Answer:
156.26N
Explanation:
The data needed are incomplete. Let the acceleration of the body be 3.5m/s²
Other given parameters
Mass = 1.35×10^1 = 13.5kg
coefficient of friction between the tires and the road = 0.850
Acceleration due to gravity = 9.8m/s²
According to Newton's second law:
Fnet = ma
Fnet = Fapp - Ff
Fapp is the applied force
Ff is the frictional force = umg
The equation becomes:
Fapp - Ff = ma
Fapp-umg = ma
Fapp - 0.85(13.5)(9.8) = 13.5(3.5)
Fapp - 109.0125 = 47.25
Fapp = 47.25+109.0125
Fapp = 156.2625N
Hence the applied force that caused the acceleration is 156.26N
Note that the acceleration of the car was assumed. Any value of acceleration can be used for the calculation.
How far does a car travel in 30.0 s while its velocity is changing from 50.0 km/h to 80.0 km/h at a uniform rate of acceleration?
Answer:
The car will travel 541.67 m
Explanation:
Constant Acceleration Motion
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
The relation between the initial and final speeds is:
[tex]v_f=v_o+a.t[/tex]
Where:
vf = Final speed
vo = Initial speed
a = Acceleration
t = Elapsed time
The acceleration can be calculated by solving for a:
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
And the distance traveled is:
[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}[/tex]
The car travels during t=30 s, and its speed changes from vo=50 Km/h to vf=80 Km/h.
Let's convert the speeds to m/s. Recall 1 Km=1000 m and 1 h= 3600 s:
vo=50 Km/h * 1000/3600 = 13.89 m/s
vf=80 Km/h * 1000/3600 = 22.22 m/s
Calculating the acceleration:
[tex]\displaystyle a=\frac{22.22-13.89}{30}[/tex]
[tex]a = 0.28\ m/s^2[/tex]
Now for the distance:
[tex]\displaystyle x=13.89*30+\frac{0.28.30^2}{2}[/tex]
[tex]x=416.67\ m+125\ m[/tex]
[tex]\boxed{x = 541.67\ m}[/tex]
The car will travel 541.67 m
According to the question,
Final speed, [tex]v_f = 80 \ km/h \ ,or \ 22.22 \ m/s[/tex]Initial speed, [tex]v_o = 50 \ km/h, or \ 13.89 \ m/s[/tex]Elapsed time, [tex]t = 30 \ s[/tex]We know the relation,
→ [tex]v_f = v_o +at[/tex]
or,
→ [tex]a = \frac{v_f-v_o}{t}[/tex]
Now,
The acceleration,
→ [tex]x = v_o.t+ \frac{a.t^2}{2}[/tex]
By substituting the values,
[tex]= \frac{22.22-12}{30}[/tex]
[tex]= 0.28 \ m/s^2[/tex]
hence,
The distance traveled be:
→ [tex]x = 13.89\times 30+\frac{0.28.30^2}{2}[/tex]
[tex]= 416.67+125[/tex]
[tex]= 541.67 \ m[/tex]
Thus the above answer is right.
Find out more information about the distance here:
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the work of warren washington would be most likely to explain which of the following phenomena
Answer:
t
Explanation:
A gradual shift of the North Magnetic Pole's position on Earth a steady rise in the planet's average temperature, a reduction in the Moon's light reflection.
What is the work of warren Washington?Dr. Washington and his team were able to recreate the impacts of physics on several components of weather, such as “how heat energy, water vapor, and chemicals travel between Earth's seas and the atmosphere,”
Through the use of climate models. 3 After that, computers projected atmospheric changes using the information from the climate models.
The 2007 Intergovernmental Panel on Climate Change (IPCC) report, which found that human behavior has a direct impact on the environment, used these models to support its findings.
As a result, Dr. Washington and his team were awarded the 2007 Nobel Peace Prize3.
Therefore, a slow change in the location of Earth’s North Magnetic Pole A gradual increase in Earth’s average temperature A decrease in the light reflection of the Moon
Learn more about warren Washington here:
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#SPJ2
A wagon moving down the road has 2000 kg m/s of momentum. Then a 500-kg bag of feed with no velocity is added to the wagon. What is the new momentum of the wagon with the feed bag in kg m/s?
Answer:
The new momentum of the wagon with the feed bag is (also) 2,000 kg·m/s
Explanation:
The given momentum of the wagon = 2000 kg·m/s
The mass of the added bag = 500 kg
The initial velocity of the bag of feed = 0 m/s
Therefore, by the conservation of momentum principle, we have;
The total initial momentum of the wagon and the bag of feed before the bad of feed is added to the wagon = The new momentum of the wagon with the feed bag
Which gives;
The total initial momentum of the wagon and the bag of feed before the bad of feed is added to the wagon = 2,000 kg·m/s + 500 kg × 0 = 2,000 kg·m/s = The new momentum of the wagon with the feed bag
Therefore, the new momentum of the wagon with the feed bag = 2,000 kg·m/s.
What do the most abundant elements in Earth’s atmosphere have in common?
Answer: Since the oceans are mostly water, the elements hydrogen and oxygen are the most common. Sodium and chlorine are found in the salt in ocean water. Earth's atmosphere is made up of a combination of gases, including nitrogen, oxygen, and carbon dioxide. Nitrogen and oxygen are the most common elements in the atmosphere..
an object on a planet has a mass of 243 Kg. what is the acceleration of the object, if the radius of the planet is 2.32 x 10^7m and the mass of the planet is 6.35 x 10^30 Kg estimate g as 6.67 x 10^-11 N (m/kg)
Answer:
7.87x10^5m/s^2
Explanation:
Which ball has larger Inertia before touching the ground when it falls freely from a
certain height (air friction is zero)
a. mass 50 kg b. mass 25 kg
c. mass 10 kg d. All of these
Answer:
[tex]Mass = 50kg[/tex]
Explanation:
Required
Which has the larger inertia
Inertia (I) is calculated as follows:
[tex]I = mr^2[/tex]
Where m represents the mass of the object and r represents the distance from the axis.
Take for instance m = 10 and r = 4
The inertia would be:
[tex]I = 10 * 4^2[/tex]
[tex]I = 160[/tex]
Assume r is constant and m is increased to 25
The inertia becomes:
[tex]I = 25 * 4^2[/tex]
[tex]I = 25 * 16[/tex]
[tex]I = 400[/tex]
Notice that there is an increment in inertia, when the mass increased.
Hence, mass of 50kg will have the larger inertia.
why can't birds fly in space
Answer:
wouldn't they be dead and there's no oxygen in space....right???
Answer:
no gravity and no oxygen
Explanation:
the bird would die of lack of oxygen before they're even able to fly, but they can't fly anyway because there's no gravity so they'd just float
A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the surface. (refractive index of water = 1.3) a) What is the angle of refraction? b) What should be the angle of incidence if we want an angle of refraction not greater than 45 ° ? c) What is the critical angle?
Answer:
a
[tex]\theta _2 = 13^o[/tex]
b
[tex]\theta _1 =32.94^o[/tex]
c
[tex]\theta_c = 53.05^o[/tex]
Explanation:
From the question we are told that
The angle of incidence is [tex]\theta_1 = 10^o[/tex]
The refractive index of water is [tex]n_1 = 1.3[/tex]
Generally Snell's law is mathematically represented as
[tex]n_1 sin(\theta_1) = n_2 sin(\theta_ 2)[/tex]
Here [tex]n_2[/tex] is the refractive index of air with value [tex]n_2 = 1[/tex]
[tex]\theta_2[/tex] is the angle of refraction
So
[tex]\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ][/tex]
=> [tex]\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ][/tex]
=> [tex]\theta _2 = 13^o[/tex]
Given that the angle should not be greater than [tex]\theta _2 =45^o[/tex] then the angle of incidence will be
[tex]\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ][/tex]
=> [tex]\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ][/tex]
=> [tex]\theta _1 =32.94^o[/tex]
Generally for critical angle is mathematically represented as
[tex]\theta_c = sin^{-1}[\frac{n_2}{n_1} ][/tex]
=> [tex]\theta_c = sin^{-1}[\frac{1}{1.3} ][/tex]
=> [tex]\theta_c = 53.05^o[/tex]
does the voice from loudspeaker also get shrill when the listener reaches the sounding source?
Answer:
Yes, cause when listener is far away from music source... the listener feels it's pleasant.... but when observer approaches near loudspeaker... sound become unpleasant..........sound becomes noise when frequency crosses 85 Decibels.....so you can say that the voice from loudspeaker also get shrill when the listener reaches the surrounding source......
A golf player hits a 0.045 kg golf ball that is initially at rest, changing its momentum by 4.2 kg times m/., What is the final speed of the golf ball?
Answer:
v₂ = 93.33 [m/s]
Explanation:
In order to solve this problem, we must apply the definition of momentum and amount of movement, which can be calculated by means of the following expression.
[tex]P=m*v[/tex]
where:
P = momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
The amount of movement before and after the impact should be analyzed.
[tex](m_{1}*v_{1})+Imp_{1-2}=(m_{1}*v_{2})[/tex]
where:
m₁ = mass of the ball = 0.045 [kg]
v₁ = velocity of the ball before the hit = 0
Imp₁₋₂ = Impulse [kg*m/s]
v₂ = velocity after the hit [m/s]
[tex](0.045*0)+4.2=(0.045*v_{2})\\4.2 = 0.045*v_{2}\\v_{2}= 93.33 [m/s][/tex]
An object has three forces acting upon it, the first force is 58.2 N to the right and the second force is 14.0 N to the left. What is the magnitude of the third force, such that the object travels at a constant velocity? Answer in units of N and be sure to include all necessary work, as well as a free body diagram.
Help Please!
Answer:
44.2 N towards left
or
-44.2 N towards right
Explanation:
pls refer attachment for detailed solution
A stunt-rider on a motorcycle rides down a ramp and into a vertical loop-the-loop. If the diameter of the loop is 7.50 m, then the slowest speed the motorcycle can have at the top of the loop if it is to remain in contact with the loop is _________ km/h.
Answer:
v = 6.06 m/s
Explanation:
In order for the rider to pass the top of the loop without falling, his weight must be equal to the centripetal force:
[tex]Centripetal Force = Weight \\\frac{mv^2}{r} = mg\\\\\frac{v^2}{r} = g\\\\v = \sqrt{gr}[/tex]
where,
v = minimum speed of motorcycle at top of the loop = ?
g = acceleration due to gravity = 9.8 m/s²
r = radius of the loop = diameter/2 = 7.5 m/2 = 3.75 m
Therefore, using these values in equation, we get:
[tex]v = \sqrt{(9.8\ m/s^2)(3.75\ m)}[/tex]
v = 6.06 m/s
A car of weight 11000 N moves with constant velocity along a horizontal road.A driving force of 5000 N acts on the car what is the force opposing the motion of the car
Answer:
5000 N
Explanation:
Since F - f = ma, where F = driving force = 5000 N, f = opposing force, m = mass of car and a = acceleration of car.
Since the car is moving with constant velocity, a = 0.
So, F - f = ma
F - f = m(0)
F - f = 0
F = f.
Since F = 5000 N and F = f, f = 5000 N.
So, the opposing force is 5000 N
The opposing force car, weight 11000 N that is moving at constant velocity is 5000 N.
The opposing force can be calculated by the formula,
F - f = ma
Where,
F = driving force = 5000 N,
f = opposing force = ?
m = mass of car = 11000 N
a = acceleration of car
Since, the car is at constant velocity, a = 0.
So,
F - f = m(0)
F - f = 0
F = f.
Therefore, the opposing force car, weight 11000 N that is moving at constant velocity is 5000 N.
To know more about opposing force,
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An object has an acceleration of 6.0 m/s/s. If the net force acting upon this object were tripled , then its new acceleration would be _____ m/s/s.
Answer:
18m/s/s
Explanation:
6 times 3= 18
Two equal forces act at the same time on the same motionless object, but in opposite directions. Which statement best describes the object’s resulting motion?
A. The object will accelerate.
B. The object will move at a constant speed.
C. The object will remain motionless.
D. The object will change direction.
Un bloque de 25 n se encuentra suspendido por un hilo al techo vitamina la tensión que aparece en el hilo
Answer:
T = 25 N
Explanation:
The question says that "A 25 n block is suspended by a wire from the ceiling vitamin the tension that appears in the wire ?"
Weight of the block, W = 25 N
Weight of a body acts in downward direction and tension acts in upward direction. It would mean that,
Tension = weight of the block
T = mg
T = 25 N
Hence, the tension in the wire is 25 N.
Dawn and Timothy are sailboating in Lake Obewon. Starting from rest near the shore, they accelerate with a uniform acceleration of 0.39 m/s/s. How far are they from the shore after 18 seconds? *
Answer:
The value is [tex]s =63.18 \ m[/tex]
Explanation:
From the question we are told that
The uniform acceleration is [tex]a = 0.39 \ m/s^2[/tex]
The time considered is [tex]t = 18 \ seconds[/tex]
Generally the from kinematic equation we have that
[tex]s = ut + \frac{1}{2} at^2[/tex]
Given that the boat started from rest , the initial velocity is u = 0 m/s
So
[tex]s = 0 * 18 + \frac{1}{2} *0.39 * 18^2[/tex]
=> [tex]s =63.18 \ m[/tex]
a student determines the circumference of a golf ball. a man pulls a sledge of mass 25 kg across level ground with a horizontal force of 60 N. a constant force of friction of 20 N acts on sledge. what is the acceleration of the sledge?
Explanation:
Fnet = Ft - Ff = 60N - 20N = 40N
a = Fnet / m = 40N / 25kg = 1.6m/s².
Calculate resultant force:
R = ma
Descomposing:
F - Fr = ma
Replacing we have:
60 N - 20 N = 25 kg * a
Resolving:
40 N = 25 kg * a
40 N / 25 kg = a
1,6 m/s^2 = a
The aceleration is 1,6 meters per second squared.
Which one produces the brightest light out put on a miniature bulb,the solar panels connected in series or parallel???
(1) A block of wood weighs 980N on earth where acceleration due to gravity is 9.8N/Kg. The
moon has acceleration times that on earth. Determine the weight of the block of wood
on the moon.
(3 marks)
Given,
Weight of a block on the Earth = 980 N
Acceleration due to gravity on Earth = 9.8 N/kg
To find,
Weight of the block on the moon.
Solution,
The moon has acceleration (1/6) times that on earth. Let m be the mass of the block on the moon.
We know that,
W = mg
[tex]m=\dfrac{W}{g}\\\\m=\dfrac{980}{9.8}\\\\=100\ kg[/tex]
As mass of an object remains the same everywhere. Weight of the block on Moon is :
W' = mg'
As g'=(g/6)
So,
[tex]W=100\times \dfrac{g}{6}\\\\=100\times \dfrac{9.8}{6}\\\\=163.33\ N[/tex]
So, the weight of the block on the moon is 163.33 N.
A tourist stands at the top of the Grand Canyon, holding a rock, overlooking the valley below. Find the final velocity and displacement of the rock after 4.0 seconds give the three initial conditions listed below:
a. the rock is dropped vertically from rest
b. the rock is tossed vertically upward with 8.0m/s of speed
c. The rock is tossed vertically downward with 8.0 m/s speed
Answer:
a. -39.2 m/s; -78.4 m
b. -31.2 m/s; -46.4 m
c. -47.2 m/s; -110.4 m
Explanation:
Part (a)We are given/can infer these variables:
t = 4.0 sa = -9.8 m/s² v_0 = 0 m/sWe want to find the displacement and the final velocity of the rock.
Δx = ?v = ?We can use this equation to find the final velocity:
v = v_0 + atPlug in the known variables into this equation.
v = 0 + (-9.8)(4.0) v = -9.8 * 4.0 v = -39.2 m/sThe final velocity of the rock is -39.2 m/s.
Now we can use this equation to find the displacement of the rock:
Δx = v_0 t + 1/2at²Plug in the known variables into this equation.
Δx = 0 * 4.0 + 1/2(-9.8)(4.0)² Δx = 1/2(-9.8)(4.0)² Δx = -4.9 * 16 Δx = -78.4 mThe displacement of the rock is -78.4 m.
Part (b)We are given/can infer these variables:
v_0 = 8.0 m/sa = -9.8 m/s²t = 4.0 sWe can use this equation to find the final velocity:
v = v_0 + atPlug in the known variables into this equation.
v = 8.0 + (-9.8)(4.0) v = 8.0 + -39.2 v = -31.2 m/sThe final velocity of the rock is -31.2 m/s.
We can use this equation to find the displacement:
Δx = v_0 t + 1/2at²Plug in known variables:
Δx = 8.0(4.0) + 1/2(-9.8)(4.0)² Δx = 32 - 4.9(16)Δx = -46.4 mThe displacement of the rock is -46.4 m.
Part (c)We are given/can infer these variables:
v_0 = -8.0 m/sa = -9.8 m/s²t = 4.0 sWe can use this equation to find the final velocity:
v = v_0 + atPlug in the known variables into this equation.
v = -8.0 + (-9.8)(4.0) v = -8.0 - 39.2 v = -47.2 m/sThe final velocity of the rock is -47.2 m/s.
We can use this equation to find the displacement:
Δx = v_0 t + 1/2at²Plug in known variables:
Δx = -8.0(4.0) + 1/2(-9.8)(4.0)²Δx = -32 - 4.9(16)Δx = -110.4 mThe displacement of the rock is -110.4 m.
What is the velocity of a car with a momentum of 100 kg*m/s and a mass of 35kg?
Answer:
2.86 m/sExplanation:
The velocity of the car can be found by using the formula
[tex]v = \frac{p}{m} \\ [/tex]
p is the momentum
m is the mass
From the question we have
[tex]v = \frac{100}{35} = \frac{50}{7} \\ = 2.857142...[/tex]
We have the final answer as
2.86 m/sHope this helps you
A force of 10N is making an angle of 30
with the horizontal. Its vertical component
will be;
Explanation:
The vertical component = 10sin30° = 5.0N.
Which celestial body would have the strongest gravitational pull on a satellite orbiting 100 km above its surface? Explain your answer.
Answer:
Jupiter
Explanation:
Since the mass of Jupiter is the greatest from the given choices, it will exert the most force on any object orbiting 100km above its surface.
This is compliance with the Newton's law of universal gravitation which states that "the force of attraction between two bodies is directly proportional to the magnitude of their masses and inversely proportional to the distances between them".
Therefore, the more the masses of two bodies, the higher the gravitational attractionSince the distance is the same, the planet with the greater mass will exert the most force on the satellite.Answer:
Sample Response: Jupiter would have the strongest gravitational pull on a satellite orbiting above its surface because gravity is directly proportional to mass and Jupiter is the most massive planet.
Explanation:
edge 2021
Which option is an element?
A. Carbon dioxide
B. Carbon
C. Air
D. Water
The variables for Part I of this experiment include whether or not the car hits the barrier and the distance that the washers travel. Use the drop-down menus to identify the independent and dependent variables.
The independent variable, the one that is intentionally manipulated, is
____.
The dependent variable, the one that you measure the response in, is
___.
Answer:
1. An encounter with the barrier
2. The distance traveled by the washer.
Explanation:
Answer:
Peep above me is correct
1.An encounter with the barrier
2.The distance traveled by the washer
Explanation:
got it right on edge2020 have a nice day peeps ^w^
Which law represents a balanced chemical equashion
Answer:
law of conservation of mass
Explanation:
hope that helped :)
Where is the epicenter of the hypothetical earthquake as shown in the illustration below?
A. Point A
B. Point B
C. Point C
D. Point D
Answer:
Point D
Explanation:
The epicenter of a hypothetical earthquake is located at the point where the earthquake begins.
(See the attached image).
Hope it helps!
We work in less time.
3.
Differentiate between:
a)
Work and energy
b)
Work and power
c) Kinetic energy
Kinetic energy and potential energy
Work from burning fuel and work from moving objects
d)
110
Answer:
1 work is scalar quantity
2 it si unit of work is joule
3The equation of caculate is work= Force× Displacement
1 power is scalar quantity
2 it si unit of power is watt
3 the equation of calculateis power= work \time
PLEASE ANSWER - WILL MARK BRAINLIEST - THANKS
List three scientific ideas, discoveries, concepts, or issues that interest you.
Answer:
Evolution, Big Bang Theory and Cell theory.
Explanation:
These theories interest me because they talk about life and how life started.