Answer:
Explanation:
If we ignore wind resistance, the time needed for the hay to drop from vertical rest is
t = √(2h/g) = √(2(300)/9.81) = 7.82 s
d = vt = 80.0(7.82) = 626 m before passing over the cow.
f. Protons and neutrons
2. TRUE or FALSE: An object that is positively charged contains all protons and no electrons
3. TRUE or FALSE: An object that is negatively charged could contain only electrons with me
accompanying protons
4. TRUE or FALSE: An object that is electrically neutral contains only neutrons.
An object which has more electron than proton is negatively charged, otherwise positively charged. Every statement is false
What is atom?Atom is the smallest unit of the matter consist of the positive charged nucleus and the electrons which moves around it. The atom can not be divided further.
The atom of a matter is made by three elements-
1) Neutron-Neutron is the element of atom, which has zero charge.2) Proton-Proton is the element of atom, which has positive charge.3) Electron-Electron is the element of atom, which has negative charge.For the Protons and neutrons, lets check all the statement wheather they are true or false.
2. An object that is positively charged contains all protons and no electrons- An object which is positively charged, has more number of proton than electron. This is a false statement.3. An object that is negatively charged could contain only electrons with me accompanying protons- An object, which is negatively charged, has more number of electron than proton. This is a false statement.4. An object that is electrically neutral contains only neutrons-The number of electron and proton is equal in an electrically neutral object. This is a false statement.Thus, every statement is false as a object which has more electron than proton is negatively charged, otherwise positively charged.
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A baseball is hit with a speed of 27.0 m/s at an angle of 47.0 ∘ . It lands on the flat roof of a 10.0 m -tall nearby building. Part A If the ball was hit when it was 1.3 m above the ground, what horizontal distance does it travel before it lands on the building?
Answer:
H = Vy t - 1,2 g t^2 formula for height of ball after t sec
H = 10 - 1.3 = 8.7 m
Vy = 27 sin 47 = 19.7 m/s vertical speed of ball
8.7 = 19.7 t - 9.8/2 t^2 height of ball after t sec
4.9 t^2 - 19.7 + 8.7 = 0 rearranging
[19.7 ± (388 - 170)^1/2] / 2 *4.9 = [19.7 ± 14.7] / 9.8 = .51 ,3.5 sec
.51 sec would be on the way up and 3.5 sec on the way down
Sx = 27 * cos 47 * 3.5 = 64.4 m around 200 ft seems reasonable
The maximum speed with which an 1000 kg car makes a 180-degree turn is 10 m/s. The radius of the circle through which the car is turning is 24 m. Determine the force of friction and the coefficient of friction acting upon
the car.
1. A roller coaster with a mass of 800 kg sits stationary at the top of a section of track, 75 m above
the ground as shown. When the brake is released, it starts to roll down the track
2. For each height represented in the diagram, calculate the gravitational potential energy using
Ep = mgh. Show ONE SAMPLE calculation in the calculations section below and fill in Table 1 for
each of the heights of the roller coaster. (6 marks)
3. Assuming there is no friction, determine the mechanical kinetic energy using Ek = Etotal - Ep.
Show ONE SAMPLE calculation in the calculations section below and fill in Table 1 for each of
the heights of the roller coaster. (6 marks)
4. For each height represented in the diagram, calculate the velocity using = �2
. Show ONE
SAMPLE calculation in the calculations section below and fill in Table 1 for each of the heights of
the roller coaster. (6 marks)
5. Use your answers to graph how gravitational potential energy, mechanical kinetic energy, and
velocity change as the roller coaster changes height. Use different colours for the three lines on
the graph. Graph paper is provided below. (3 marks)
6. Repeat steps 1 – 5 above for a roller coaster cart that has a mass of 300 kg and enter your
results in Table 2.
Calculations:
800 kg roller coaster cart:
Sample calculation for gravitational potential energy:
Sample calculation for Mechanical kinetic energy:
Sample calculation for velocity:
300 kg roller coaster cart:
Sample calculation for gravitational potential energy:
Sample calculation for mechanical kinetic energy:
Sample calculation for velocity:
Results:
Table 1: Potential energy, kinetic energy, total energy, and velocity of the 800 kg roller coaster cart
Table 2: Potential energy, kinetic energy, total energy, velocity of the 300 kg roller coaster cart.
Graphs:
It’s graphing time. These graphs are a bit different than the ones you did in the
data analysis assignment at the beginning of the course. In this case you have
three things to graph on each graph. (One graph for the 800 kg roller coaster cart
and one graph for the 300 kg roller coaster cart.) You need to graph the
gravitational potential energy with respect to height, the mechanical kinetic
energy vs height, and the velocity vs height.
Let’s look at the energy graphs first. In this case both kinetic energy and
mechanical energy cover the same range of values. This means they can use the
same scale on the y-axis. So, you will use the left y-axis and the x-axis to graph
the kinetic energy vs height and the potential energy vs height. You will need a
legend to explain which line is which. Colour coding is a nice way to highlight this.
The velocity values are much different than the energy values. This means you
need a totally different scale. So, your left y-axis won’t work. You need to make a
second scale on the right y-axis for your velocity values. You will plot the points
the same way as normal, but you will use the numbers on the right-hand scale
instead. Again, be sure to add your velocity line to the legend with a separate
colour code.
Discussion Questions:
1. Describe the relationship between the gravitational potential energy and the mechanical kinetic
energy of the roller coaster on your graph. (2 marks)
2. Describe the shapes of each of the three lines in the graph. Explain why the velocity is different.
(4 marks)
3. Describe how mass affects the speed at the bottom of the roller coaster. (2 marks)
4. Describe how mass affects the gravitational potential energy and the mechanical kinetic energy
of the roller coaster. (2 marks)
5. At what point does the roller coaster have a maximum value for the following? Justify your
answer by explaining why. (2 marks each)
a. Gravitational potential energy
b. Mechanical energy
c. Velocity
6. In your calculations, you assumed that the roller coaster was frictionless. All real roller coasters
encounter friction. Describe how the actual values of the variables would differ, or not differ,
from your calculated values for a real roller coaster. (Hint: what form of energy would some of
the total energy be converted to if there was friction in the system?) (4 marks)
How you will be graded:
Grades will be based on answering questions to demonstrate an understanding of the material covered
in this unit. Point form answers are okay if ideas are complete and use vocabulary (Word Bank)
provided. For questions out of 4 marks, there are 4 responses expected.
Answer:
Give me some hint please
Based on the calculations, potential energy of this roller coaster at a height of 75 meters is equal to 588,000 Joules.
How to calculate potential energy?Mathematically, potential energy is calculated by using this formula:
P.E = mgh
Where:
P.E represents potential energy.m is the mass.h is the height.g is acceleration due to gravity.Note: Acceleration due to gravity is equal to 9.8 m/s².
At a height of 75 m, we have:
P.E = 800 × 9.8 × 75
P.E = 588,000 Joules.
At a height of 60 m, we have:
P.E = 800 × 9.8 × 60
P.E = 470,400 Joules.
At a height of 45 m, we have:
P.E = 800 × 9.8 × 45
P.E = 352,800 Joules.
At a height of 30 m, we have:
P.E = 800 × 9.8 × 30
P.E = 235,200 Joules.
At a height of 15 m, we have:
P.E = 800 × 9.8 × 15
P.E = 117,600 Joules.
In conclusion, we can deduce that the potential energy of this roller coaster decreases with a decrease in height.
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a
A person throws a ball up into the air, and the ball falls back towar
would the kinetic energy be the lowest? (1 point)
at a point before the ball hits the ground
when the ball leaves the person's hand
o when the ball is at its highest point
at a point when the ball is still rising
Answer:
when the ball is at its highest point
Explanation:
Provided the ball returns to where it was thrown. The velocity, and therefore kinetic energy, will be momentarily zero at the highest point of the throw.
Need help asap
A scientist has invented a robot to work on the seabed. According to his calculation, the armour of the robot can withstand a maximum pressure of 10⁵ Pa exerted by the sea water. If the density of the sea water is 1025 kg/m3, what is the maximum depth of the seabed that this robot can work? [Given g = 9.81 m/s2 and rho water = 1000 kg/m3]
Answer:
Explanation:
Well which is it ? ρ = 1000 kg/m³ or ρ = 1025 kg/m³?
Obviously the sea is salt water so we can ignore ρ = 1000 kg/m³
1025 kg/m³(d m)(9.81 N/kg) = 1 x 10⁵ N/m² = Pa
d = 9.9450535...
d = 10 meters
That's if we only account for the pressure due to the water. On top of that pressure would be atmospheric pressure which is about 101000 Pa
so the robot would be a hair above its pressure limit before it even got in the water.
How many valence electrons do most stable atoms have?
A) one
B) two
C) four
D) eight
Answer:
D) 8
Explanation:
Due to the octet rule the most stable atoms will have 8 valence electrons.
I need help with parts a and B of this question
Answer:
Explanation:
Let x be the leaf spring compression distance
F = kx
5.20 x 10⁵ = (5.45 x 10⁵)x + (3.80 x 10⁵)(x - 0.500)
5.20 x 10⁵ = (5.45 x 10⁵)x + (3.80 x 10⁵)(x) - 1.90 x 10⁵
7.10 x 10⁵ = (9.25 x 10⁵)x
x = 0.76756...
x = 0.768 m
W = ½(5.45 x 10⁵)0.768² + ½(3.80 x 10⁵)(0.768 - 0.500)²
W = 174,148.648648...
W = 174 KJ
is xenon a pure substance
[tex]\large\huge\green{\sf{Yes}}[/tex]
Why is it important that we learn about our solar system, the sun, and planets in our solar system other than Earth?
Answer:
To understand all the different kind of elements and how they work. Nature is a beautiful thing and it is important we understand how they survive and how we can help. The solar system is a mandatory nature. Learning about other planets also helps us feel safe and more knowledgable.
a uniform thin rod of length l and mass m is allowed to rotate on a frictionless pin passing through one end. The rod is released from rest in the horizontal position. a.) What is the speed of the center of gravity when the rod reaches its lowest position? b.) What is the tangential speed of the lowest point of the rod when the rod reaches its lowest position?
Answer:
Explanation:
Potential energy gets converted to rotational kinetic energy
a) ½Iω² = mgh
½(mL²/3)ω² = mgL/2
(L/3)ω² = g
ω = [tex]\sqrt{3g/L}[/tex]
v(CG) = (L/2) [tex]\sqrt{3g/L}[/tex]
Not sure if you wanted angular speed or tangential speed of the CG so I gave both.
b) v = Rω = L [tex]\sqrt{3g/L}[/tex]
A box of mass 7.7 - kg is accelerated from rest across a floor at a rate of 2.6 m/s2 for 18.5 s. Find the net work done on the box
Answer:
Explanation:
The net work will change the kinetic energy
W = ½mv² = ½m(at)² = ½ma²t²
W = ½(7.7)2.6²(18.5²) = 8907.3985 = 89 kJ
An LED is useful because when a current passes through it, it gives out... what?
Enter your answer
An LED is useful because when a current passes through it, it gives out light.
An LED is useful because when a current passes through it, it gives out Light.
What is an LED?
LED, in full light-emitting diode, in electronics, a semiconductor device that emits infrared or visible light when charged with an electric current.
A light-emitting diode (LED) emits light by applying a forward current to the pn junction of a compound semiconductor.
When forward current is passed through the light-emitting diode, carriers (electrons and holes) move. The holes in the p-type region move to the n-type region and the electrons in the n-type region move to the p-type region. The injected carriers recombine, and the energy difference before and after recombination is released as light. The emitted light depends on the energy band gap (Eg) of the compound semiconductor.
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A 25.0kg boy is sliding on a frictionless frozen lake at 5.00m/s to the north when he is struck by a 1.00kg
snowball moving at 15.0m/s from the west. If the snowball sticks to him, how fast, and in what direction,
does the boy move after the collision?
The final velocity of the boy after the collision with snowball is 4.84 m/s at 18.4⁰ north-east
The given parameters;
Mass of the boy, m₁ = 25 kgSpeed of the boy, u₁ = 5 m/sMass of the snowball, m₂ = 1.0 kgSpeed of the snow ball, u₂ = 15 m/sThe initial momentum of the boy is calculated as follows;
[tex]P_y = m_1 u_1\\\\P_y = 25 \times 5\\\\P_y = 125 \ kgm/s \ \ north[/tex]
The initial momentum of the snowball is calculated as follows;
[tex]P_x = m_2 u_2\\\\P_x = 1 \times 15 \ \\\\P_x = 15 \ kgm/s \ \ west[/tex]
The resultant momentum of the boy and the snowball after collision is calculate as follows;
[tex]P_f = \sqrt{P_y^2 + P_x^2} \\\\P_f = \sqrt{125^2 + 15^2} \\\\P_f = 125 .9 \ kgm/s[/tex]
The final velocity of the system boy-snowball system is calculated as;
[tex]v_f(m_1 + m_2)= P_f\\\\v_f = \frac{P_f}{m_1 + m_2} \\\\v_f = \frac{125.9}{25 + 1} \\\\v_f = 4.84 \ m/s[/tex]
The direction of the boy after the collision is calculated as follows;
[tex]\theta = tan^{-1}(\frac{v_y}{v_x} )\\\\\theta = tan^{-1} (\frac{5}{15} )\\\\\theta = 18.4 \ ^0[/tex]
Thus, the final velocity of the boy after the collision with snowball is 4.84 m/s at 18.4⁰ north-east
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How is Compression Force Measured?
don't just copy theanswer please
Compression force can be measured with a force gage or load cell.
4. Protons and neutrons are held together to form this _________
Answer:
strong nuclear force.
Explanation:
hope this helps you!!
A 4.0 kilogram projectile is fired horizontally from a 500 kilogram cannon initially at rest. The momentum of the projectile after being fired is 600 kilogram-meters per second to the north
(neglecting friction).
What is the speed of the cannon after firing?
0.83 m/s
1.2 m/s
o
3.3 m/s
150 m/s
The speed of the cannon after firing is 1.2 m/s
This can be solved using the law of conservation of momentum.
From the law of conservation of momentum,
⇒ The momentum of the projectile is equal to the momentum of the cannon.
MV = P................ Equation 1⇒ Where :
M = mass of the cannon,V = velocity of the cannonP = momentum of the projectile.⇒ make V the subject of the equation
V = P/M.................. Equation 2From the question,
⇒ Given:
P = 600 kgm/sM = 500 kg⇒ Substitute these values into equation 2
V = 600/500V = 1.2 m/sHence, The speed of the cannon after firing is 1.2 m/s
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One indicator NOT related to domestic violence is
violent attitudes
good nutrition
violations of court orders
substance abuse
Answer:
good nutrition
Explanation:
I think that one's pretty self explanatory
Which statement is true for a series circuit
Answer: they have one path to flow
Explanation: share the same current
when the tides are especially weak it is called a _____ tide
Answer:
When the tides are especially weak, it is called a NEAP tide.
Explanation:
Hope this helps :)
Answer:
Neap
Explanation:
Answered!
the distance between student home and school is 1.5 km what is the distance traveld bythe students in a week
Answer:
10.5k
Explanation:
1.5x7=10.5
True or False. What makes faith genuine is its object.
Answer:true
Explanation:
What are the two types of force
Answer:
Forces can be divided into primarily into two types of forces:
Contact Forces. Non-contact Forces.Explanation:
This is physics and it says collision and elastic/inelastic i need help
The initial velocity of the 3250 Kg mass is 2.1 m/s. The distance covered by the larger mass in 5s is 4.7 cm.
In this problem, we have to apply the law of conservation of linear momentum. Note that;
Momentum before collision = Momentum after collision
m1u1 + m2u2 = m1v1 + m2v2
(2150 × 10) + (3250u1) = (2150 + 3250)5.22
21500 + 3250u1 = 5400 × 5.22
3250u1 = 28188 - 21500
u1 = 28188 - 21500/3250
u1 = 2.1 m/s
2) Again from the principle of conservation of linear momentum;
(0.40 × 3.5) + (0.60 × 0) = (0.40 × 0.70) + (0.60 × v2)
1.4 = 0.28 + 0.60v2
1.4 - 0.28 = 0.60v2
v2 = 1.87 cm/s
Using;
s = 1/2 ( u + v)t
s = 1/2(0 + 1.87) × 5
s = 4.7 cm
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A 2.40 kg ball is attached to an unknown spring and allowed to oscillate. The figure below shows a graph of the ball’s position �� as a function of time . What are the oscillation’s: a. period; b. frequency; c. angular frequency; d. amplitude; and e. What is the force constant of the spring?
(a) The period of the oscillation is 0.8 s.
(b) The frequency of the oscillation is 1.25 Hz.
(c) The angular frequency of the oscillation is 7.885 rad/s.
(d) The amplitude of the oscillation is 3 cm.
(e) The force constant of the spring is 148.1 N/m.
The given parameters:
Mass of the ball, m = 2.4 kgFrom the given graph, we can determine the missing parameters.
The amplitude of the wave is the maximum displacement, A = 3 cm
The period of the oscillation is the time taken to make one complete cycle.
T = 0.8 s
The frequency of the oscillation is calculated as follows;
[tex]f = \frac{1}{T} \\\\f = \frac{1 }{0.8} \\\\f = 1.25 \ Hz[/tex]
The angular frequency of the oscillation is calculated as follows;
[tex]\omega = 2\pi f\\\\\omega = 2\pi \times 1.25\\\\\omega = 7.855 \ rad/s[/tex]
The force constant of the spring is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\ k = \omega ^2 m\\\\k = (7.855)^2 \times 2.4\\\\k = 148.1 \ N/m[/tex]
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A textbook weighs 34 N on the surface of the Earth. What is the book’s mass on Earth’s surface?
Answer:
About 3.47kg
Explanation:
Recall that weight is equal to mass times acceleration.
In this case, our acceleration is due to gravity which on earth is about 9.8m/s/s
So we have 34N=9.8 *mass, divide both sides by 9.8 we get mass is equal to about 3.47kg.
Two forces act on a block as shown in the picture. What is the net force of the block?
30 N to the right
30 N to the left
10 N to the left 10
10 N to the right
Answer:
10 N to the left.
Explanation:
Since the forces are acting in opposite directions, you need to calculate the difference.
20 N - 10 N = 10 N
More force is being exerted to the left. Therefore, the net force is 10 N to the left.
Which of the following is NOT a function of the lens in the eye?
W
It can perform minor adjustments for distance.
It flattens when light rays from distant objects are to be focused.
It is a light receptor that generates nerve signals that are sent to the brain
It maintains its spherical shape to view nearby objects.
Answer:
the last one
Explanation:
What happens to the gravitational force between two objects if the distance between them triples?
A. The force increases by a factor of 9
B. The force decreases by a factor of 9
C. The force decreases by a factor of 3
D. The force increases by a factor of 3
Two Blocks are connected by a massless rope over a massless,
frictionless pulley, as shown in the figure. The mass of block 2
is m2 = 10.1 kg, and the coefficient of kinetic friction
between block 2 and the incline is Mk = 0.200. The angle 0 of
the incline is 27.5º. If block 2 is moving up the incline at
constant speed, what is the mass mi of block 1?
The mass of block 1 will be 1.99 kg.The tension force is applied along the whole length of the wire, pulling energy equally on both ends.
What is tension force?The tension force is described as the force transferred through a rope, string, or wire as it is pulled by opposing forces.
Given that,
Mass of block 1=? kg
The coefficient of the kinetic friction,μ=0.200
Now consider the weight component in the uphill direction.The weight is found as;
[tex]\rm W=m_1gsin \theta[/tex]
The force is balanced in the vertical direction as;
[tex]\rm T=F_f-W[/tex]
When the force of friction is;
[tex]\rm F_F=\mu_k N[/tex]
[tex]\rm F_f=(m_1 gcos \theta)[/tex]
Substitute the value in the vertical balanced equation;
[tex]\rm T=m_1gsin(27.5)^0-\mu_kmgcos27.5^0[/tex]
[tex]\rm T-m_2g=0\\\\T=m_2g[/tex]
[tex]\rm (10.1) g=m_1g(0.699-0.2 \times(-0.714) ) \\\\ (10.1) g=m_1g (0.699+0.1428) \\\\\ (10.1) g= m_1 \times 0.8418 \\\\ m_1 =11.99 \ kg[/tex]
Hence the mass of block 1 will be 1.99 kg.
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