A 50n couch is lifted off the ground. How much work is done on the couch
The work done in lifting the couch to the given height is 250 J.
What is the work done on the couch?
The work done in lifting the couch is the product of the applied force and the displacement of the block.
Based on the law of conservation of energy, the work done in lifting the couch to the given height is equal to change in the gravitational potential energy of the couch.
P.E = mgh
P.E = Wh
where;
m is the mass of the couchg is acceleration due to gravityh is the height through which the couch was liftedW is the weight of the couchP.E = 50 N x 5 m
P.E = 250 J
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The complete question is below;
A 50n couch is lifted off the ground. How much work is done on the couch if it was lifted to a height of 5 m?
Answer:
25 J
Explanation:
A man walking through a field travelled 30 m south and then 20 m west.
Calculate the distance he travelled and his displacement
The net distance and displacement traversed is 50m and 36.0555 m respectively.
Given - A man has walked 30m and 20m south and west respectively.
To find distance and displacement .
principle- Distance means path covered and a vector quantity. Displacement means shortest distance covered and state function. It is also scalar quantity. Displacement can be zero but distance can't be zero.
Total distance traversed= 30 + 20 = 50m
Total displacement traversed (s) ⇒√(x₁² + x₂²)
⇒√(30² + 20²) ⇒√(900 + 400)
=√1300
=36.0555 m
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A car travels at a constant speed of 408 km/h, how long does it take to travel 80 m?
We first apply the data to the problem.
Data:
[tex] \bold{\: \: \: \: \: \: \: \: \: \: \:V = 408km/h}[/tex]
[tex] \bold{ \: \: \: \: \: \: \: \: \: \: \: D = 80m} \\ \bold{T = ?}[/tex]
Now, we convert km/h to m/s.
Conversion:
[tex] \bold{408km/h * (1000m/1km) * (1h/3600s) }[/tex]
[tex] \bold{V = 113.3m/s}[/tex]
Then, we apply the formula that is.
Formula:
[tex] \bold{T = D/V}[/tex]
Finally we develop the problem.
Developing:
[tex] \bold{T = 80m / 113.3m/s}[/tex]
[tex] \bold{T = 0.706s}[/tex]
In a time of 0.706 seconds.
At which point does the bungee jumper’s kinetic energy begin to change into potential energy?
At the bottom of the “ride,” when the jumper momentarily stops
What is curved space?
a bolder of mass 45kg is pushed on a surface with a coefficient of kinetic friction of 0.85. what force has to be applied to produce an acceleration of 2 m/s^2
Answer: The answer is 386 N
Explanation:
A wooden block of mass m is placed on a horizontal aluminum ramp. The angle of
the ramp is slowly increased to 10.8° above the horizontal, at which point the
mass slides down the ramp with constant speed.
Determine the coefficient of kinetic friction, %, between the wooden block and
the aluminum ramp.
Write your answer to 2 significant figures.
MAIN ANSWER- [tex]5/\sqrt{21} sec[/tex]
supporting answer-When a body is moving, it experiences resistance because it interacts with its surroundings. This resistance is a friction force. Friction opposes relative motion between systems in contact but also allows us to move, a concept that becomes obvious if you try to walk on ice. Friction is a common yet complex force, and its behavior still not completely understood. Still, it is possible to understand the circumstances in which it behaves.
body of the answer- a block of mass m with 10kgs is sliding on a inclined plane with frictionless surface
final answer-5/\sqrt{21} sec
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A 0.80-m-tall barrel is filled with water (with a weight density of 9800 N/m³). Find the water
pressure on the bottom of the barrel.
Express your answer to two significant figures and include the appropriate units.
Pressure on the bottom of the barrel = 1.78 * 10⁵Pₐ
a) The force applied perpendicular to the object's surface and dispersed over the area is referred to as the pressure.
Given,
height of barrel = 0.8mP
density = 9800N/m²
atmospheric pressure = Pₙ = 1.013 * 10⁵Pₐ
From the relation,
Pressure at the bottom
Pₓ = Pₙ + pgh
Pₓ = 1.013 * 10⁵ + 9800 * 9.8 * 0.8
Pₓ = 1.78 * 10⁵Pₙ
b)
Given,
mass = 1kg
area, a =1.2cm² =1.2 * 10⁻⁴ m²
Gravitational acceleration g = 9.8m/s²
We know that
Pressure on the finger,
Pf = force/area
Pf = (mass*g) / a
Pf = (1 * 9.8)/ (1.2 * 10⁻⁴)
Pf = 8.17 * 10⁴ N/m²
a) pressure on the bottom = 1.78 * 10⁵Pₐ
b) pressure on the finger = 8.17 * 10⁴ N/m²
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a train travels at 36 km/hr and accelerates uniformly to 108km in 10seconds. what is the acceleration?
Answer:
Below
Explanation:
Acceleration = change in velocity / change in time
(108 km/s - 36 km/s) / (10 s) = 7.2 m/s^2
Will give Brainliest!
A thin 2.50 kg box rests on a 5.50 kg board that hangs over the end of a table, as shown in (Figure 1).
How far can the center of the box be from the end of the table before the board begins to tilt?
Based on the principle of moments. the distance from the end of the table should the box be placed before the board begins to tilt is 8.6 cm.
What distance from the end of the table should the box be placed before the board begins to tilt?The distance from the end of the table should the box be placed before the board begins to tilt is determined from the principle of moments as follows:
sum of clockwise moments = sum of anticlockwise momentsThe block is 30 cm on the table and 20 cm outside it.
The downward force acting on the left-hand side of the box = 3/5 x 5.50 = 3.3kg.
This force acts at the center of gravity, 15 cm or 0.15 m away.
Therefore, anticlockwise moments o the left side = 3.3 x 0.15 = 0.495 J
Also, the clockwise moment on the right side = Force * distance
Force = 2/5 x 5.5 = 2.2 N
Distance from the center of gravity = 10 cm or 0.10 m away.
the clockwise moment on the right side due to the board = 2.2 x 0.1 = 0.22.
The moment due to the box with a weight of 2.5 kg at a distance of x meters will be:
the total clockwise moment on the right side = 0.28 + 2.5 * x.
When the board is just about to tilt:
0.495 = 0.28 + 2.5 * x
2.5x = 0.495 - 0.28
x = 0.086 m or 8.6 cm
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A race car leaves the starting line and travels 36,000 m in the first 600 seconds of the race. They are then forced to take a pit stop and don't go anywhere for 250 seconds. After the pit stop, they finish the race, going 24,500 m in 350 seconds.
a. What is the car's average speed during the first part of the race? (before the pit stop)
b. What is the car's average speed during the pit stop?
c. What is the car's average speed after the pit stop?
d. What is the car's average speed for the whole trip?
Answer:s= A. 60 m/s
B. s=0 m/s
c. s=70 m/s
D. s=50.4 m/s
Explanation:
Sorry if its wrong
The half-life of chromium-51 is 28 days. If the sample contained 510 grams, how much chromium would remain after 56 days?
The quantity of 128 grams of chromium would be remaining.
The time required for half the reaction to change is called half life period.
This concept is useful for understanding carbon dating in Nuclear Physics. The term is used more to characterize any type of exponential decay. Nuclei that decay easily have shorter half life.
Given :- Half Life of Cr = 28 days
m= 510 grams
To find :- quantity of Cr after 56 days.
Formula : Nₙ= N₀/(2)ⁿ
Nₙ= 510/2²
Nₙ= 128g
Therefore 128 grams will be available after 56 days.
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During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +30.0m/s at an angle of 50.0°, how long does it take the stone to hit the ground? What is the maximum distance that the trebuchet can be from the castle wall to be in range? How high will the stones go? Show all your work.
(a) The time of motion of the stone is 4.69 seconds.
(b) The maximum distance the trebuchet can be from the castle wall to be in range is 90.44 m.
(c) The maximum height reached by the stone is 26.95 m.
What is the time of motion of the stone?
The time of motion of the stone is calculated as follows;
t = 2u sinθ / g
where;
u is the initial velocity of the stoneθ is the angle of projection of the stoneg is acceleration due to gravityt = (2 x 30 x sin50) / (9.8)
t = 4.69 seconds
The maximum distance the trebuchet can be from the castle wall to be in range is calculated as;
R = u²sin(2θ) / g
R = (30² sin(2 x 50))/(9.8)
R = 90.44 m
The maximum height reached by the stone is calculated as follows;
H = u² sin²θ / 2g
H = (30² x (sin 50)²) / (2 x 9.8)
H = 26.95 m
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Question 3 (2 points)
Of the following exercise stimuli, ordered from the lightest to the heaviest, select the
intensity that has been shown to significantly improve affect for most people (albeit
perhaps for a relatively short period of time)
a 10-minute walk at a self-selected intensity
a 20-minute session of a cycle ergometer at 60% VO2max
a 30-minute treadmill run at 75% VO2max
a 60-minute session of aerobics at 75% VO2max
multiple 30 second intervals at 150% VO2max
A 20-minute cycle ergometer exercise at 60% VO2max has been demonstrated to dramatically improve impact for the majority of persons.
Intensity and an example are what?the property of being strongly felt or having a powerful impact: The explosion was so loud that it could be heard from five kilometres distant. Measures of luminosity. [C or U] the strength of being something that may be measured, such as sunlight, sound, etc.
The measurement of intensity:A measure that is produced from a randomised measure is known as an intensity measure in probability theory. Since the randomised measure of a set's expected value is the intensity measure, which is a non-random measure, it equates to the average volume that the random measure assigns.
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A cable exerts a constant upward tension of magnitude 1.38 ✕ 104 N on a 1.20 ✕ 103 kg elevator as it rises through a vertical distance of 2.50 m.
(a)
Find the work done by the tension force on the elevator (in J).
J
(b)
Find the work done by the force of gravity on the elevator (in J).
J
a) Work done by the tension force is obtained as 3.45 * 10^4 J
b) Work done by gravity is obtained as 2.94 * 10^4 J
What is the work done?In physics, the work done is referred to as the product of the force and the distance that have been covered by the force. We know that the work done by the tension force is the work that is done as the string is stretched across the distance as we have been given.
The work done by gravity is the work that is done as the mass is lifted across a height. Thus the work done is the work done when we lift the mass from one point to another.
In order to find the work done in each case;
a) Work done by the tension force = 1.38 ✕ 10^4 N * 2.50 m
= 3.45 * 10^4 J
b) Work done by the gravity = 1.20 ✕ 10^3 kg * 9.8 m/s^2 * 2.50 m
= 2.94 * 10^4 J
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2. What is the force on a 1 kg bal that is falling freely due to the pull of gravity?
Answer: 9.8 N
Explanation:
The pressure acting on a diver in a take is 5 bar. The height of water column above him is nearly
A) 9 m B) 30 m C) 40 m D) 50 m E) 60 m
Answer:
Diver's dive depth 50 m
Explanation:
Given:
p = 5 bar = 5·10⁵ Pa - Pressure
ρ = 1000 kg/m³ - Water density
g = 10 m/s²
______________
H - ? Diver's dive depth
p = ρ·g·H
H = p/ ( ρ·g)
H = 5·10⁵ / (1000·10) = 50 m
Questions
1. a. Explain why you need to use half the time taken for an echo to be
detected when you are doing calculations of distance.
b. Sound travels at 330 m/s in air. It takes 1.5s to detect the echo
from a building. Calculate the distance to the building.
(Hint: Remember that the time is for the journey there and back).
2. Explain why dolphins can find food at greater distances than bats.
3. A woman is having an ultrasound scan. Ultrasound travels at
1500 m/s inside the body. The ultrasound transmitter sends out a
pulse of ultrasound waves, which are reflected from a tissue surface
7.5 cm away. Calculate the length of time before the echo is received
(Hint: Be careful with units!)
1.a. The reason of using half time is an echo refers to the Reflection of sound after hitting the target just like a basket ball bouncing back as it hits its target i.e. earth.
So parameters like total energy , distance, amplitude of waves,.etc.. refers to the both action and reaction journey.
Like for example if a rubber ball travels 5m to hit the wall and returns back . so the net distance = incident distance + reflection distance⇒ 10m
But in Physics to compute the accurate result of incoming signals or energy perceived we just have to take either of the 2 journeys i.e. incidence or reflection. Second perception is the concept of taking average of multiple data. When there is 2 times i.e. one time to take incidence and other one is reflection , so in such cases average of the time for close vicinity of result.
1.b . The net distance propagated is 247.5 m.
Given - Sound travels 330m/s taking time 1.5s
To find the distance travelled -
distance travelled in 1 sec= 330 m
distance for 0.75 sec= 330 x 0.75 = 247.5m
Since 1.5 sec is the total time of the process of echo.
[tex]echo = incidence + reflection[/tex]
so taking time of one journey = 1.5 sec/ 2 = 0.75 sec
Hence the distance traversed is 247.5 m .
2. Dolphins uses echolocation as an ability or tool to perceive the incoming waves in atmosphere that includes sound waves.
They generate a special kind of sound that makes its target i.e. food to come to it.
Now talking about bats , although they also have similar ability of echolocation , but its mechanism to capture a food is time consuming comparatively and dolphins use higher frequencies compared to bats.
3.The length of time before the echo is received is 0.0001 sec.
Given - Speed = 1500m/s , distance =7.5cm ⇒0.075m
To find how long it will take before the receptance of incoming sound signal :-
[tex]2d=speed * time \\2d=vt[/tex]
[2 x 0.075]/ 1500= t
t= 0.0001 sec
so it takes 0.0001 sec to traverse before the echo is achieved.
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b) Robert pulls at a steady speed for
2m/s how far will she travel in 10 minutes
10mins?
I need help someone pleaseeeee
Tysm
Robert pulls at a steady speed for 2m/s. She travelled 1200m in 10 minutes.
Distance : Path travelled by an object is the total distance.speed : distance travelled in an interval of time.Time : It is inversely proportional to the time.Speed = 2m/s
time taken = 10 minutes
(1 minute = 60 seconds )
then,
10 mint = 10 x 60 = 600 seconds
To calculate how far she travel we use :
distance travelled = speed x time
distance = 2 x 600
distance = 1200m
Robert pulls at a steady speed for 2m/s. She travelled 1200m in 10 minutes.
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A
where light does not strike.
A) mirror
C) wave
is an area
B) transparent
D) shadow
q
Answer:
d
Explanation:
because when an object stand in front of light, light bends around the other and doesn't go through it
Units and magnitudes 1. Make the following transformations 500 cm to m:
Answer:
5 m
500 cm to m is 5 meters
To convert one length to another, a conversion is applied, but first we will propose some data to serve us when applying the conversation.
Data:
100 cm is the same as a meter.& 1 meter is the same as 100 cm.Now, we convert centimeters to meters.
Conversion:
500cm • (1m / 100cm)5mExplanation:
They are 5 meters because multiplying 500 by 1 will give you the same result of 500, then we divide the 500 with the last amount that we have left, which is 100, thus giving us a result of 5 meters.
how can an object have the same speed and mass but change its momentum?
Answer: It can change its direction
Explanation:
Momentum is a vector, which has magnitude and direction. An object can have the same speed and mass (magnitude) but if it changes direction, it just changed its momentum
A rocket blasts off from rest and attains a speed of 33.9 m/s in 12.7 s. An astronaut has a mass of 65.1 kg. What is the astronaut's
apparent weight during takeoff?
The apparent weight of the astronaut of mass 65.1 kg moving with a speed of 33.9 m/s in 2.7 s is 811.797 N.
What is weight?Weight is the force with which a body is attracted toward the earth or a celestial body by gravitation and which is equal to the product of the mass and the local gravitational acceleration.
To calculate the astronaut's apparent weight, we use the formula below.
Formula:
W = m{[(v-u)/t]+g}............ Equation 1Where:
W = The apparent weight of the astronautm = Mass of the astronautv = Final speedu = Initial speedt = Timeg = Acceleration due to gravityFrom the question,
Given:
m = 65.1 kgv = 33.9 m/su = 0 m/s (from rest)t = 12.7 sg = 9.8 m/s²Substitute these values into equation 1
W = 65.1{[(33.9-0)/12.7]+9.8]W = 65.1×12.47W = 811.797 NHence, the apparent weight of the astronaut is 811.797 N.
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URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!
Answer: B
Explanation:
You will have air resistance as a type of friction, and since there isn't a normal force opposing it, gravity will bring the ball toward the ground.
2. A runner whose initial speed is 8.06 m/s increases her speed to 8.61 m/s in order to win a race. If the
runner takes 5.00 seconds to complete this increase in speed, what is her acceleration?
Answer:
Explanation:
Given:
V₀ = 8.06 m/s
V = 8.61 m/s
t = 5.00
________
a - ?
Acceleration:
a = (V - V₀) / t
a = (8.61 - 8.06) / 5.00 ≈ 0,11 m/s²
Answer:0.11
Explanation:
Four park rangers are chasing an elephant, and from the GPS collar, they get the following vector readings: 720.4 m, 32.0° east of north; 570.3 m, 36.0° south of west; 170.8 m straight south. Instead of following the path, they decide to calculate the vector sum, and go directly to the animal, what do they calculate?
A(n) _____ bond is one in which electrons are shared equally.
Answer:
Non-polar covalent bond
Explanation:
This type of covalent bond where electrons are shared equally between atoms is called a non-polar covalent bond.
Calculate the work done by a 40 N force pushing a 0.5 kg toy 2.3 m.
The work done calculated is 92J.
What is work done and how to find it?Work is considered completed when a force produces motion. There are numerous aspects that affect how well a force performs. One of the aspects is the object's displacement in the force's direction. The second component is force. Work is referred to as the result of an object's displacement plus a force applied in the force's direction. Work is equal to F×S, where F represents force and S represents displacement. When a body is moved by a distance while being subjected to a force, the work done can be calculated by:
W= F×S
Given that,
Force applied= 40N
Displacement= 2.3m
So, on putting the given values in above equation,
W= 40×2.3
W= 92 J
Therefore, the answer is 92 Joule.
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[100 points]a force of 330 newtons is required to lift a crate up a ramp a distance of 5 meters in 2 seconds. how much power is used?
Answer:
17,482.36 J
Explanation:
Ans. W = F •d cos = 120 N • 165 m cos 28.0 = 17,482.36 J
Answer:
Explanation:
Given:
F = 330 N
h = 5 m
t = 2 s
_________
N - ?
N = A / t = F·h / t = 330·5 / 2 = 825 W
Find the quantity of heat required to melt completely 200g of lead initially at 27°C given that for lead melting point is 327°C. The specific heat capacity of lead is 0.14J/gK and specific latent heat of fusion for lead is 270J/g.
The qhantity of heat required to melt completely 200 g of lead is 62400 J.
What is heat?Heat can be defined as a form of energy that brings about the sensation of warmth.
To find the quantity of heat required, we use the formula below.
Formula:
Q = cm(t₂-t₁)+Cm.......... Equation 1Where:
Q = Quantity of heatc = Specific heat capacity of leadm = Mass of leadt₂ = Final temperaturet₁ = Initial temperatureC = Specific latent heat of fussion for leadFrom the question,
Given:
c = 0.14 J/gKC = 270 J/gm = 200 gt₁ = 27 °Ct₂ = 327 °CSubstitute these values into equation 1
Q = 0.14×200×(327-27)+270×200Q = 8400+54000Q = 62400 JHence, the quantity of heat required is 62400 J.
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