The starting principal sum is multiplied by both of these variables, and the annual interest rate is increased to the number of compound periods minus one.
The full original amount of the loan is then reduced by the resulting value. Compound interest is the term for interest that is earned on interest. This may be shown using simple math: if you start with $100 and it earns 5% interest every year, you will have $105 at the end of the first year. At the end of the second year, you will have $110.25. On the initial principal sum, simple interest is calculated each time. Compound interest is calculated using the sum of the principal and interest. It is equal every year on a particular principal.
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Sam pushes a 10.0 kg sack of bread flour on a frictionless horizontal surface with a constant horizontal force of 2.0 N starting from rest. (a) What is the kinetic energy of the sack after Sam has pushed it a distance of 35 cm? (b) What is the speed of the sack after Sam has pushed it a distance of 35 cm?
10 kg sack of bread flour is given a horizontal force (2 N) and move about 35 cm distance.
Kinetic energy is formed and the speed is changed. The value of kinetic energy and the speed after pushed is calculated as follows :
Change the unit of distance (s = 35 cm = 0,35 m)
• Work = Δ EK
F × s = EK (after pushed) – EK (before pushed)
2 × 0,35 = EK (after pushed) – 0
0,7 Joule = EK (after pushed)
*EK before pushed is zero because the sack does not move*
• Speed after pushed
EK (After pushed) = ½ × m × v²
0,7 = ½ × 10 × v²
0,7 = 5 × v²
v² = 0,7/5 = 0,14
v (after pushed) = 0,37 m/s
So, kinetic energy after pushed is 0,7 joule and the speed is changed to 0,37 m/s.
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Exercise 2.42
A hot-air balloonist, rising vertically with a constant velocity of magnitude v=5.00 m/s, releases a sandbag at an instant when the balloon is a height h=40.0m above the ground (Figure 1). After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive.
A) Compute the position of the sandbag at a time 0.245 s after its release.
B) Compute the magnitude of the velocity of the sandbag at a time 0.245s after its release.
C) Determine the direction of the velocity of the sandbag at a time 0.245 s after its release.
D) Compute the position of the sandbag at a time 1.15 s after its release.
E) Compute the magnitude of the velocity of the sandbag at a time 1.15 s after its release.
F) Determine the direction of the velocity of the sandbag at a time 1.15 s after its release.
The position and the velocity of the sandbag at a time 0.245 s after its release is 44.7 m and 2.599 m/s.
The position and the velocity of the sandbag at a time 1.15 s after its release are 38.52 m and -6.25 m/s.
What is the equation of motion?The equations of motion can describe the relationship between the velocity, time, acceleration, and displacement of a moving object.
The equations of motions can be represented as:
[tex]v = u + at\\S = ut +\frac{1}{2}at^2\\ v^2-u^2=2aS[/tex]
Given, the maximum height of the sandbag, h = 40 m
The velocity of the sandbag moving upward, v = 5 m/s
Calculate the time taken by the ballon:
v = u -gt
0 = 5 -9.8×t
t = 0.51 s
The position of the sandbag at 0.245 s after its release:
[tex]y = h +v_ot-\frac{1}{2}gt^2[/tex]
[tex]y = 40 +5 -0.5\times 9.8\times (0.245)^2[/tex]
y = 44.70 m
The velocity of the sandbag at a time 0.245 s after its release:
v = v₀ - gt
v = 5 - (9.8)(0.245)
v = 2.599 m/s
The position of the sandbag at a time 1.15 s after its release:
[tex]y = h +v_ot-\frac{1}{2}gt^2[/tex]
[tex]y = 40 +5 -0.5\times 9.8\times (1.15)^2[/tex]
y = 38.52 m
The velocity of the sandbag at a time 1.15 s after its release:
v = v₀ - gt
v = 5 - (9.8)(1.15)
v = -6.25 m/s
The velocity is negative so the sandbag is going down.
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Write the slope-intercept form of the given line. Include your work in your final answer.
The slope intercept form of given line is y = 1x+-1.
Solution:
y = mx+b
m = slope which is 1/1 because it is rise over run (rising 1, running 1)
b = y-intercept form which is -1 because that is where the line meets the y-axis.
The slope is the first number appearing in the equation. The slope is the coefficient of x, regardless of order. The intersection point is the point on the y-axis through which the slope of the line passes. The y-coordinate of the point where the line or curve intersects the y-axis. This is shown by writing the equation for the straight line y = mx+c. where m is the slope and c is the y-intercept.
The slope indicates how much the vertical changes for a given horizontal change. The shape of the slope intercept is actually equal to y plus MX plus b. It's called a slope-intercept shape because it has a slope and an intercept. The slope is defined as the ratio of the vertical change between two points to the horizontal change between the same two points.
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Disclaimer:- Your question is incomplete,please see below for the complete question.
Write the slope-intercept form of the given line. Include your work in your final answer.
Line is Y = mx +c
where m=1
c = -1
Question 5
The state of a sample of water depends on (location, temperature).
O No answer text provided.
O location
O Temperature
O No answer text provided.
Yes, the state of a sample of water depends on location and temperature.
What are the states of water and how does water's condition change as a result of temperature?Pure water has no flavour, no aroma, and no colour. There are three different states of water: solid (ice), liquid, and gas (vapour). Ice is frozen water or solid water. Because the molecules of water expand as it freezes, ice is less dense than water. As ice will weigh less than the same volume of water, it will float in the liquid. At 32° Fahrenheit, or 0° Celsius, water freezes. Water that is liquid is moist and fluid. This is the type of water that we are most accustomed to. We use liquid water in a variety of ways, such as drinking and washing.
The air all around us is constantly filled with the vapour of water. It is not visible. When water is boiled, it transforms from a liquid to a gas or water. At ambient temperature, water is a liquid; however, when cold, it solidifies to form ice. If heated, the same water transforms into a gas known as water vapour. Only when the substance reaches a specific temperature do the modifications take place. Ice formation occurs at 32°F (0°C). It requires more energy to raise and lower the temperature of water because a large body of water has a higher heat capacity than land. As a result, cities near water typically have a more limited range of temperatures throughout the year.
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Tina’s teacher is demonstrating the effect of a sheet of paper on nuclear radiation. He uses an Americium-241 radioactive source close to a detector. There is a count rate of 130 counts per second without the paper. When he puts the sheet of paper between the radioactive source and the detector, the count rate falls almost to 0.
Give
the type of nuclear radiation that the Americium-241 source is producing
The type of nuclear radiation that the Americium-241 source is producing is
an alpha emitter, but also emits some gamma rays.
What is a nuclear radiation?Nuclear radiation (also known as ionising radiation) is energy that is released in the form of high-speed charged particles or electromagnetic waves. Radiation can come from a variety of sources, both natural and man-made. Every living thing is constantly bombarded with low-level radiation from rocks, sunlight, and cosmic rays.
Am-241 emits mostly alpha rays but also some gamma rays. If ingested (swallowed) or inhaled, it poses a greater risk. Americium-241 is created in nuclear power plants by neutron activation of 239Pu and 240Pu, followed by beta decay of 241Pu (T1/2 = 14.35 years). This means that 241Am is produced in burned-up fuel by the decay of 241Pu a long time after the fuel is removed from a nuclear reactor.
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What is the change in internal energy if 60 J of heat is released from a system
and 30 J of work is done on the system? Use AU = Q-W.
OA. 30 J
OB. -30 J
OC. 90 J
OD. -90 J
Answer:
30J
Explanation:
Δq = Δu + Δw
⇒Δu = Δq - Δw
here, Δq = 60J & Δw = 30J
∴ Δu = 60 - 30 = 30 J
The length of a simple pendulum is 0.80 m and the mass of the "bob" at the end is 0.31 kg. The pendulum is pulled away from equilibrium at an angle of 6.8 degrees and released from rest.
a. What is the angular frequency of the motion?
b. Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth.
c. What is the bob's speed as it passes through the lowest point of the swing?
a) angular frequency of the motion is ω = 3.5s⁻¹
b) total mechanical energy of the bob is T.M.E = 0.029J
c) bob's speed at the lowest point is V= 0.43m/s.
A mechanical device that oscillates or sways is known as a simple pendulum. The gravitational force is primarily responsible for this motion, which takes place in a vertical plane.
Given,
Length of a simple pendulum, L = 0.80m
Mass of the bob, m = 0.31 kg
h= L- LcosΘ = L (1 - cosΘ)
a) Angular frequency, ω = 2π / T
where, T = 2π √(L/g)
ω = 2π / 2π √(g/L)
ω =√(9.8 / 0.80) s⁻¹
ω = 3.5s⁻¹
b) Total mechanical energy, T.M.E = (K.E.)ₐ + (P.E.)ₐ
As (K.E.)ₐ = 0
(P.E.)ₐ =mgh
T.M.E = 0 + mgh, where h = L (1-cosΘ)
T.M.E = ngL (1-cosΘ)
= 0.31*9.8*0.80(1-0.988)
= 0.029J
Thus, T.M.E = 0.029J
c) bob's speed at the lowest position, V
TMEb = 0.029J
(KE)b + (PE)b = 0.029J
1/2mV² + 0 = 0.029J
V =√(2*0.029/m)
V = 0.43m/s
So, the speed at the lowest point is V= 0.43m/s.
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You and a friend are astronauts. You push off eachother with a force of 50N. You both have a weight of 0 in space, but your mass is 110kg, and your friend’s is 80kg. Calculate the accelerations of both you and your friend
The force acting on a body in space is zero, hence the acceleration of the friends in space will be;
acceleration for the 110 kg person is 0.acceleration for the 80 kg person is 0.What is the acceleration of a body?The acceleration of a body is the rate of change of the velocity of the body.
The acceleration of a body, the mass of the body, and the force acting on the body are related by the formula below:
Mathematically:
Force = mass * accelerationHence,
acceleration = force / massThe force acting on a body in space = 0 N
For a person having a mass of 110 kg;
acceleration = 0 / 110
Acceleration = 0
For a person having a mass of 110 kg;
acceleration = 0 / 80
Acceleration = 0
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After completing a workout/activity, it is best to use this type of stretching.
After completing a workout/activity, it is best to use Static stretching.
Static stretching is most effective at the quit of your workout. It includes stretches that might be held for a period of time to help extend and unfasten your muscle mass and connective tissue. that is special from a dynamic warmup due to the fact you hold your body nonetheless.
It is a good concept to do dynamic stretching before a workout, and static stretching after a workout, irrespective of what type of exercise you're doing. What you need to differ based on your workout is which muscle groups you are running, on account that it's crucial to warm up the muscle mass you are certainly going to be using.
A dynamic stretching heat-up gets you prepared for something exercising you're approximate to accomplish this is the first-rate form of stretch before an exercising while static stretching is quality for the stop of your workout.
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The position as a function of time of a mass at the end of a spring that is undergoing
SHM is given by x(t) = A sin (ωt + θ) and the velocity is given by ν(t) = A ω cos
(ωt, +θ). At time t = 0.00 seconds, the oscillating mass/spring system has a displace-
ment x = 2.83 cm and a velocity ν= 3.25 cm/s. It is oscillating with an angular
frequency of 2.64 radians per second. Determine the constants A and θ
The value of constants A and θ will be θ = 66.5° or 1.161 radians and A = 3.1 cm
Simple Harmonic Motion or SHM is defined as a motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of this restoring force is always towards the mean position.
x(t) = A sin (ωt + θ)
time = 0.00 seconds
2.83 = A sin ( 0 + θ)
A sin θ = 2.83 `equation 1
ν(t) = A ω cos (ωt+θ)
3.25 = A * 2.64 * cos (0 + θ )
3.25 = 2.64 A cos θ
A cos θ = 3.25 / 2.64 = 1.23 equation 2
dividing equation 1 and 2 , we get
tan θ = 2.30
θ = 66.5° or 1.161 radians
from equation 1
A sin θ = 2.83
A sin ( 66.5 ) = 2.83
A = 3.1 cm
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How could you demonstrate a magnetic field using your body?
Help pls
With the use of a magnetometer, the magnetism or magnetic field of the human body can be measured or demonstrated.
What is a Magnetometer?A magnetometer is a device used to measure the magnetic field or the magnetic dipole moment. Magnetometers of various sorts measure the direction, intensity, or relative change of a magnetic field at a specific area.
A magnetometer is a scientific tool that measures the intensity of magnetic fields. Magnetometers may be used on land to locate iron ore reserves for mining.
Magnetometers are instruments that can detect the amount or direction of a magnetic field. They may be found practically anywhere in electronics. They might be as basic as the one used by your smartphone to identify whether it is upright or as complicated as the one used by NASA to assess Mars' magnetic field.
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The gravitational force between two objects that are 2.1x10‐1 m apart is 3.2x10‐6 N. If the mass of one
object is 55 kg what is the mass of the other object? What will be their acceleration if they are released
from their position?
The mass of the other object is 18.11 kg and their acceleration will be respectively 5.8 × 10⁻⁸ m/s² and 17.66× 10⁻⁸ m/s².
Given parameter:
The gravitational force between two objects : F = 3.2×10⁻⁶ N.
Distance between two objects : r = 2.1x10⁻¹ m.
The mass of one object m₁ = 55 kg.
The mass of other object m₂ =?
From universal gravitational law, we know that, force acting between two objects of mass m₁ and m₂ separated by a distance r is:
F = G m₁ m₂ / r²
Where, G = universal gravitational constant = 6.67 × 10⁻¹¹ N m⁻² kg⁻².
So, we can write,
3.2×10⁻⁶ N = (6.67 × 10⁻¹¹ N m⁻² kg⁻²) (55 kg) m₂/( 2.1x10⁻¹ m)
⇒ m₂ = (3.2×10⁻⁶ N × 2.1x10⁻¹ m )/ { (6.67 × 10⁻¹¹ N m⁻² kg⁻²) (55 kg)}
= 18.11 kg.
Hence, the mass of the other object is 18.11 kg.
So, acceleration of the object of mass 55 kg is:
a₁ = F/m₁ = 3.2×10⁻⁶/55 m/s² = 5.8 × 10⁻⁸ m/s².
The acceleration of the object of mass 18.11 kg is:
a₂ = F/m₂ = 3.2×10⁻⁶/18.11 m/s² = 17.66× 10⁻⁸ m/s².
Hence, their acceleration if they are released from their position will be respectively 5.8 × 10⁻⁸ m/s² and 17.66× 10⁻⁸ m/s².
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With an average acceleration of -0.50 m/s^2, how long will it take a cyclist to bring a bicycle with an initial speed of 13.5 m/s to a complete stop?
The time taken for the bicycle to come to a complete stop with an initial velocity of 13.5 m/s is 27 seconds.
What is time?Time can be defined as the calculable measure of motion with respect to before and afterness.
To calculate the time it takes the cyclist to bring the bicycle to complete stop, we use the formula below.
Formula:
t = (v-u)/a.......... Equation 1Where:
t = Timev = Final velocityu = Initial velocitya = AccelerationFrom the question,
Given:
v = 0 m/s (to stop)u = 13.5 m/sa = -0.5 m/s²Substitute these values into equation 1
t = (0-13.5)/-0.5t = -13.5/-0.5t = 27 seconds.Hence the time taken for the bicycle to stop completely is 27 seconds.
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n his fishing trip, Justin rides in a boat 10 km south. The fish aren’t biting so they go 4 km west. They then follow a school of fish 1 km north. What distance did they cover? What was their displacement?
The displacement covered will be 9.85 km
The distance covered will be = 10 + 4 + 1 = 15 km
using Pythagoras theorem
[tex]H^{2} = P^{2} + B^{2}[/tex]
H = [tex]\sqrt{4^{2} + 9^{2} }[/tex]
H = [tex]\sqrt{97}[/tex]
H = 9.85 km
The displacement covered will be 9.85 km
The distance covered will be = 10 + 4 + 1 = 15 km
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11. If a body on which three forces are acting is in equilibrium which of the following-can
be correctly inferred?
I. The vector sum of the three forces is zero
II. Two of the forces act in the same direction
III. Each of the forces has the same magnitude
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
If a body has the three forces acting upon it and is in equilibrium then the vector sum of the three forces is zero. Thus, (A) I only is the correct answer.
I. The vector sum of three forces is zero means that the vector sum of all the three forces is equal to resultant zero vector.
II. If two forces act in the same direction then the body experiences more force on that side and hence is not in equilibrium. For the body to be in equilibrium, then the resultant of any two forces should be opposite, equal and collinear with the third force.
III. For the body to be in equilibrium, magnitude along with direction should be mentioned. If the three forces are in the same plane, convergent (cross the same point) and non-parallel.
Hence, the body on which three forces is acting will be in equilibrium when the vector sum of three forces is zero.
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M
6
3310
M
WWW
Find Req between a and b
The answer is 5.6 ohms.
The electrical resistance of a circuit component or device is defined as the ratio of the applied voltage to the current through it. If the resistance is constant over a considerable voltage range. This is shown below. To calculate the total resistance of multiple resistors connected in this way, add the individual resistances.
calculation:-
V = iR
R = V/I
= 33.10/6
5.6.
Wire resistance is proportional to resistance and length and inversely proportional to wire cross-sectional area. To determine the condition of a circuit or component. The higher the resistance, the lower the current flow, and vice versa. Resistance is a measure of resistance to the flow of current in an electrical circuit. Resistance is measured in ohms and represented by the Greek letter omega.
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What is the most common cause to food born illnesses?
1.Biological hazards (people working while ill, not washing hands, improper temperature control)
2.Chemical hazards (poisonous substances like soaps, sanitizers, or pesticides)
3.Physical hazards (broken glass, jewelry, bandages, or fingernails)
Option 1. Biological hazards (people working while ill, not washing hands, improper temperature control), is the most common cause to food born illnesses.
Infectious organisms or their toxins are the maximum commonplace reasons for food poisoning. Food poisoning symptoms may additionally include cramping, nausea, vomiting, or diarrhea. Maximum meal poisoning is slight and resolves without treatment. ensuring ok hydration is the maximum vital element of treatment.
Foodborne infection (food poisoning) is a result of consuming contaminated meals, beverages, or water and may be a diffusion of microorganisms, parasites, viruses, and/or pollution. lots of those pathogens can be received thru extra than simply food, drinks, or water.
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What is fluid friction
Fluid friction is an opposition to the ease movement of fluids known as viscosity.
What is fluid friction?
Fluid friction describes the friction between layers of a viscous fluid that are moving relative to each other.
Typically, fluid friction is a type of friction that exist in fluids and it is known as viscosity.
Fluid viscosity can be defined as the type of friction that act on fluid. It can also be defined as the measure of the resistance of a fluid to gradual deformation by shear or tensile stress.
Thus, the fluid friction is equal to fluid viscosity and it is the type of friction that exists only in fluids.
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About how much greater is the gravitational force exerted by the moon on the near side of the earth than on the far side?.
Difference in Gravity of 7%
The near side of the Earth is 12740 km closer to the Moon than the far side. As a result, the near side experiences a gravitational force that is 7% stronger than the far side.The force that pulls items toward the centre of a planet or other entity is called gravity. All of the planets are kept in orbit around the sun by gravity.Differential gravitational forces, or gravitational forces that are not equal across a body's finite size, are the fundamental mechanism.
Radius of the earth, r = 6.37*10^6
by solving ((Fn-Ff)/Fn)*100 answer will be of 7%
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A 5 kg body rests on a horizontal plane where it is missing frictional force. A force of 10N is applied at the angle of 37 ° over the plan, for a time of 5 s. How far does the body move during this time?
Answer:
the formula of acceleration
A 1600kg car is traveling over a hill that has a radius of curvature of 25m. The car is slowing down as it goes over the hill. It slows down at a constant rate from a speed of 25ms to a speed of 10ms over a distance of 50m ending at the top of the hill. The net acceleration of the car at the top of the hill is most nearly.
The correct answer is net acceleration of the car at the top of the hill is 6.6 m/s²
distance traveled by the car, d = 50 m mass of the car, m = 1600 kg radius of the hill's curvature, r = 25 m beginning speed of the car, u = 25 m/s
The third kinematic equation is applied to get the car's tangential acceleration, as shown below;
v2 = u2 + 2as, where a represents the vehicle's acceleration and 2as = v2 - u2 equals a=5.25 m/s
A net equals a2 + aca net = (5.25) 2 + 42a net = 6.6 m/s. 2
Consequently, the vehicle's net acceleration at the summit of the hill is 6.6 m/s2.
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A small 10.0 g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50.0 g and is 90 cm in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 cm/s relative to the table.
What is the angular speed of the bar just after the frisky insect leaps?
Express your answer in radians per second.
the angular speed of the bar just after the bug leaps is 0.044 ras/sec.
Given parameter:
Mass of the bug, m= 10.0g
Mass of the bar, M= 50.0 g.
Length of the bar, l = 90 cm.
Speed of the bug ,v = 20.0 cm/s.
Speed of the bar, V = ?
From conservation of momentum;
momentum of the bar = momentum of the bug
⇒ MV = mv
⇒ V = mv/M
= (10.0× 20.0)/50 cm/s.
= 4 cm/s.
Then, angular speed of the bar will be = V/l
= 4/90 rad/s.
= 0.044 rad/s.
Hence, the angular speed of the bar just after the frisky insect leaps is 0.044 rad/sec.
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A block with an unknown mass is on a smooth tabletop attached to a string that is attached to a hanging block of mass of 3.6kg. If the blocks accelerate at 2.1 m/s^2 What is the mass of the block on the table? What is the Tension in the rope?
The mass of the block on the table is 2.33 kg.
The Tension in the rope is 4.9 N.
What is the mass of the block on the table?
The mass of the block on the table is calculated by applying Newton's second law of motion.
Newton's second law of motion states that the force applied on an object is directly proportional to the product of mass and acceleration of the object.
F = ma
where;
m is the mass of the blocka is the acceleration of the blockF(net) = ma
m₂g - m₁g = a(m₁ + m₂)
where;
m₂ is the mass of the hanging blocka is the acceleration of the blockm₁ is the mass of the block on tablem₂g - m₁g = m₁a + m₂a
m₂g - m₂a = m₁a + m₁g
m₂g - m₂a = m₁(a + g)
3.6(9.8) - 3.6(2.1) = m₁(2.1 + 9.8)
27.72 = 11.9m₁
m₁ = 27.72 / 11.9
m₁ = 2.33 kg
The tension in the rope is calculated as follows;
T = m₁a
T = 2.33 kg x 2.1 m/s²
T = 4.9 N
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Bumper car A (282 kg) moving +2.82 m/s makes an elastic collision with bumper car B (210 kg) moving +1.72 m/s. What is the velocity of car A after the collision?
The velocity of Car B after the collision is 4.856 m/s.
What is elastic collision?
A collision is considered elastic if it does not result in a net decrease in the system's kinetic energy. In an elastic collision, both momentum and kinetic energy are conserved.
In an elastic collision, the kinetic energy is essentially unchanged before and after the contact and is not changed into another type of energy.
Either one dimension or two dimensions are possible. Perfectly elastic collision is not feasible in reality since there will always be some energy exchange, no matter how tiny.
m1v1 + m2v2 = m1u1 + m2u2; where
m1= mass of object 1
m2= mass of object 2
v1= initial velocity object 1
v2= initial velocity object 2
u1= final velocity object 1
u2= final velocity object 2
m1v12/2 + m2v22/2 = m1u12/2 + m2u2/2
281·2.82 - 209·1.72 = 281u1 + 209u2;
281·2.822 + 209·(- 1.72)2 = 281u12 + 209u22;
281u1 + 209u2 = 432.94 and 281u12 + 209u22 = 2852.93
u1 = 1.05407 - 0.74377u2;
u1 = (10.15278 - 0.74377u2)1/2
u2 = 4.856 m/s
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In the figure, the light, taut, unstretchable cord B joins block 1 and the larger-mass block 2 . Cord A exerts a force on block 1 to make it accelerate forward.
(1)
(a) How does the magnitude of the force exerted by cord A on block 1 compare with the magnitude of the force exerted by cord B on block 2 ? Is it larger, smaller, or equal? larger:
smaller
equal
(b) How does the acceleration of block 1 compare with the acceleration (if any) of block 2 ?
larger
smaller
equal
(c) Does cord B exert a force on block 1 ? If so, is it forward or bsckward? Is it larger, smaller, or equal in magnitude to the force exerted by cord B on biock 2?
Yes; backward and equal
Yes; backward and less than
Yes; forward and equal
Yes; forward and less than
Yes; forward and greater than
Yes; backward and greater than,
No
a) The magnitude of force exerted by cord A on block 1 is equal to magnitude of the force exerted by cord B on block 2
b) The acceleration of block 1 is equal to that if block 2
c)The force is forward and equal to the force of cord B on block 2.
What is the force exerted?We know that the force that acts along a cord can be regarded as the tension force. The magnitude of the tension force is equal to the force that is applied on the cord. If an object is hanged on the cord then the tension force would equal the weight of the object.
a) We can see that block 1 is connected to block 2 by the use of cord B and the force on cord B must necessarily be the force on cord A since the weight of the blocks are equal.
b) Since the force on cord A is equal to the force on cord B, the acceleration of block 1 must be equal to the acceleration of block 2.
c) Cord B exerts a force on block 1 in the forward direction and this force is equal to the force that is exerted on block 2.
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The period of Jupiter is known to be 11.9 years. Use this data to determine its average distance from the Sun.
The period of Jupiter is known to be 11.9 years then the average distance of Jupiter to the Sun is given to be 5.21 AU.
Kepler's laws of planetary motion :
According to this law, the semi-major axes of the planets' orbits are directly proportional to the squares of the planets' orbital periods.
period of the Jupiter = 11.9 years
radius of the earth = 1 AU
orbital period of the earth = 1 year
According to the Kepler's 3rd law of planetary motion :
T² ∝ R³
T² = k R³
k = T²/ R³
[tex]k = \frac{T^2_j}{R^3_j} = \frac{T^2_e}{R^3_e}[/tex]
T² /(R)³ = (1)² /(1)³
Rj = [tex]T^{\frac{2}{3}}[/tex]
T = 5.21 AU
The period of Jupiter is known to be 11.9 years then the average distance of Jupiter to the Sun is given to be 5.21 AU.
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Please help with these questions
3) unit of the following are :
a) force = Newton
b) distance = meter
c) work = Joule
4) 90 J work has been done by her
5) 51 J work was done
6) He have done [tex]10^{4}[/tex] J of work
7) no work is done
8) no work is done
9) 700 J is the work done on the car
10) 10.68 N force is applied
3 ) unit of the following are :
a) force = Newton
b) distance = meter
c) work = Joule
4 ) work done = force * displacement
= 75 * 1.2 = 90 J
5 ) work done = force * displacement
= 15 * 3.4 = 51 J
6 ) work done = force * displacement
= 100 * 100 = [tex]10^{4}[/tex] J
7) no work is done because no displacement occurred
8) no work is done because no displacement took place
9) work done = force * displacement
= 12.5 * 56 = 700 J
10) force = work done / displacement
= 142 / 13.3 = 10.68 N
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A bullet of mass m and speed v is fired at, hits and passes completely through a pendulum bob of mass M on the end of a stiff rod of negligible mass and length L. The bullet emerges with a speed of v/2 and the pendulum bob just makes it over the top of the trajectory without falling backward in its circular path. Determine an expression (V=?) for the minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle. (Use any variable or symbol stated above as necessary.)
Required value of initial speed of the bullet be ( 4M/m)√(gL).
Given parameters:
Mass of the bullet =m.
Mass of the bob of the pendulum = M.
speed of the bullet before collision = v
Speed of the bullet after collision = v/2.
Length of the pendulum stiff rod = L.
Let speed transmitted to the pendulum be u.
Using principle of conservation of momentum:
mv = Mu + mv/2
⇒ Mu = mv/2
⇒ u = (m/M)v/2
We know that: to make the bob over the top of the trajectory without falling backward in its circular path, required speed be = √(4gL). [ where g = acceleration due to gravity]
To be minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle:
u = √(4gL)
⇒ (m/M)v/2 = √(4gL)
⇒ v =( 4M/m)√(gL).
Hence, minimum required speed of the bullet be ( 4M/m)√(gL).
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A person throws a ball up into the air, and the ball falls back toward Earth. At which point would the kinetic energy be the lowest?(1 point)
Responses
at a point before the ball hits the ground
at a point before the ball hits the ground
when the ball is at its highest point
when the ball is at its highest point
when the ball leaves the person’s hand
when the ball leaves the person’s hand
at a point when the ball is still rising
at a point when the ball is still rising
Answer:
highest point
Explanation:
at highest point , ball stops moving, velocity = 0
KE = 1/2 m v^2
= 1/2 m ( 0)^2 = 0 at highest point
Why is the sun not the major factor in the tides?
Answer: the moon is closer.
Explanation:
The moon is closer to the earth than the sun, which results in a larger gravitational gradient for the moon even though the sun is considerably more massive and has stronger overall gravity than the moon.
