16 m north
Final position, b = 21 m east
Resultant displacement, d = √a² + b²
d = √(16² + 21²)
d = 26.4 m
The direction of the resultant displacement is calculated as follows;
tan θ = 16 / 21
where;
θ is the direction of the displacement obtained by forming right triangle from the initial and final positiontan θ = 0.762
θ = arc tan(0.762)
θ = 37.3⁰
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Newton's Law with Friction (Force on Angle) please help
The Acceleration of the system as depicted in the diagram is 1.38 m/s².
What is acceleration?Acceleration is the rate of change of velocity. The S.I unit of acceleration is m/s²
To calculate the acceleration of the system, we use the formula below.
Formula:
a = (Fcos∅-mgμ)/m.......... Equation 1Where:
a = Acceleration of the systemm = Mass of the systemF = Force applied to the system at a certain angleg = Acceleration due to gravity∅ = Angle at which the force was appliedμ = Coefficient of static frictionFrom the question,
Given:
m = 1.1 kgF = 5.4 N∅ = 54.6°μ = 0.15g = 9.8 m/s²Substitute these values into equation 1
a = {(5.4cos54.6°)-(1.1×9.8×0.15)}/1.1a = (3.13-1.617)/1.1a = 1.38 m/s²Hence, the acceleration of the system is 1.38 m/s².
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An astronaut jumped on Mars. His initial velocity
was 1.88 m/s. If the gravitational acc. is -3.5-m/s²
on the moon,
a) What is the TOTAL time the astronaut is in the
atmosphere?
b) How high did the astronaut jump?
Due to the law of gravity, or gravitation, which draws all objects with mass in its vicinity toward it, Mars' gravitation is a natural occurrence.
.53 s the astronaut is in the atmosphere.
astronaut jump form .49 m.
What is Mars famous for?The rusty red color of Mars, the fourth planet from the sun, is well known. The atmosphere of the cold, arid Red Planet is extremely thin. But the dusty, (as far as we can tell) lifeless planet is far from boring.The closest planets to Earth are Venus and Mars, though in different ways. Venus and Earth are remarkably similar in terms of size, average density, mass, and surface gravity. Mars, however, resembles Earth the most in other respects.Due to the law of gravity, or gravitation, which draws all objects with mass in its vicinity toward it, Mars' gravitation is a natural occurrence. Due to the planet's lesser mass, its gravity is weaker than that of Earth.Mars only has 0.38g of gravity, which is around one-third of Earth's. Despite the fact that any visitors would need to wear protective spacesuits, this indicates that moving around on the planet's surface will be more easier than on Earth.from motion Equation : v= u + at
then we can find t = [tex]\frac{v-u}{a}[/tex] = -1.88/ 3.5 = .53 sec
s = ut + ½at²
put u = 1.88 a= -3.5 and t= .53
s = .49 m
.53 s the astronaut is in the atmosphere.
astronaut jump form .49 m.
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A rifle is aimed horizontally at a target 53 m away. The bullet hits the target 2.3 cm below the aim point.
The time of flight of bullet is 0.0685 seconds and the speed when it emerges out of the gun is 775.
As the bullet is fired horizontally from the gun at a distance of 53m away. But the bullet hits the target 2.3cm below the target.
So, the vertical displacement of the bullet is 2.3cm. The accelerations in the vertical direction will be g (acceleration due to gravity).
(a) As we know, in case of gravity, we can use equations of motion,
So, using equation,
S = ut + 1/2gt²
Where, u is the initial velocity of the bullet which is 0m/s in this case,
S is the vertical displacement of the bullet,
g is acceleration due to gravity and,
t is time of flight of the bullet.
Putting all the values,
S = 1/2gt²
2.3/100 = 1/2(9.8)(t²)
0.023 = 4.9t²
t² = 0.00469387
t = √0.0046987
t = 0.0685 seconds.
(b) The velocity as it emerges from the gun is equal to the muzzle velocity of the gun,
So, we can write,
Muzzle velocity V = Horizontal distance covered/time taken.
V = 53/0.0685
V = 773.72 m/s.
One thing that should be noted here is that the initial velocity that we took as zero in part (a) is taken with respect to the vertical motion and nit the Horizontal motion. So, one should not confuse the muzzle velocity in part (b) with the initial velocity mentioned in the part(a)
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Complete question - A rifle is aimed horizontally at a target 53 m away. The bullet hits the target 2.3 cm below the aim point. Find
(a) the bullets time of flight and (b) its speed as it emerges from the rifle?
it takes 4.0 μj of work to move a 15 nc charge from point a to b. it takes -5.0 μj of work to move the charge from c to b. part a what is the potential difference vc−va ? express your answer in volts.
W is the work done by charge from A to B and from C to B then the potential difference (vc−va)=552.62V.
The difference between the energy levels that charge carriers have at two places in a circuit is known as the potential difference.
Voltage, commonly known as potential difference (p.d. ), is a unit of electrical potential. The charge carriers in a circuit pass through the electrical components, transferring energy to them. We gauge the potential difference using a voltmeter (or voltage).
V = I x, the potential difference formula
The quantity of current times the resistance equals the potential difference (also known as voltage), which is equal to that. When a charge travels between two locations in a circuit, a potential difference of one Volt is equivalent to one Joule of energy being used by that charge.
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a 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. it accelerates upward at 30 m/s2 for 34 s, then runs out of fuel. ignore any air resistance effects.
A rocket with 200kgs is loaded extra 100kgs then the Rocket maximum altitude is 93005 m and total Time-of-Flight is 294 s
The Time-of-Flight concept (ToF) is a technique for determining distances between sensors and objects based on the amount of time that passes between a signal being sent and returning to the sensor after being reflected by an object. The Time-of-Flight principle can be employed with a variety of signals (also known as carriers), with sound and light being the most popular ones. The time it takes for photons to travel between two points—from the sensor's emitter to a target and back—in order to calculate distance for all Time-of-Flight (ToF) sensors.
Direct and indirect ToF both have unique advantages in particular situations. Each pixel in a scene can be simultaneously measured for distance and intensity.
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the moderate temperatures of islands throughout the world has much to do with water's vast supply of internal energy. high evaporation rate. absorption of solar energy. high specific heat. poor conductivity.
High specific heat of the water. Option (c)
What is Specific heat?
The amount of heat required to increase the temperature of one gram of a substance by one degree Celsius is referred to as the substance's specific heat. Typically, calories or joules are used per gram and degree Celsius when referring to the units of specific heat.
The moderate temperature of islands has much to do with the water's high specific heat. The typical off-water is more significant than this clear land or soil. Due to this fact, water absorbs and releases eat more slowly. In comparison to the land.
Hence, the water has high specific heat.
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a rotor rotating about a fixed axis through its mass center, the rotor has a mass of 60 kg, radius of gyration of 70 mm, and angular acceleration of 9 rad/s2. determine the moment the fixed axis.
Given
60 kg of mass, 70 mm or 0.07 meters of gyroscopic radius, and 9 rad/sec2 of angular acceleration
Utilized conceptInertia, It is an object's resistance to any change in motion. The amount of inertia is directly related to an object's mass; in fact, as we frequently observe, the heavier an object is, the harder it is to change its direction of motion.
The moment of inertia must be identified. EM = I > I > MK2
I stands for inertia mass moment.
I = 60 kg/m2 M = I = 0.216 kg/m2 = 11 kg/m2 M = 2.376 nm
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Answer:
Explanation:
Given:
m = 60 kg
R = 70 mm = 0.070 m
ε = 9 rad/s²
__________________
M - ?
The moment:
M = J·ε = m·R²·ε
M = 60·0.070²·9 ≈ 2.65 N·m
a plane accelerates from 4 m/s/s for 28.7 s until it finally lifts on the ground. Assuming the plane started from the rest, what was the planes displacement before takeoff?
There were 1647 before liftoff. 38 m moving planes.
A Straight Line Motion is defined as?An object is considered to be in motion if its location throughout time changes in relation to its surroundings. It is the gradual change in an object's location. The only kind of motion that exists is linear motion.
Which three motion equations are there?The three equations are as follows: S = ut + 12at2, v = u + at, v2 = u + 2as.
The airplane's initial speed is U=0 m/s.
The plane accelerates at a 4 m/s rate.
with a time stamp of 28.7 seconds.
Distance traveled using the formula S=ut+a
⇒ S=0(28.7)+0.5(4) (4)
=1647.38 ~ 1647m
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a race car driver is driving his car at a constant speed of 51.0 m/s on a circular track with a radius of 235 m. (a) what is the angular speed (in rad/s) of the car? rad/s (b) what are the magnitude (in m/s2) and direction of the car's acceleration? magnitude m/s2 direction
0.19362 rad/s is the angular speed and 8.8096 m/s2 is car's acceleration.
[tex]angular speed=\omega=v/r=45.5m/s/235m=0.19362rad/sec[/tex]
[tex]car acceleration =a=v^2/r=45.5^2m/s^2/235m=8.8096m/s^2[/tex]
The definition of angular speed is the rate of change of angular displacement, which is the angle a body travels along a circular route. The ratio of the number of rotations or revolutions performed by a body to the time taken is used to compute angular speed. The Greek letter Omega, stands for angular speed. Rad/s is the angular speed unit in the SI.
As is common knowledge, angular speed is defined as the rate of change of displacement divided by the time. Therefore, the equation for angular speed is = /t. Despite the aforementioned method, there is a different and more popular formula for calculating angular speed when it comes to competitive tests. As ω = θ/t
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A cliff jumper runs off a 15\,\text m15m15, start text, m, end text high cliff. Rocks extend 3.0\,\text m3.0m3, point, 0, start text, m, end text past the cliff bottom. We can ignore air resistance.
1.7 m/s is the horizontal velocity that is required to clear the rocks.
As the jumper runs off a 15 m high cliff.
And according to the question, starts with 0 velocities in the start direction thus he covers 15 m.
Given,
x = 15 m
a = g = 9.8 m, which is the acceleration due to gravity.
iv is the initial velocity which is 0, vo = 0
vf is the final velocity,
Using 2nd equations of motion,
x = iv × t + (1 ÷ 2) a × t²
15 = 0 + ((1 ÷ 2) × ( 9.8) × t²)
t = 1.75
Thus, the time is 1.75 seconds.
Using the same equation, horizontal velocity will be:
x = iv × t + (1 ÷ 2) a × t²
3 = vf × (1.75) + 1 ÷ 2 × (0) × (1.75)
vf = 1.7
Therefore, The horizontal velocity required to cover the rocks is 1.7 m/s.
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the trees in a wooded area near a factory look like burnt toothpicks. what might be causing the problem?
Isaac is practicing his volleyball skills by volleying a ball straight up and down, over and over again. His teammate
Marie notices that after one volley, the ball rises 3.6 m above Isaac's hands. What is the speed with which the ball left
Isaac's hand? (8.4 m/s) ANSWER FAST PLEASE!
The action of volleying a ball straight up and down comes under the free-fall motion of the ball. Free-fall is defined as the motion of a body which falls freely due to the gravitational pull of the earth.
The velocity (v) of the ball undergoing free-fall can be calculated using the free-fall formula,
[tex]v^{2}[/tex] = 2gh
where, g is the acceleration due to gravity and h is the height of the freely falling body.
Given that, the height to which ball rises, h = 3.6m. (g = 9.8 m/[tex]s^{2}[/tex])
Substituting h and g in the velocity formula,
[tex]v^{2}[/tex] = 2 × 9.8 m/[tex]s^{2}[/tex] × 3.6m
[tex]v^{2}[/tex] = 70.56 [tex]m^{2}/s^{2}[/tex]
The velocity of volley ball is,
v = [tex]\sqrt{70.56m^{2}/s^{2} }[/tex]
∴ v = 8.4 m/s
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Which of the following scenarios would most likely be considered pseudoscience?
A team of scientists replicating another experiment.
Generating a claim based on empirical evidence.
The theory of relativity.
The belief that the stars determine our fate.
The belief that the stars determine our fate is considered to be pseudoscience .
Large celestial bodies called stars are mostly made of hydrogen and helium, and in their cores, churning nuclear forges produce light and heat. Except for our sun, all of the other stars in the sky that we can see are light-years from the our planet i.e. earth.Celestial bodies, also referred to as heavenly bodies, are celestial objects in space, including the sun, moon, planets, and stars. They are a component of the vast universe that we exist in and are normally fairly far away from us.Pseudoscience, which is defined as an activity that purports to be scientific but is actually not, is defined as seems science. Science does not engage in extraterrestrial affairs it simply studies the natural world. Science cannot provide an answer to morality, immorality, or questions of meaning or purpose, but it does require ethical behavior. In other hand astronomy is regarded as pseudoscience since no link between the locations of the stars and human behavior can be demonstrated using evidence from science.
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Answer: The answer is The belief that the stars determine our fate or C
Explanation:
The diagram shows a position-time graph.
Position vs Time
What is the displacement of the object?
O-2 m
O-1 m
O 3m
O4m
Answer:
0_2 is 0.2 with 0_1 0.1 will 0_3 is 0.3 and with 0_2 0.
Explanation:
2
A 75 kg student drives a 650 kg car. What is the applied force on the car if the car
starts from rest and accelerates to 60 m/s in 8 seconds?
Answer:
5437.5
Explanation:
I took the quiz.
two small, identical steel balls collide completely elastically. initially, ball 1 is moving with velocity ????1 directly toward ball 2, and ball 2 is stationary. what are the final velocities of ball 1 and ball 2, respectively?
When two small ball collide elastically in which first ball is moving with velocity, v with second ball which is in rest than after collision the second ball will move with velocity, v and first ball will come to rest.
What is elastic collision?An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In elastic collisions, momentum and kinetic energy are both conserved.
An elastic collision is one in which the momentum, p and the kinetic energy, KE are both conserved.
[tex]p_{i} =p_{f}[/tex]
[tex]KE_{i} = KE_{f}[/tex]
Given in the question two identical steel ball that means have same mass and one is moving with velocity, v another is in the rest initially, after collision the momentum of first ball will transfer to second ball so second ball will move with velocity, v and first ball will come to rest.
So final velocity of first ball is zero and second ball is v.
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arunner jack 12km north then turns and runs 16km in 3hours calculate his average velocity
A runner jack 12km north then turns and runs 16km in 3hours Therefore his average velocity is 1.3 km/h.
What is average velocity?
The change in position or displacement (x) divided by the time intervals (t) during which the displacement happens is the definition of average velocity. Based on the direction of the displacement, the average velocity might either be positive or negative. Meters per second (m/s or ms-1) is the SI measure for average velocity.
Given:
A runner jack 12km north then turns and runs 16km in 3hours .
Therefore average velocity= (16 - 12)/ 3 km/h
=1.3 km/h
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According to Newton's first law of motion, what will an object in motion do when no external force acts on it?
come to a stop
move at the same velocity
speed up
change direction
Mark this and return
Answer:
Move at the same velocity
Explanation:
According to newton's 1st law of motion, an object at rest remains at rest or if in motion, it will remain in motion at constant velocity unless acted upon by a net external force
A merry-go-round spins freely when janice moves quickly to the center along a radius of the merry-go-round. It is true to say that.
No, its fake the instant of inertia of the machine decreases and the angular pace will increase.
A merry-go-spherical spins freely while janice actions fast to the middle alongside a radius of the merry-go-spherical. Moment of inertia is a amount withinside the rotational movement which performs a position analogous to position performed with the aid of using mass withinside the linear movement. Hence, it's also referred to as angular mass or the rotational momentum.
Janice decreases her contribution to the instant of inertia of the machine with the aid of using transferring towards the rotation axis. This decreases the whole second of inertia of the machine. The angular momentum L of the machine does now no longer extrade seeing that no outside torque is applied. The angular pace ω will increase sinceω = L/I.
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The largest and the smallest balls used in the experiment are with diameter 9. 52 mm, and 2. 38 mm respectively. For a glycerin with viscosity 1. 0 pa. S, what is the time necessary for each ball to reach a velocity 95% of the terminal velocity? density of the ball material is given in the text. Round the result to three decimal places.
The time necessary for the larger ball is 66.904 x 10⁻³ s and for the smaller ball is 4.166 x 10⁻³ s.
We have ,
density of balls(ρ) = 1.42 g/cm³
density of glycerin(σ) = 1.30 g/cm³
acceleration due to gravity = 9.8 m/s²
Diameter of larger ball = 9.52 mm
diameter of smaller ball = 2.38 mm
we know that the formula for terminal velocity is given by
[tex]V_{t}[/tex] =( [tex]\frac{2r^{2}g}{9n}[/tex] ) (ρ - σ)
now for larger ball terminal velocity is given by
[tex]V_{1}[/tex] = (2 [9.52/2 x 10⁻³]² x 9.8 x(1.42 x 10⁻³ - 1.30 x 10⁻³ )) / 9 x 1.0
∴ [tex]V_{1}[/tex] = 5.92 x 10⁻³ m/s
now for smaller ball terminal velocity is given by
[tex]V_{s}[/tex] = (2 [2.38/2 x 10⁻³]² x 9.8 x(1.42 x 10⁻³ - 1.30 x 10⁻³ )) / 9 x 1.0
[tex]V_{s}[/tex] = 0.37 x 10⁻³ m/s
Now 95 % of the terminal velocity of larger ball will be = 5.62 x 10⁻³ m/s
95 % of the terminal velocity of smaller ball will be = 0.35 x 10⁻³ m/s
Also the acceleration = 0.084 m/s⁻²
using the equation of motion
v= u + at ( v= 95% velocity and u = initial velocity (0) )
as u = 0
∴ v = at
t = v/a
so we can calculate time as follows
t (larger ball) = 5.62 x 10⁻³/0.084 = 66.904 x 10⁻³ seconds
t (smaller ball)= 0.35 x 10⁻³/ 0.084 = 4.166 x 10⁻³ seconds
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Can anyone give me the step by step solution for this question?
Answer:
ans a
Explanation:
a block is given an initial velocity of 5.00 m/s up a frictionless incline of angle 20 degrees. the magnitude of the block’s acceleration is 2.5 m/????2 while it is on the ramp. how far up the incline does the block get before sliding down again?
The block will incline up with projectile motion until 0.085 m before sliding down.
We need to know about the projectile motion to solve this problem. The projectile motion is known as parabolic motion and the velocity is divided by 2 axes.
vox = vo cosA
voy = vo sinA
where vox is initial velocity of x axis, voy is initial velocity of y axis, vo is initial velocity and A is the angle.
From the question above, the given parameters are
vo = 5 m/s
A = 20⁰
By using the uniform motion, where vty = 0 m/s. Hence,
vty² = voy² - 2gh
0 = 5.sinA - 2 . 10 . h
20h = 5 . sin20⁰
h = 1/4 . 0.34
h = 0.085 m
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7) You push on box G that is next to box H, causing both boxes to slide along the floor, as shown in the figure.
The reaction force to your push is
Push
GH
A) the push of box G against you. B) the acceleration of box G.
C) the push of box H on box G. D) the push of box G on box H.
Answer:
c
Explanation:
because whenever you push g box into h box and then h displaces according to third law of motion h box will apply the force on g box so that's why
a 2000 kg elevator with broken cables in a test rig is falling at 4.00 m/s when it contacts a cushioning spring at the bottom of the shaft. the spring is intended to stop the elevator, compressing 2.00 m as it does so. during the motion a safety clamp applies a constant 17,000 n friction force to the elevator.
A. The speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring is 3.65m/s
B. The acceleration when the elevator is 1.00 {\rm m} below point where it first contacts a spring is 4m/s²
In calculating the speed of the elevator and acceleration, first we have to find the force of gravity F on the elevator, which is the force pulling the elevator in downward direction. Using the equation for force of gravity which is:
F = mg
Where:
Mass of the elevator; m= 2000kg
Acceleration due to gravity; g = 9.8m/s
2000kg × 9.8m/s²= 19600N
F = 19600
Force of opposing friction clamp of gravity = 17000N
Net force on the elevator = force of gravity - Force of opposing friction clamp
Net force on the elevator = 19600 - 17000
Net force on the elevator = 2600 N
We will also find the kinetic energy K.E; of the elevator at the point of contact with the spring using:
K.E = 1/2 mv²
Where
Mass of the elevator; m = 2000kg
Velocity of the elevator = 4.00m/s
K.E = (1/2)*2000kg*(4m/s)²
K.E = 16000J
The kinetic energy and energy gained will be absorbed by the spring across the next 2m
Therefore,
Energy; E = K.E + P.E
Where:
Kinetic energy K.E = 16000J
Potential Energy P.E = ?
P.E of spring = net force absorbed × distance at compression
Where:
Net force absorbed = 2600N
Distance at compression = 2.0m
P.E = 2600*2
P.E = 5200J
E = 16000J + 5200J
E = 21200J
Spring constant = k
To find k
Using:
E = (1/2)*k*(x)²
Where:
E = 21200J
k = ?
x = 2m
21200J = (1/2)*k*(2m)²
21200J*2 = (4m)k
K = 42400J/4m
K = 10600N/m
Therefore,
Acceleration at 1m compression = ?
Using:
F = K*X
Where
F is force provided by the spring = 10600N/m,
K = 10600 N/m
X = 1m
F = 10600N/m * 1m
F = 10600N (upward)
A. The speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?
Using:
Original Kinetic energy + net force on the elevator = final kinetic energy + spring energy
16000N + 2600N = (1/2)mv² + (1/2)k x²
18600 = (1/2)(2000)(v²) + (1/2)(10600N)(1²)
18600 = 1000(v²) + 5300
18600 - 5300 = 1000(v²)
13300 = 1000(v²)
V² = 13.300
V =3.65m/s
B. The acceleration of the elevator is 1.00m below point where it first contacts a spring
Spring constant = net force on the elevator + resultant force
Where:
Spring constant = 10600N
Net force on the elevator = 2600N
Resultant force = ?
10600N = 2600N + resultant force
Resultant force = 10600N - 2600N
Resultant force = 8000N
Using the equation for Newton's 2nd law where F = ma,
a = F/m
Where:
Resultant force; F =8000N
Mass of the elevator; m =2000kg)
a = 8000 / 2000
a = 4m/s²
Here's the complete question:
In a "worst-case" design scenario, a 2000kg elevator with broken cables is falling at 4.00m/s when it first contacts a cushioning spring at thebottom of the shaft. The spring is supposed to stop the elevator,compressing 2.00m as it does so. During the motion a safety clampapplies a constant 17000N frictional force to the elevator.
1. What is the speed of the elevator after it has moved downward 1.00m from the point where it first contacts aspring?
2. When the elevator is 1.00m below point where it first contacts a spring, what is its acceleration?
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What do we call the illusion of movement that results from two or more stationary, adjacent lights blinking on and off in quick succession?.
In 1912 Wertheimer discovered the phi phenomenon, an optical illusion in which stationary objects shown in rapid succession, transcending the threshold at which they can be perceived separately, appear to move.
What is phi phenomenon ?When two neighbouring optical stimuli are presented in alternation with a relatively high frequency, an apparent motion is seen that is known as the phi phenomenon. At higher frequencies, beta movement is visible, but the stimuli themselves don't seem to move.
The concept of the phi phenomenon is crucial to the field of psychology known as Gestalt psychology. This field focuses on perception and is interested in comprehending how various components of a whole affect perception. The phi phenomenon tricks the brain into thinking that objects that aren't moving are.Learn more about Phi phenomenon here:
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if the parachutist comes to rest with constant acceleration over a distance of 0.660 mm , what force does the ground exert on her?
The force exerted by the parachutist on the ground is 471.24 N when she comes to rest at a distance of 0.660 m
Let t = Time taken
u = Initial velocity , v = Final velocity
s = Displacement, a = Acceleration
As given in the problem initial velocity u = 3.85 m/s
Final velocity v = 0 (rest)
s = 0.660 m
From the equation of motion v² = u² + 2as
we have a = v²-u² / 2s
⇒a = 0-3.85² / (2* 0.660)
⇒a = -11.22 m/s² (negative sign shows deceleration)
we know force F = mass m * acceleration a
Given mass of parachutist m = 42 kg
∴Force exerted by parachutist on ground
F = 42 * 11.22 = 471.24 N
If the distance is shorter, acceleration increases in magnitude and so does the force exerted by the parachutist.
The question is incomplete and the full question is probably " A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant acceleration over a distance of 0.660 m, what force does the ground exert on her? "
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Bode made a rocket using an empty plastic bottle and fins made out of cardboard. After partially filling the bottle with water, he pumps air into the bottle using a pump attached to the rocket with plastic tubes. When the pressure in the bottle becomes high enough, the water gushes out and launches the rocket into the air, as shown in the diagram below.
How does the force that causes the water to leave the rocket compare to the force that causes the rocket to launch?
A.
They are unequal and act in opposite directions.
B.
They are equal and act in the same direction.
C.
They are equal and act in opposite directions.
D.
They are unequal and act in the same direction.
The force that causes the water to leave the rocket compared to the force that causes the rocket to launch is that they are unequal and act in opposite directions.
According to Newton's third law of motion, every action has a corresponding and opposing response. In the instance of Bode's rocket, the pressured air in the bottle causes the water to rush downward out of the bottle, causing the rocket to fly upward into the air. The upward force on the rocket is equivalent to the force of the water streaming down. Thus, the two forces are equivalent and exert opposing forces.
When you pump, more air is pumped into the bottle while maintaining the same volume. By doing this, you raise the pressure inside the bottle, increasing the force acting on the interior of the bottle. Eventually, when the bottle can no longer withstand the pressure, the cork will be violently ejected from the bottle's bottom by the force of the pressure.
The air will be forced out of the bottle as rapidly as possible by the abrupt pressure drop caused by the new opening. When the air leaves the bottle, the pressure that was previously holding it under pressure is used to drive all of the water out with it.
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Answer:
They are equal and act in opposite directions.
Explanation:
I just answered the question on study island and got it right :P
help
What causes polar jet streams to form? (1 point)
A.rapid changes in wind direction
B.air masses moving away from each other
C.warm and cold air masses
D.high-pressure systems
Polar jet streams are formed because of the warm and cold air masses.
What is polar jet streams?
Polar front jet stream, also called polar front jet or mid-latitude jet stream stream, a belt of powerful upper level winds that sits at top the polar front.
The winds are strongest in the troposphere, which is the upper boundary of the troposphere, and move in a generally westerly direction in mid-latitudes.
The vertical wind shear which extends below the core of this jet stream is associated with horizontal temperature gradients that extend to the surface. As a consequence, this jet manifests itself as a front that marks the division between colder air over a deep layer and warmer air over a deep layer.
Polar jet streams are formed due to warm and cold air masses.
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Answer:
C
Explanation:
While on the moon, Buzz Aldrin carried on his back a support system that would
weigh over 1,775 N on Earth. What did the backpack weigh on the moon? (gravity
on the moon is 1.62 m/s^2)
Answer:
W' = 287.55 N
Explanation:
We know that weight force is defined as follows:
W = m × g
So:
On Earth (W = 1775 N):
1775 = m × 9.8
m = 1775 / 9.8
m = 181.12 kg
On the Moon (g' = 1.62 m/s²):
W' = m × g'
W' = 181.12 × 1.62
W' = 293.42 N