Answer:
.............................
A child is stationary on a swing.(a)The child is given a push by his brother to start him swinging.His brother applies a steady force of 84 N over a distance of 0.25 m.(i) Calculate the work done by this force.(2)..............................................................................................................................................(ii) State how much energy is transferred by this force.(1)..............................................................................................................................................(iii) After several more pushes, the child has a kinetic energy of 71 J.The mass of the child is 27 kg.Show that the velocity of the child at this point is about 2.3 m/s.(2)(iv) Which one of these quantities changes in both size and direction while he is swinging?Put a cross ( ) in the box next to your answer.(1)Ahis gravitational potential energyBhis momentumCthe force of gravity acting on himDhis kinetic energy
Answer:
Work done = Fs
Explanation:
(i) W = Fxs
= 84N x0.25m
= 21 Joules
(ii) Ek = 1/2mv^2
71 = 1/2×27×v^2
71×2/27 = v^2
v = 23m/s
The work done by this force is 21 Joule.
Energy transferred by the force is 21 Joule.
The boy's momentum changes periodically, that is, in both size and direction.
What is force?An external force is an agent that has the ability to change the resting or moving condition of a body. It has a direction and a magnitude. So, it is a vector quantity. Newton is the SI unit of force (N).
(i) Given parameter:
Steady force applied by his brother, F = 84 N.
Displacement, d = 0.25 m.
The work done by this force, W = F.d = 84×0.25 = 21 Joule.
(ii) Amount of work done is transferred by this force as an energy. So, amount of energy transferred by the force is 21 Joule.
(iii) Given parameter:
Mass of the child, m= 27 kg.
Kinetic energy, E = 71 J.
Let, velocity of the child = v, then kinetic energy is,
E = 1/2 × mv²
⇒ v = √(2E/m)
= √(2×71/27)
= 2.3 m/s.
Hence, the velocity of the boy is 2.3 m/s. (proved).
(iv) During swing, his velocity changes periodically. That's why, his momentum changes periodically, that is, in both size and direction.
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#SPJ2
100N force is applied to a ball with a mass of 0.25kg. Calculate the ball acceleration.
Please find attached photograph for your answer.
Hope it helps.
Do comment if you have any query.
List examples of how the Bill of Rights protects you:
.
.
.
.
.
guaranteeing freedom of speech, press, assembly, and exercise of religion
Imagine you have a ball tied to the end of a string. You hold the other end of the string and swing it around. Suppose the string breaks, what direction will the ball travel
A 22 kg body is moving through space in the positive direction of an x axis with a speed of 190 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 13 kg, moves away from the point of explosion with a speed of 130 m/s in the positive y direction. A second part, with a mass of 2.2 kg, moves in the negative x direction with a speed of 460 m/s. What are the (a) x-component and (b) y-component of the velocity of the third part
Answer: Our notation is as follows : the mass of the original body is M=20.0kg ; its initial velocity is
ν
0
=(200m/s)
i
^
; the mass of one fragment is m
1
=10.0kg ; its velocity is
ν
1
=(100m/s)
j
^
; the mass of the second fragment is m
2
=4.0kg ; its velocity is
ν
2
=(−500m/s)
i
^
; and , the mass of the third fragment is m
3
=6.00kg . Conservation of linear momentum requires
M
ν
0
=m
1
ν
1
m
2
ν
2
+m
3
ν
3
.
The energy released in the explosion is equal to ΔK , the change in kinetic energy .
(a) Using the above momentum -conservation equation leads to
ν
3
=
m
3
M
ν
0
−m
1
ν
1
−m
2
ν
2
=
6.00kg
(20.0kg)(200m/s)
i
^
−(10.0kg)(100m/s)
j
^
−(4.0kg)(−500m/s)
i
^
=(1.00×10
3
m/s)
i
^
−(0.167×10
3
m/s)
j
^
.
The magnitude of
ν
3
is
ν
3
=
(1000m/s)
2
+(−167m/s)
2
=1.01×10
3
m/s
It points at θ=tan
−1
(−167/1000)=−9.48
∘
(that is at 9.5
∘
measured clockwise from the +x axis) .
(b) The energy released is ΔK :
ΔK=K
f
−K
i
=(
2
1
m
1
ν
1
2
+
2
1
m
2
ν
2
2
+
2
1
m
3
ν
3
2
)−
2
1
Mν
0
2
=3.23×10
6
J
Explanation:
URGENT NEED GIVING POINTS question is how do i find R and the angle THANK YOU
Answer:
Theta = 26.93°
R = 0.38m
Explanation:
Theta = cos^-1(0.74m/0.83m)
Theta = 26.93°
R = sqrt(0.83²-0.74²)
R = 0.38m
The Maximum and Minimum resultant of two forces are 35N and 51N. What is the possible Value of the two forces ?
hplhevsdvakxjb1zonsvaka shdv
please help me guys please
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Solution is in attachment ~
I hope that you got what you were looking for, and if there's different data then go through the same procedure, using same formula with different values and you will get your answer ~
[tex]\mathrm{✌TeeNForeveR✌}[/tex]
Answer:1.9 × 10³⁰
Explanation:
IF THE VECTOR COMPONENTS OF VECTOR A ARE MULTIPLIED BY 9 THEN THE MAGNITUDE OF VECTOR A IS INCREASES BY A FACTOR OF .........
Answer:
it increased by 29 sorry hope this helps
Rory uses a force of 25 N to lift her grocery bag while doing 50 J of work. How far did she lift the
grocery bag?
Answer:
[tex]W=Fd \\ 50 = 25d \\ d = \frac{50}{25} \\ \color{yellow} \boxed{d = 2m}[/tex]
Jefferson refers to
_____ rights in the Declaration of Independence.
Explain what these rights are:
Answer: natural
Explanation:
Answer:
natural i just copy tha answer in there i just explain it(づ￣ 3￣)づ
Explanation:
The meaning of the term “Pursuit of Happiness.” In the Declaration of Independence, Thomas Jefferson announced that every human being has “certain unalienable rights,” among which are those to “life, liberty, and the pursuit of happiness.” What did he mean by “the pursuit of happiness”?
what is the average speed of the toy car during the two trials to the nearest tenth of m/s
wait nvm the answer is D
Answer:
what is the average speed of the toy car during the two trials to the nearest tenth of m/s
wait nvm the answer is B
Explanation:
CARRY ON LEARNING
If F = 4.0 N and m = 2.0 kg, what is the magnitude a of the acceleration for the block shown below? The surface is frictionless. *
The magnitude of the acceleration of the block is 3.5 [tex]\bold{m/s^2.}[/tex]
The correct option is (C) 3.5 [tex]\bold{m/s^2.}[/tex]
What do you mean by acceleration?Acceleration is the rate of change of velocity with time.
By Newton’s Second law
This law establishes the relationship between the net applied force on an object with the resultant acceleration. For a constant magnitude of force, the more is the mass of the object, the lesser will be its acceleration.
Given,
The mass of the block is 2.0 kg.
The force in the horizontal direction on the block is [tex]\bold{4.0 N + 4.0 N \cos 40{}^\circ.}[/tex]
a is the acceleration in the horizontal direction.
The acceleration for the block can be determined using Newton’s Second law,
[tex]\begin{aligned} \Sigma F&=ma \\ F+F\cos 40{}^\circ &=ma \\ a&=\frac{F\left( 1+\cos 40{}^\circ \right)}{m} \\ &=\frac{4.0\text{ N}\left( 1.766 \right)}{2.0\text{ kg}} \\ &=3.5\text{ m/}{{\text{s}}^{2}} \end{aligned}[/tex]
Thus, the acceleration is 3.5 [tex]\bold{m/s^2.}[/tex]
Learn more about acceleration, here:
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A car accelerates from 20m/s to 30m/s in 10 sec. Find the cars acceleration using v=u+at
Answer:
1
Explanation:
because when we use v=u+at we derived v-u÷t
Asap
A cyclist needs to spin 100 meters with a velocity of 5 m/s. How long will the cyclist take?
Answer:
[tex]v = \frac{d}{t} \rightarrow t= \frac{v}{d} \\ = 100m \div 5m. {s}^{ - 1} = \frac{100}{5} \\ \color{green} \boxed{ t = 20s}[/tex]
To minimize signal distortion, at each end of the J-1939 CAN-bus there is a(n)_____________resistor.
Explain when acceleration remains constant.
Explanation:
acceleration remains constant when velocity does not increase or decrease
If a person is pushing a cart with a force of 40 Newtons and it
accelerates at 0.5 m/s², what is the mass of the cart?
Answer:
80kg
Explanation:
A = Fm
0.5 = 40m
m = 40/0.5
m = 80kg
2. Given what you know about the acceleration of Earth's gravity (g = 9.8 m/s2), is this number accurate?
accurate. If not explain why you think it is not accurate. Pleaseee help mee
Answer:
it is correct
Explanation:
Though no rounded numbers can be defined as accurate, if we were going by people's discovery, and research, we can define that the number, g = 9.8m/s^2, is accurate
Listed below are four activities. In which activity is friction useful?
a. climbing up a cliff
b. running on the ground
c. opening a bottle
d. catching a ball
Answer:
a. climbing up a cliff
Explanation:
How to find resistance
Answer:
R = V/I
Explanation:
Example:
V = IR
R = V/I
If we were to use V = 120 and I = 0.25
R = 120/0.25
R = 480 Ohms
Your resistence would equal Ohms
Mjchddjnfndndnddnndnddn
Answer:b
Explanation:
which acts as a transverse wave with particle motion perpendicular to wave motion?
Answer:
transverse wave, motion in which all points on a wave oscillate along paths at right angles to the direction of the wave's advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves.
a comet or asteroid impact on the earth is a frequently suggested cause for many of the mass extinctions. for which of the major mass extinctions does the most evidence exist that this was the main trigger for the event?
Answer:
Cretacious-Tertiary
Explanation:
I hope this helps!
A person who sits on the right-hand seat of the car that is making a left turn slides over to the right. What is the possible reason for this?
Answer:
zsjgbjidasngwhugbhwuabvhuvbwhuebewghvwev
Explanation:
11111111111111111111111111111111111111111111111111111
1 point
What is the potential energy of a bird flying, if the bird has 1000 J of total
energy and 450 J of kinetic energy? (only put the number, no units or
commas)
Answer:
the bird has 550J of potential energy
Explanation:
PEtotal=PE+KE
1000J=PE+450J
subtracting the kinetic energy from the total we get:
1000J-450J=PE
550J=PE
Does the mass of an object always equal the sum of its parts?
yes
no
What is the weight, on Earth, of a book with a mass of 1. 5 kg? 1. 5 N 6. 5 N 11. 3 N 14. 7 N.
Answer:
The weight, on Earth, of a book with a mass of 1.5kg is 14.7N.
Explanation:
Hi there!
Weight of an object = mass (kg) × acceleration due to gravity (N/kg)
Plug in the given values using the g constant g = 9.8 N/kg:
W = 1.5 × 9.8 = 14.7 N
Find the gravitational potential energy of a body of mass 25kg,kept at a height of 4m
Answer:
Massm=2.5kg
Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .
Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mgh
Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mghU=2.5×10×15
Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mghU=2.5×10×15U=25/10×10×=375j
Find the number of moles in a 28.0g sample of NH3.
PLS URGENTLY
Answer:
Molar mass of NH3=14+3=17g
[tex]mole = \frac{given \: mass}{molar \: mass} \\ mole = \frac{28}{17} \\ mole = 1.65[/tex]
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