Answer:
parasites are creatures the gain benefit off of other animals usually harming them eg:ticks on dogs
A caterpillar climbs up a one-meter a wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the
caterpillar to climb to the top. What is the distance traveled? (Round the number to the nearest hundred.)
cm
Answer:
1.5 m is ur answer
Explanation:
A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s. (a) What is the moment of inertia of the system
Complete Question
A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s.
(a) What is the moment of inertia of the system
(b) What is its rotational kinetic energy? axis 50 cm 30 cm 20 cm
Answer:
a) [tex]M.I=0.32kg.m^2[/tex]
b) [tex]K.E=621.8J[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=2.0kg[/tex]
Disk Radius [tex]r_d=50cm=0.5m[/tex]
Mass of annular cylinder [tex]M_c=1.0kg[/tex]
Inner Radius of cylinder [tex]R_i=20cm=0.2m[/tex]
Outer Radius of cylinder [tex]R_o=0.3m[/tex]
Angular Velocity [tex]\omega=10rev/s=2 \pi*10=62.83[/tex]
Generally the equation for moment of inertia is mathematically given by
[tex]M.I=0.5M r_d^2+0.5M_c(R_i^2+R_o^2)[/tex]
[tex]M.I=0.5(2*0.50)^2)+0.5*1*(0.2^2+0.30^)[/tex]
[tex]M.I=0.32kg.m^2[/tex]
Generally the equation for Rotational Kinetic Energy is mathematically given by
[tex]K.E=0.5M.I \omega^2[/tex]
[tex]K.E=0.5 *0.32*62.83[/tex]
[tex]K.E=621.8J[/tex]
Một điện tích q = 6.10-6 C đặt tại tâm của một hình lập phương. Tính thông lượng điện trường gửi qua mỗi mặt hình lập phương.
b. Một điện tích q = 4.10-8 C đặt tại tâm của một hình vuông. Tính thông lượng điện trường gửi qua mỗi mặt của hình vuông.
Answer:
mmmlbdhdjdkekkdnxnfjkckckcklclglglvkglglvkgkvkvkvkvkvkvkvkvkvkvkbkkbkbkkbbkkbkbkbkkbkbkblbkkbkb
How many feet does a No. 2 pencil last?
Answer:45,000
Explanation:
:TIL the average pencil holds enough graphite to draw a line about 35 miles long or to write roughly 45,000 words
Which one is the dependent variable in distance, force, or work
Answer:
Distance
Explanation:
Work can be defined as the energy transferred to a physical object by exertion of a force on the object to cause a displacement of the object. Thus, work is typically done when a person or simple machine move an object over a distance through the application of a force.
Mathematically, work done is given by the formula;
[tex] W = F * d[/tex]
Where,
W is the work done
F represents the force acting on a body.
d represents the distance covered by the body.
A dependent variable is the event expected to change when an independent variable is manipulated.
Hence, distance is the dependent variable because its value changes with respect to the amount of force exerted on an object.
The angle between the reflected ray and the normal line is called the________________
Answer:
Angle of reflection
Explanation:
The angle between the reflected ray and the normal line is called the angle of reflection
Answer:
Activity 2 Directions: Solve the following problems using Charles' Law,
1. A 3.8 mL of a gas in a container has a temperature of 2.0K. What would the volume be if the temperature becomes 8.0K?
2. If at 27°C the volume of a dry gas is 3.0 ml. What is the new temperature if the volume increased to 7.5 ml?
3. The temperature of the gas is 460,0 mL at 3.5°C. Calculate the new volume at 1.5°C
A electron gains electric potential energy as it moves from point 1 to point 2. Which of the following is true regarding the electric potential at points 1 and 2?
a. V1 = V2
b. V1 > V2
c. V1 < V2
Answer:
We know that the change in electric potential energy is defined as:
q*ΔV = ΔP
So, the change in the electric potential energy is the charge times the change in the electric potential.
For the case of an electron gas, we have:
q = -e
where -e is the charge of an electron (remember that is negative).
So, if the electron gains electric potential then:
ΔP > 0
this means that the final potential energy is larger than the initial one, then we have:
-e*ΔV > 0
This means that ΔV must be negative.
V₂ = electric potential at point 2, so it is the final electric potential
V₁ = electric potential at point 1, so it is the final electric potential
Then we should get:
ΔV = V₂ - V₁ < 0.
This means that:
V₂ < V₁
The correct option is b.
A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.255 m while bringing a 830 kg car to rest from an initial speed of 1.4 m/s.
Answer:
the magnitude of the average force on the bumper is 3189.8 N
Explanation:
Given the data in the question;
In terms of force and displacement, work done is;
W =[tex]F^>[/tex] × [tex]x^>[/tex]
W = [tex]Fxcos\theta[/tex] ------- let this be equation 1
where F is force applied, x is displacement and θ is angle between force and displacement.
Now, since the displacement of the bumper and force acting on it is in the same direction,
hence, θ = 0°
we substitute into equation 1
W = [tex]Fxcos([/tex] 0° [tex])[/tex]
W = [tex]Fx[/tex] ------- let this be equation 2
Now, using work energy theorem,
total work done on the system is equal to the change in kinetic energy of the system.
[tex]W_{net[/tex] = ΔKE
= [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu² --------- let this be equation 3
where m is mass of object, v is final velocity, u is initial velocity.
from equation 2 and 3
[tex]Fx[/tex] = [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu²
we make F, the subject of formula
F = [tex]\frac{m}{2x}[/tex]( v² - u² )
given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s
so we substitute
F = [tex]\frac{830}{(2)(0.255)}[/tex]( (0)² - (1.4)² )
F = 1627.45098 ( 0 - 1.96 )
F = 1627.45098 ( - 1.96 )
F = -3189.8 N
The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.
Therefore, the magnitude of the average force on the bumper is 3189.8 N
Which type of balance is key to sitting?
dynamic
static
bosu
level
Explanation:
bosu
here is your answer
Ocean waves of wavelength 26 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.
Required:
How far from the wall is she? What is the period of her up-and-down motion?
Answer:
a) the distance between her and the wall is 13 m
b) the period of her up-and-down motion is 6.5 s
Explanation:
Given the data in the question;
wavelength λ = 26 m
velocity v = 4.0 m/s
a) How far from the wall is she?
Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;
x = λ/2
we substitute
x = 26 m / 2
x = 13 m
Therefore, the distance between her and the wall is 13 m
b) What is the period of her up-and-down motion?
we know that the relationship between frequency, wavelength and wave speed is;
v = fλ
hence, f = v/λ
we also know that frequency is expressed as the reciprocal of the time period;
f = 1/T
Hence
1/T = v/λ
solve for T
Tv = λ
T = λ/v
we substitute
T = 26 m / 4 m/s
T = 6.5 s
Therefore, the period of her up-and-down motion is 6.5 s
can someone please help me
F = 153 N
[tex]\theta = 11.3°[/tex]
Explanation:
Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components [tex]F_{x}[/tex] and [tex]F_{y}[/tex]:
[tex]F_{x}[/tex] = 350 N - 200 N = 150 N
[tex]F_{y}[/tex] = 180 N - 150 N = 30 N
The magnitude of the resultant force F is given by
[tex]F = \sqrt{F_{x}^{2}+F_{y}^{2}}[/tex]
[tex]\:\:\:\:\:\:= \sqrt{(150\:N)^{2}+(30\:N)^{2}}[/tex]
[tex]\:\:\:\:\:\:=153\:N[/tex]
To find the direction [tex]\theta[/tex], we use
[tex]\tan \theta = \dfrac{F_{y}}{F_{x}}=\dfrac{30\:N}{150\:N}=0.2[/tex]
or
[tex]\theta = \tan^{-1}(0.2) = 11.3°[/tex]
In 1913 Niels Bohr formulated a method of calculating the different energy levels of the hydrogen atom.
a. True
b. False
How much time will it take for a person to walk the length of a football field (100 yards)
at a constant speed of 5 ft/s ?
The speed is in feet per seconds so change the length of the field from yards to feet.
1 yard = 3 feet
100 yards x 3 = 300 feet
The field is 300 feet long
Time = distance / speed
Time = 300 feet / 5 feet per second = 60 seconds = 1 minute
It will take 1 minute
Answer:
A person will take 60 Seconds to walk the distance of 100 yards.
Explanation:
Data Given ;
Distance ( d ) = 100 yards = 300 Ft
Speed ( v ) = 5 Ft/s
Time ( t ) = ?
What is speed ?The distance travelled in unit time is called speed.
formula ; [tex]v = \frac{d}{t} \\[/tex]
On putting values,
[tex]5 = \frac{300}{t}[/tex]
[tex]t =\frac{300}{5}[/tex]
[tex]t = 60 sec[/tex]
Hence the time taken by the person is 60 sec.
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Can someone help me
Btw the last one say current
Alice's friends Bob and Charlie are having a race to a distant star 10 light years away. Alice is the race official who stays on Earth, and her friend Darien is stationed on the star where the race ends. Bob is in a rocket that can travel at 0.7c; whereas Charlie's rocket can reach a speed of 0.866c. Bob and Charlie start at the same time. Draw space- diagrams from each perspective.
Required:
Estimate how long it takes Bob and Charlie to finish the race from each perspective.
Solution :
The distance between the starting point and the end point, [tex]L_0[/tex] = 10 light years
But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.
For Bob,
Speed of Bob's rocket with respect to Alice, [tex]L_b = 0.7 \ c[/tex]
So the distance appeared to Bob due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.49} \ Ly$[/tex]
[tex]$=7.1 \ Ly$[/tex]
Therefore, the time required to finish the race by Bob is
[tex]$t_b = \frac{L_b}{V_b}$[/tex]
[tex]$=\frac{7.1 \ c}{0.7 \ c}$[/tex]
= 10.143 year
For Charlie,
Speed of Charlie's rocket with respect to Alice, [tex]L_c = 0.866 \ c[/tex]
So the distance appeared to Charlie due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.75} \ Ly$[/tex]
[tex]$=5 \ Ly$[/tex]
The time required to finish the race by Charlie is
[tex]$t_b = \frac{L_c}{V_c}$[/tex]
[tex]$=\frac{5 \ c}{0.866 \ c}$[/tex]
= 5.77 year
Please help, only answer if your 1000% correct im in summer school and need to pass this class
Action reaction forces never cancel each other out because .............................? *
1 point
they do not act on objects.
they act on the same objects
they act on different objects.
none of the above
A force can be described as? *
1 point
push or pull action
pulling direction
direction of change
push action only
Using EquatIO calculate the resultant force for the horizontal (x-axis) component. Show your calculation step. *
2 points
Captionless Image
A player hits a baseball with a bat. The action force is the impact of the bat against the ball. The reaction force is .......................? *
1 point
Captionless Image
the grip of the player's hands on the ball
the weight of the ball
the air resistance of the ball
the force of the ball against the bat
Which law is this? "An object at rest tends to stay at rest and an object in motion tends to stay in motion unless acted upon by an unbalanced force." *
1 point
Newtons third law
Newtons second law
Newtons first law
None of the above
Which law is this? "If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A." *
1 point
None of the above
Newtons second law
Newtons first law
Newtons third law
What can be said about a force on an object that is not stationary? *
1 point
Captionless Image
there are no forces acting on the object
there are unequal forces acting on the object
there is only one force acting on the object
there are equal and opposite acting on the object
Using EquatIO calculate the resultant force for the vertical (y-axis) component. Show your calculation step. *
2 points
Captionless Image
Identify the different force acting on a moving vehicle shown in the mage below. *
5 points
Captionless Image
A B C D E
Reaction force
Weight
Friction
Air resistance
Thrust
Reaction force
Weight
Friction
Air resistance
Thrust
Which law is this? "The acceleration of an object is directly related to the net force and inversely related to its mass." *
1 point
None of the above
Newtons third law
Newtons first law
Newtons second law
If forces acting on an object are unbalanced, what can occur to the object? *
1 point
acceleration
all of the above
change of shape
deceleration
change in direction
If a player hits a baseball with a force of 870N Calculate the acceleration (state the units) of the ball . Show your calculation step using only EquatIO. *
2 points
Captionless Image
Action reaction forces never cancel each other out because .............................? *
1 point
they act on different objects.
none of the above
they act on the same objects
they do not act on objects.
because they work on the same object
Explanation:
A push and pull factor
What will be resultant force ?
Answer:
It will go to the right way 500N.
Explanation:
Because, up and down is 300 and 300 so it minus each other and get zero, that made the resultant force will never go up or down. And now we are focusing on the left and right. 900 is more than 400, so it will go to the right. But we also pull to the left too!! That make the resultant force is 500 to the right.
Water from a fire hose knocks over a wooden shed. Compared with the pressure within
the water, the pressure exerted against the shed is
a) less.
b) the same.
c) more.
d) nonexistent
Consider a 92.0 kg ice skater who is spinning on the ice. What is the moment of inertia of the skater, if the skater is approximated to be a solid cylinder that has a 0.140 m radius and is rotating about the center axis of the cylinder.
Answer:
[tex]I=0.902kg*m^2[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=92.0kg[/tex]
Radius [tex]r=0.140m[/tex]
Generally the equation for moment of Inertia is mathematically given by
[tex]I = 0.5*m*r^2[/tex]
[tex]I=0.5*(92)*0.14^2[/tex]
[tex]I=0.902kg*m^2[/tex]
Match each planet to an accurate characteristic
Answer:
venus - 2
earth - 3
mars - 4
mercury - 1
An object is positively charged if it has more what
Answer:
An Object is positively charged if it has more Positive Electrons in that object
A 3.0 kg pendulum swings from point A of height ya = 0.04 m to point B of height yb = 0.12 m, as seen in the diagram below.
Answer:
3.0−0.12=2.88 or 2.88÷0.04=72
0.04×3.0=0.12 and 0.04+3.0=3.04
Answer: 2.4 J
Explanation: Khan Academy
Monica pulls her daughter Jessie in a bike trailer. The trailer and Jessie together have a mass of 25 kg. Monica starts up a 100-m-long slope that's 4.0 m high. On the slope, Monica's bike pulls on the trailer with a constant force of 9.0 N. They start out at the bottom of the slope with a speed of 5.0 m/s.
What is their speed at the top of the slope?
Answer:
Explanation:
From the given information:
total mass = 25 kg
distance d = 100 m
height = 4.0 m
Force F = 9.0 N
The speed at (bottom) u = 5.0 m/s
Using the concept of energy conservation;
[tex]\dfrac{1}{2}mu^2 + W = \dfrac{1}{2}mv^2 + mgh[/tex]
divide both sides by m
[tex]\dfrac{1}{2}u^2 + \dfrac{W }{m}= \dfrac{1}{2}v^2 + gh[/tex]
multiply both sides by 2
[tex]\dfrac{1}{2}u^2\times 2 + \dfrac{W }{m}\times 2= \dfrac{1}{2}v^2\times 2 + gh\times 2[/tex]
[tex]u^2 +2 \dfrac{W }{m}=v^2 + 2gh[/tex]
[tex]v^2 =u^2 - 2gh+ 2 \dfrac{W }{m}[/tex]
Recall that:
W = Fd
∴
[tex]v^2 =u^2 - 2gh+ 2 \dfrac{Fd }{m}[/tex]
[tex]v^2 =(5.0 \ m/s)^2 - 2(9.81 \ m/s)(4.0 \ m)+ 2 \dfrac{( 9.0 \ N \times 100 \ m) }{25 \ kg}[/tex]
[tex]v^2 =(25.0 ) - 78.48 +72[/tex]
[tex]v^2 = 18.52 \ m^2/s^2[/tex]
[tex]v = \sqrt{18.52 \ m^2/s^2}[/tex]
v = 4.30 m/s
If it were to snow in Phoenix in July, which type of Earth scientist would be most
surprised?
Answer:
Economists, I guess.
Explanation:
Three resistors (16 ohm, 16 ohm and 8 ohm) are connected in parallel. The equivalent
resistance (Re)
Answer:
4 ohm
Explanation:
The equivalent resistance (Re) of three resistors in parallel is given by;
1/Re = 1/R1 + 1/R2 + 1/R3
Where; R1 = 16 ohm, R2 = 16 ohm, R3 = 8 ohm
1/Re= 1/16 + 1/16 + 1/8
1/Re= (0.0625) + (0.0625) + (0.125)
Re= 4 ohm
Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 24 N, and F3 = 13 N. The radial distances are R2 = 0.22 m and R3 = 0.10 m. The forces F2 and F3 are perpendicular to the radial lines labeled R2 and R3. The moment of inertia of the cylinder is 1/2 MR 2/2. Find the magnitude of the angular acceleration of the cylinder about the axis of rotation.
Answer:
α = 13.7 rad / s²
Explanation:
Let's use Newton's second law for rotational motion
∑ τ = I α
we will assume that the counterclockwise turns are positive
F₁ 0 + F₂ R₂ - F₃ R₃ = I α
give us the cylinder moment of inertia
I = ½ M R₂²
α = (F₂ R₂ - F₃ R₃) [tex]\frac{2}{M R_2^2}[/tex]
let's calculate
α = (24 0.22 - 13 0.10) [tex]\frac{2}{12 \ 0.22^2}[/tex]2/12 0.22²
α = 13.7 rad / s²
A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that is perpendicular to the platform at its center. The angular speed of the merry-go-round is constant, and a child at a distance of 1.4 m from the axis has a tangential speed of 2.2 m/s. What is the tangential speed of another child, who is located at a distance of 2.1 m from the axis?
(a) 1.5 m/s
(b) 3.3 m/s
(c) 2.2 m/s
(d) 5.0 m/s
(e) 0.98 m/s
Answer:
[tex]V_2=3.3m/s[/tex]
Explanation:
From the question we are told that:
Distance [tex]d_1=1.4m[/tex]
Tangential speed [tex]V=2.2m/s[/tex]
Distance 2 [tex]d_2=2.1m[/tex]
Generally the equation for Angular velocity is mathematically given by
[tex]w=\frac{v}{r}[/tex]
Therefore
[tex]\frac{v_1}{r_1}=\frac{v_2}{r_2}[/tex]
[tex]V_2=\frac{2.2*2.1}{1.4}[/tex]
[tex]V_2=3.3m/s[/tex]
A 75.0 kg diver falls from rest into a swimming pool from a height of 5.10 m. It takes 1.34 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
Answer:
559.5 N
Explanation:
Applying,
v² = u²+2gs............. Equation 1
Where v = final velocity,
From the question,
Given: s = 5.10 m, u = 0 m/s ( from rest)
Constant: 9.8 m/s²
Therefore,
v² = 0²+2×9.8×5.1
v² = 99.96
v = √(99.96)
v = 9.99 m/s
As the diver eneters the water,
u = 9.99 m/s, v = 0 m/s
Given: t = 1.34 s
Apply
a = (v-u)/t
a = 9.99/1.34
a = -7.46 m/s²
F = ma.............. Equation 2
Where F = force, m = mass
Given: m = 75 kg, a = -7.46 m/s²,
F = 75(-7.46)
F = -559.5 N
Hence the average force exerted on the diver is 559.5 N
A wheelbarrow can be used to help lift a load, such as a pile of dirt, and then push the load across a distance.
A man pushes a wheelbarrow.
Which simple machines make up a wheelbarrow?
a pulley and an inclined plane
a wheel and axle and a lever
a pulley and a wheel and axle
a lever and a wedge
Answer:
a wheel and axle and a lever
The wheelbarrow has a wheel and the then the barrow. The wheel is an wheel and axle machine while the other part is a lever.
What is a simple machine?A simple machine is a device that is used to make work easier. It ensures that less effort is applied to overcome a large load.
The wheelbarrow has a wheel and the then the barrow. The wheel is an wheel and axle machine while the other part is a lever.
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a motor car reaches a velocity of 15m/s in 6s from rest on a perfect test track . what is the average acceleration
Answer:
[tex]{ \tt{initial \: velocity, \: u = 0}} \: (at \: rest) \\ { \tt{final \: velocity, \: v = 15 { {ms}^{ - 1} }}} \\ { \tt{time, \: t = 6s}} \\ { \bf{from \: first \: newtons \: equation \: of \: motion : }} \\ { \bf{v = u + at}} \\ { \tt{15 = 0 + (a \times 6)}} \\ { \tt{6a = 15}} \\ { \tt{acceleration, \: a = 2.5 \: {ms}^{ - 2} }}[/tex]