5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley to a 0.0150 kg mass.
Draw a free body diagram showing all the forces acting on each object
Calculate the acceleration of the masses
Calculate the tension force in the cord
How long will it take the block to get to go 1.2 m to the edge of the table.
What will the velocity be as soon as it gets to the edge?
Answers
Answer:
1) Please find attached, created with Microsoft Visio
2) The acceleration of the masses connected by the light string is 0.00735 m/s²
3) The tension in the cord is 0.147 N
4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s
5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s
Explanation:
1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio
2) The acceleration of the masses connected by the light string is given as follows;
F = Mass, m × Acceleration, a
The mass of the truck, M = 20.0 kg
The mass attached to the string, hanging rom the pulley, m = 0.0150 kg
The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N
The pulling force acting on the cart, F = M × a
∴ F = 0.147 N = 20.0 kg × a
a = 0.147 N/(20.0 kg) = 0.00735 m/s²
The acceleration of the truck = a = 0.00735 m/s²
3) The tension in the cord = F = 0.147 N
4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²
Where;
s = The distance to the edge of the table = 1.2 m
u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)
a = The acceleration of the cart = 0.00735 m/s²
t = The time taken
Substituting the known values, gives;
s = u·t + 1/2·a·t²
1.2 = 0 × t + 1/2 ×0.00735 × t²
1.2 = 1/2 ×0.00735 × t²
t² = 1.2/(1/2 ×0.00735) ≈ 326.5306
t = √(1.2/(1/2 ×0.00735)) ≈ 18.07
The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s
5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;
v² = u² + 2·a·s
u = 0 m/s
v² = 0² + 2 × 0.00735 × 1.2 = 0.001764
v = √(0.001764) = 0.042
The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.
Answer:
There's no answer I'm just taking points like you did me, so thank you for your points I'll put them to good use ;)
Related Questions
A train travels with a speed of 115km/hr. How much time does it take to cover a distance of 470km.
Answers
It would take approximately 4 hours for the train to cover a distance of 470km.
4 hours 5 minutes 13.04 seconds
WILL MARK BRAINLY!! QUICK HELP PLEASE!
A 40.0 kg rocket produces a
764 N upwards force ("thrust").
What is the net force acting on
the rocket?
Answers
Answer:
372Explanation:(40.0)×(9.8)=392
764N-392=372
Using Velocity vs Time Graphs to Find Acceleration
A graph titled velocity versus time has horizontal axis time (seconds) and vertical axis velocity (meters per second). A line has 4 straight segments. Line segment A runs from 0 seconds 0 meters per second to 1 seconds 15 meters per second. Then segment B runs to 2 seconds 20 meters per second. Then segment C runs to 4 seconds 20 meters per second. Then segment D runs to 5 seconds 0 meters per second.
The acceleration of segment D is m/s2.
Rank segments A, B, and C from least acceleration to greatest acceleration.
Least:
Greatest:
Answers
The acceleration at segment D is -20m/s²
The rank of the acceleration from the least to the greatest is -20m/s² < 0m/s² < 5m/s² < 10m/s² (D<C<B<A)
Acceleration is the change in velocity with respect to time.
a = v-u/t
Acceleration at segment A:
Aa = 15-0/1-0
Aa = 15m/s²
Acceleration at segment B:
Ab = 20-15/2-1
Ab = 5m/s²
Acceleration at segment C:
Aa = 0-0/4-2
Aa = 0m/s²
Acceleration at segment D:
Ac = 0-20/5-4
Ac = -20m/s²
Hence the acceleration at segment D is -20m/s²
The rank of the acceleration from the least to the greatest is -20m/s² < 0m/s² < 5m/s² < 10m/s² (D<C<B<A)
Learn more here: https://brainly.com/question/18398656
A 1500 kg car has an applied forward force of 5000 N and experiences an air resistance of 1250 N. What is the car's acceleration?
Answers
Answer:
[tex]2.33\ m/s^2[/tex]
Explanation:
Net Force
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
Fn = m.a
Where a is the acceleration of the object. The net force is the sum of the individual vector forces applied to the object.
The m=1500 Kg car has two horizontal forces applied: the forward force of 5000 N that causes the movement and the air resistance force of 1250 N that opposes motion.
The net force is Fn = 5000 N - 1500 N = 3500 N
To find the acceleration, we solve the equation for a:
[tex]\displaystyle a=\frac{Fn}{m}[/tex]
[tex]\displaystyle a=\frac{3500}{1500}[/tex]
[tex]\boxed{a = 2.33\ m/s^2}[/tex]
The car's acceleration is [tex]a = 2.33\ m/s^2[/tex]
n Olympic diver is on a diving platform 3.60 m above the water. To start her dive, she runs off of the platform with a speed of 1.3 m/s in the horizontal direction. What is the diver's speed, in m/s, just before she enters the water
Answers
Answer:
8.5m/s
Explanation:
Using the equation of motion
v² = u² + 2gH
v is the final speed of the diver
u is the initial speed of the diver
g is the acceleration due to gravity
H is the height of the object
Given the following
V = 1.3m/s
H = 3.60m
g = 9.8m/s²
Required
Initial speed of the diver u
Substitute the given values into the formula:
v² = u² -2gH
1.3² = u² - 2(9.8)(3.60)
1.69 = u²-70.56
u² = 1.69+70.56
u² = 72.25
u = √72.25
u = 8.5m/s
Hence the diver's speed, in m/s, just before she enters the water is 8.5m/s
Which one of the following is not a member of the solar family: (a) Morning Star (b) Evening Star (c) Pole Star (d) Tailed Star plzzz answer this...
Answers
Answer:
Tailed star
Explanation:
Tailed star
An object is falling from a height of d and reaches a final velocity of vf. Calculate velocity when the object has fallen a distance d/2.
Answers
Answer:
The final velocity, [tex]v_{f}[/tex], is gd.
Explanation:
The condition here is a free falling object. Thus from the third equation of motion under free fall, we have;
[tex]v_{f}[/tex] = [tex]v_{i}[/tex] + 2gs
where [tex]v_{f}[/tex] is the final velocity of the object, [tex]v_{i}[/tex] is the initial velocity of the object, g is the gravitational force and s is the height.
Since the object falls from a height of d, then [tex]v_{i}[/tex] = 0 m/s, and s = d.
So that;
[tex]v_{f}[/tex] = 0 + 2gd
= 2gd
[tex]v_{f}[/tex] = 2gd
When the distance is [tex]\frac{d}{2}[/tex], [tex]v_{i}[/tex] = 0 m/s.
Then;
[tex]v_{f}[/tex] = 2g[tex]\frac{d}{2}[/tex]
[tex]v_{f}[/tex] = g x d
When the object falls through the height [tex]\frac{d}{2}[/tex], then the final velocity is gd.
Which best compares radiation and conduction?
A. Both transfer thermal energy from warm objects to cooler objects.
B. Both can transfer energy through empty space.
C. Both transfer thermal energy by electromagnetic waves.
D. Both involve the transfer of energy through direct contact.
And also who likes Rick and Morty OwO
Answers
Answer:
A
Explanation:
If you could change one property of a substance ( solid, liquid, and gas) you have to use in your daily life, what would it be and why?
Please hurry
I will give you BRAINLIEST!!!!!!
And pleaseee write 5 sentences why??
Answers
Answer
Soap into a liquid
Explanation:
If 200 ml of a gas at 27 degrees celsius is cooled to -33 degrees Celsius at a constant pressure , the volume will be
A 250ml
B 204ml
C 196ml
D 160ml
Answers
Given that,
Volume of gas, V₁ = 200 mL
Initial temperature, T₁ = 27°C
Final temperature, T₂ = -33°C
To find,
New volume of the gas.
Solution,
Initial temperature, T₁ = 27°C = 300 K
Final temperature, T₂ = -33°C = 240 K
Let V₂ is the new volume.
The relation between volume and temperature is given by :
[tex]\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\=\dfrac{200\times 240}{300}\\\\=160\ mL[/tex]
Hence, the volume will be 160 mL.
an object weighs 150N on earth. what is the mass of the object?
Answers
Answer:
450 g =0.9921 lb
Explanation:
1. A man is running on the straight road with the uniform velocity
3 m/s. Calculate acceleration produced by him.
Answers
Describe the last three years of your life. Include positive and negative experiences.
For a middle school girl
Answers
Answer:
my middle school life was horrible and still is sadly ); could you please do my recent brainless <333
Explanation:
My middle school life was eh.i allways got assumed that i was a elementry kid when i was and 6 and 7 grade.until 8th when i hit my grow spurt.my voice end up getting really deep in 8th grade too.tho in 8th i got sent to a certain type of scholl for kids who did bad because of some dum teacher who lied that i hit her when i actually just went to her desk to get my phone sence she saw it in my pocket and took it away.i got sent there for 3 months and was worst thing ever.the treated us like prisoners in that school and made us tacky uniforms that i refused to where.i believe it was a school for 7th-12th.after my expierence at that school my expierence in 8th grade was almost done.i end up being popular at school for being that kid who hit the teacher.ofc that was like 2 years ago im 10th now.i dont really remember my 6th and 7th grade year.only thing i rember is i was a trouble maker and still am a trouble maker.
hope that helped you:)
please make my answer brainiest as im trying get 4 brainiest answers
26. For this table of data, which column is the independent variable?
radius of circle (m) area of circle (m2)
3.4
36.317
4.5
63.617
4.6
66.476
6.2
120.763
7.6
181.458
7.7
186.265
8.6
232.352
9.7
295.592
area of circle
o
radius of circle
O
None of these is correct.
Answers
Answer:
radius of a circle is measured by an instrument, so it is the independent variable
Explanation:
The independent variable corresponds to the magnitude controlled by the experimenter and in general can be measured by an instrument.
The dependent variable is the variable used to perform the calculations, obtained from a mathematical transformation of the independent variable.
The radius of a circle is measured by an instrument, so it is the independent variable and the area is calculated by
A = π r²
Therefore the area corresponds to the dependent variable
A washing machine is being dragged with a starting force of 380.8 N to the West, but this is reduced by friction. If the mass of the washing machine is 37 kg
and the coefficient of friction is 0.54, what was the Net acceleration experienced by the washing machine?
A 10.27 m/s^2 South
B 5 m/s^2 South
C 19.01 m/s^2 South
D 7.6 m/s^2 South
Answers
Answer:
B. [tex]5\,\frac{m}{s^{2}}[/tex] South
Explanation:
The correct statement is:
A washing machine is being dragged with a starting force of 380.8 N to the South, but this is reduced by friction. If the mass of the washing machine is 37 kg and the coefficient of friction is 0.54, what was the Net acceleration experienced by the washing machine?
The net force of the washing machine moving on the horizontal ground is equal to the external force minus the friction force, where the normal force from the ground on the washing machine is equal to its weight. The free body diagram of the washing machine is presented below. The equation of equilibrium associated with the acceleration of the washing machine is:
[tex]\Sigma F = P -\mu_{k}\cdot m \cdot g = m\cdot a[/tex] (1)
Where:
[tex]P[/tex] - External force exerted on the washing machine, measured in newtons.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]a[/tex] - Net acceleration of the washing machine.
If we know that [tex]P = 380.8\,N[/tex], [tex]m = 37\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\mu_{k} = 0.54[/tex], then the net acceleration experienced by the washing machine is:
[tex]a = \frac{P}{m}-\mu_{k}\cdot g[/tex]
[tex]a = \frac{380.8\,N}{37\,kg}-(0.54)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]a = 4.996\,\frac{m}{s^{2}}[/tex]
The correct answer is B.
An object has a force of 29.43 N acting on it. Its acceleration is 3.5 m/s2. What is the mass of this object?
Answers
Answer:
The mass of the object is 8.41 Kg.
Explanation:
Mechanical Force
The second Newton's law states that the net force exerted by an external agent on an object of mass m is:
F = m.a
Where a is the acceleration of the object. The SI unit for the force is the Newton: [tex]1\ Nw = 1~Kg.m/s^2[/tex]
We are given the net force of F=29.43 N acting on an object and producing an acceleration of a=3.5~m/s^2.
To calculate the mass of the object, we solve the above equation for m:
[tex]\displaystyle m=\frac{F}{a}[/tex]
[tex]\displaystyle m=\frac{29.43}{3.5}[/tex]
Calculating:
m = 8.41 Kg
The mass of the object is 8.41 Kg.
The force of attraction between two oppositely charged pith is 5mx 10 to the -6th power newtons. If the charge on the two is 6.7 x 10 to the -9th power coulombs what is the distance between the two charges
Answers
Answer:
0.28 m
Explanation:
The following data were obtained from the question:
Force (F) = 5×10¯⁶ N
Charge 1 (q₁) = 6.7×10¯⁹ C
Charge 2 (q₂) = 6.7×10¯⁹ C
Electrical constant (K) = 9×10⁹ Nm²C¯²
Distance apart (r) =?
Thus, the distance between the two charges can be obtained as follow:
F = Kq₁q₂/r²
5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²
5×10¯⁶ = 4.0401×10¯⁷ / r²
Cross multiply
5×10¯⁶ × r² = 4.0401×10¯⁷
Divide both side by 5×10¯⁶
r² = 4.0401×10¯⁷ / 5×10¯⁶
Take the square root of both side
r = √(4.0401×10¯⁷ / 5×10¯⁶)
r = 0.28 m
Therefore, the distance between the two charges is 0.28 m
Which of the following does not illustrate a complete circuit?
Answers
Answer:
I think A
Explanation:
because it dosn't have enough tools
Calculate the kinetic energy of a 50 kg cart moving at a speed of 18.6 m/s.
Answers
Answer:
8649 J
Explanation:
KE = 1/2mv^2
1/2(50)(18.6)^2
1/2(50)(345.96) = 8649 J
Ek = ½mv^2
This is for calculate the kinetic energy
A loudspeaker produces a sound wave of frequency 50hz.
The amplitude of sound wave is increased.
What is heard?
A) a louder sound of higher pitch.
B) a louder sound of the same pitch.
C) a sound of higher pitch but the same loudness.
D) a sound of the same pitch and loudness as before.
Answers
Here is your answer
Option A)
Explanation:
The larger the amplitude of the waves, the louder the sound. Pitch (frequency) – shown by the spacing of the waves displayed. The closer together the waves are, the higher the pitch of the sound.
Pls mark as brainliest
Alex wants to learn how to surf, but he is not a strong swimmer. He knows he needs to increase his ability to paddle out in order to catch the best waves. Which piece of advice would you give to help him get started on reaching his goal? He should swim at least three times a week at the community pool to build stamina. He should tread water every day to get more comfortable in the water. He should purchase the best surfboard he can afford because it will help him paddle faster. He should watch your friend who is a competitive surfer practice to learn her technique.
Answers
Answer:
He should swim at least three times a week at the community pool to build stamina.
Explanation:
A sound wave passes from air into water and then into steel. With each change in medium, the velocity of this wave will a(increase,then decrease b(decrease c(increase d( decrease,then increase ?
Answers
Answer:
a
Explanation:
because it can increase anything
Two skaters A and B. having masses 50 kg and
70 kg respectively, stand facing each other 6 m
apart on a horizontal smooth surface. They pull
on a rope stretched between them. How far does
each move before they meet?
(A) both move 3 m
(B) A moves 2.5 m and B moves 3.5 m
(C) A moves 3.5 m and B moves 2.5 m
(D) both move 4 m
(E) none of the above
Answers
Answer:
Explanation:
Hell im good dude
How long will it take an object traveling at 7 m/s to reach a distance of 26 meters
Answers
Answer:
3 hours 71 minutes
Explanation:
As because Speed= distance/time taken
so time taken= 3.71
explain why an inflated balloon is likely to burst if you leave it near a window with the sun streaming it?
Answers
Answer:
the sun light will heat up the balloon causing it to burst
A merry go round exerts a force of 1000 N on a rider on the
outer ring of animals when it takes 15 seconds to make a
complete revolution. If the person weighs 750 N, the radius
of the circle he is making is m. Round your answer to
the nearest tenth.
Answers
Answer:
The radius of the circle made by the person on the merry go round is 74.55 meters
Explanation:
The given parameters are;
The force the merry go round exerts on the rider = 1000 N
The time it takes the merry go round to make one complete revolution = 15 seconds
The weight of the person = 750 N
The radius of the circle made by the person on the merry go round = r
We have;
[tex]F_c = \dfrac{m \cdot v^2}{r} = m \cdot \omega ^2 \cdot r[/tex]
Where;
m = The mass of the person
v = The velocity of the person
[tex]F_c[/tex] = The centrifugal force acting on the person = 1,000 N
r = The radius of the circle made by the person on the merry go round
ω = Angular velocity = 2·π/15 rad/s
We have;
The mass of the person = The weight/(The acceleration due to gravity, g)
∴ The mass of the person = 750/9.81 ≈ 76.45 kg
By substituting the calculated and known values into the equation for the centripetal force, we have;
[tex]F_c[/tex] = m × ω² × r
1000 = 76.45 × (2·π/15)² × r
r = 1000/(76.45 × (2·π/15)²) = 74.55 m
The radius of the circle made by the person on the merry go round = r = 74.55 m.
Directions: Summarize the main ideas of this lesson by answering the question below.
How might a person acquire a fear of or aversion to something even though it has no negative effect
on him or her?
Answers
Answer:
Explanation:
Trypophobia is a fear or disgust of closely-packed holes. People who have it feel queasy when looking at surfaces that have small holes gathered close together.A person may start by imagining what they fear, then looking at pictures of the fear object, and then finally being near or even touching the source of their anxiety. In the case of trypophobia, a person with symptoms may start by simply closing his eyes and imagining something such as a honeycomb or seed poThey can help you find the root of the fear and manage your symptoms. Last medically reviewed on July 20, 2017 Medically reviewed by Timothy J. Legg, Ph.D., CRNP — Written by Annamarya Scaccia ...
the diagram shows the apparent motion of the sun across the sky?
Answers
Answer:
The earth rotates west to east
Explanation:
Considering that the sun doesn't really move and the earth is rotating around the sun, and it is moving from east to west with us (the earth) facing it in which means the earth rotates around the sun west to east!
Which of the following is NOT one of the essential components of an exercise program?
Answers
Answer:
i dont know dude but ask someone who is in your grade lol
Explanation:
A teacher pushes a cart, while a student pushes a second, heavier cart
with the same amount of force. If the student's cart has three times the
mass as the teacher's, which of the following is true? *
Answers
Answer:
frictional force acting on the cart 450 N.
Given:
mass of cart = 50 kg
acceleration = 3
To find:
Frictional force = ?
Formula used:
F - = 600 N
F = ma
= frictional force
Solution:
Force on the block is given by,
F - = 600 N
F = ma
= frictional force
= 600 - m a
= 600 - (50 × 3)
= 600- 150
= 450 N
According to the given data the frictional force acting on the cart 450 N.