Answer:
Explanation:
a )
Hoverfly will fall with acceleration equal to g .
Initial velocity of fall of hoverflies u = 0
displacement ( vertical ) h = ?
time t = 0.2 s
acceleration due to gravity g = 9.8 m / s²
h = ut + 1/2 g t²
= 0 + .5 x 9.8 x .2²
= .196 m
= 19.6 cm
b )
Time taken to fall by 50 cm or 0.5 m under free fall from initial position .
.5 = 0 + .5 x 9.8 t²
t² = .1020
t = .319 s = 319 ms .
a 6.25-gram bullet traveling at 365 ms strikes and enters a 4.50-kg crate. The crate slides 0.15 m along a wood floor until it comes to rest. What is the change in kinetic energy of the system after the collision
Answer:
the change in kinetic energy of the system is 0.577 J
Explanation:
Given;
mass of the bullet, m₁ = 6.25 g = 0.00625 kg
initial velocity of the bullet, u₁ = 365 m/s
mass of the crate, m₂ = 4.5 kg
initial velocity of the crate, u₂ = 0
distance moved by the system after collision, d = 0.15 m
Determine the final velocity of the system after collision;
m₁u₁ + m₂u₂ = v (m₁ + m₂)
0.00625 x 365 + 4.5 x 0 = v(0.00625 + 4.5)
2.2813 + 0 = v(4.5063)
2.2813 = v(4.5063)
v = 2.2813 / 4.5063
v = 0.506 m/s
The change in kinetic energy of the system after collision is calculated as;
ΔK.E = ¹/₂ (m₁ + m₂)v²
ΔK.E = ¹/₂ (4.506) x 0.506²
ΔK.E = 0.577 J
Therefore, the change in kinetic energy of the system is 0.577 J
A 14.0-g wad of sticky clay is hurled horizontally at a 90-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact
Answer:the speed of the clay immediately before impact =72.58m/s
Explanation:
Given that
mass of the stick clay, M₁= 14.0 g = 0.014 kg
mass of the block ,M₂= 90 g = 0.09 kg
Therefore the total mass= (M₁+M₂) = 104g = 0.104 kg
Also, distance, s = 7.50 m
coefficient of friction μ= 0.650
Acceleration due to gravity ,g = 9.8 m/s²
Using the Work- Energy theorem,
change in kinetic energy = work done
final kinetic energy(K₂) - initial kinetic energy(K₁) = force, F x coefficient of friction, μ x distance,s
The final kinetic energy is zero because after the impact, the block with the clay comes to a stop after 7.50m
kinetic energy =Work done
0.5 x m x v²=coefficient of friction, μ x force(F) x distance,s(Since force = m g )
0.5 x m x v²= μ x m x g x s
0.5 x 0.104 x v² = 0.650 x 0.104x 9.8 x 7.5
v²= 0.650 x 0.104x 9.8 x 7.5 / 0.5 x 0.104
v²==95.55
V = 9.77 m/s
Using the conservation of momentum formulae where
M₁ V₁ + M₂ V₂ = (M₁ + M₂ ) V
Since V₂ which is the velocity of block is zero as the block is initially at rest, We now have that
M₁ V₁ = (M₁ + M₂ ) V
0.014 kg x V₁ = 0.104 x 9.77
V₁=0.104 x 9.77 / 0.014
V=72.58m/s
Which part of the water cycle is where vapor from plants leaves the plants as they breath?
condensation
Transpiration
evaporation
Answer:
I think it is transpiration
Answer:
transpiration is the right answer
difine precision and accuracy
Jack weighs 170 lbs and is 72 inches tall. He is pulling horizontally on a door handle situated at his shoulder height. Actually, it is his body weight and lean that creates this pulling action (a hint). His center of mass while standing erect is 61 percent of his body height, measured from the floor upwards. The door handle is 60 inches above the ground, and again he is pulling purely horizontally on this handle.
If Jack's lean angle is 20 degrees and he is leaning back - pivoting about his heels, how much force does he apply to the door handle?
Include units in your answer, lbs.
Express your answer to the nearest 0.1 lbs.
Answer:
He is pulling horizontally on a door handle situated at his shoulder height. ... His Center Of Mass While Standing Erect Is 61 Percent Of His Body Height, Measured ... Actually, it is his body weight and lean that creates this pulling action (a hint).
72ibs
A block of mass m1 = 19.5 kg slides along a horizontal surface (with friction, μk = 0.35) a distance d = 2.6 m before striking a second block of mass m2 = 8.25 kg. The first block has an initial velocity of v = 6.5 m/s.
(a) Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
(b) How far does block two travel, d2 in meters, before coming to rest after the collision?
Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
Explanation:
By using conservation of linear momentum and also by equating work done to kinetic energy, [tex]V_{2}[/tex] = 15.36 m/s and [tex]d_{2}[/tex] = 4.32 meters
Parameters given are :
[tex]m_{1}[/tex] = 19.5 kg
friction, μk = 0.35
distance d = 2.6 m
mass [tex]m_{2}[/tex] = 8.25 kg.
initial velocity of [tex]U_{1}[/tex] = 6.5 m/s.
a.) Since we assumed that the block one stops after it collides with block two, the final velocity for block one will be zero. That is, [tex]V_{1}[/tex] = 0 so its final momentum = 0
Let us also assume that block two was initially at rest. Therefore, it initial velocity and its momentum will be equal to zero.
The formula to use will be :
[tex]m_{1}U_{1} = m_{2}V_{2}[/tex]
Substitute all the parameters into the formula above
19.5 x 6.5 = 8.25[tex]V_{2}[/tex]
Make [tex]V_{2}[/tex] the subject of formula
[tex]V_{2}[/tex] = 126.75/8.25
[tex]V_{2}[/tex] = 15.36 m/s
b.) Let us first calculate the work done in by block one.
The K.E = [tex]1/2mU^{2}[/tex]
substitute its mass and velocity into the formula
K.E = 1/2 x 19.5 x [tex]6.5^{2}[/tex]
K.E = 411.94 Joule
The work done = Kinetic energy
But the resultant Force F = force f - friction
where Frictional force = 0.35 x 19.5 x 9.8
Frictional force = 66.89N
Work done will be the product of resultant Force F and the distance travelled
(F - 66.89) x 2.6 = 411.94
F - 66.89 = 411.94/2.6
F - 66.89 = 158.44
F = 225.3 N
The second block will experience the same force which is equal to 225.3N
Find the kinetic energy of the second block.
K.E = [tex]1/2mV^{2}[/tex]
K.E = 0.5 x 8.25 x 15.36^2
K.E = 973.2
Using The work done = Kinetic energy
225.3[tex]d_{2}[/tex] = 973.2
[tex]d_{2}[/tex] = 973.2/225.3
[tex]d_{2}[/tex] = 4.32 meters
Learn more here : https://brainly.com/question/20658056
Specify whether the boiling point, as determined in the miniscale boiling-point apparatus, is the temperature a.of the liquid at the timebubbles first emerge slowly from the liquid. b.at the vapor-liquid interface above the surface of the boiling liquid while a drop of liquid c.is suspended from the thermometer. d.of the liquid at the timebubbles emerge rapidly from the liquid. e.of the heating source at the timebubbles emerge rapidly from the liquid.
Answer:
a. of liquid at the time bubbles first emerge slowly from the liquid.
Explanation:
Boiling point of liquid happens due to heat energy. This is an exothermic reaction as heat is released in to the environment. The initial boiling vapors slowly move away from the liquid and as the temperature increases the vapors start moving quickly.
what is permittivity
Answer:
Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and electric field intensity.
Any change in the cross section of the vocal tract shifts the individual formant frequencies, the direction of the shift depending on just where the change in area falls along the standing wave. Constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency; expansion of the tract at those same places raises the frequency. Three other major tools for changing the shape of the tract in such a way that the frequency of a particular formant is shifted in a particular direction are the jaw, the body of the tongue and the tip of the tongue. Moving the various articulatory organs in different ways changes the frequencies of the two lowest formants over a considerable range [18].
One way to increase formant frequency is to ________ the vocal tract at a place where the standing wave of a formant frequency exhibits minimum-amplitude pressure oscillations.
a. Stretch
b. Vibrate
c. Contract
d. Expand
Answer:
The correct answer is option D.
Explanation:
It is stated in the question that constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency so to increase formant frequency, the vocal should expand where the standing wave of a formant exhibits minimum-amplitude pressure oscillations. The answer is D.
I hope this helps.
Animals conduct_______.
A. cellular respiration
B. photosynthesis
C. both cellular respiration and photosynthesis
For a given substance, the molecules
move fastest when the substance is
Answer:GAS
Explanation:
Q5. Use Superposition to V. in the circuit below? (5 points)
4 mA
12V
2 ΚΩ
2 mA
1 ΚΩ
2 ΚΩ
Answer:
4va
12va
2jk
1jk
2jk
Which one of Newton’s Laws best explains a bottle flip?
You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 5.00 s after it was thrown. What is the speed of the rock just before it reaches the water 22.0 m below the point where the rock left your hand? Ignore air resistance.
Answer:
Explanation:
for vertical movement , time to reach the top = time to reach the hand = 2.5 s
v = u - gt
At the top , v = 0 , time t = 2.5 s
0 = u - g x 2.5
u = 2.5 x 9.8 = 24.5 m /s
velocity of throw = 24.5 m /s
So , when it passes the hand on its way down , it will have velocity equal to 24.5 m /s and it will accelerate downwards . Let its velocity down by 22 m be v
v² = u² + 2 g s
= 24.5² + 2 x 9.8 x 22
= 600.25 + 431.2
= 1031.45
v = 32.11 m /s .
A battery has an emf of ε = 15.0 V. THe terminal voltage of the battery is Vt = 11.6 V when it is delivering P = 20.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance r of the battery?
AnswerHM???
Explanation:
I dONT KNOW
A woman accidentally drops a flowerpot from a windowsill at a height d above the street towards a man of height h standing below. The woman calls out to the man in just enough time for the man to move out of the way. If the man needs a time interval of Δt to respond to the warning, at what height above the street will the flowerpot be when the woman calls out the warning? (Use the following as necessary: d, h, Δt, v for the speed of sound, and g for gravitational acceleration.)
Answer:
h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0
The correct result is that of a positive height
Explanation:
For this exercise we use the kinematic relations, let's start by finding the time it takes for the sound to reach the man
v_s = y / t
t = [tex]\frac{y}{ v_s}[/tex]
this height is y = h
t = \frac{h}{ v_s}
the man has a response time of t = t₀, therefore
time to move is
t' = t - t₀
the initial height of flower pot is
y = y₀ + v₀ t' - ½ g t'²
when it reaches the floor the height is zero y = 0 and as the pot is dropped its initial velocity is zero v₀ = 0
0 = y₀ +0 - ½ g (t -t₀)²
if the initial height is i = h,
h = ½ g ([tex]\frac{h}{v_s}[/tex] - t₀)²2
[tex]\frac{2}{g} h[/tex] = [tex]\frac{h^2}{v_s^2}[/tex] - [tex]\frac{2t_o }{v_s} h[/tex] + t₀²
[tex]\frac{h^2}{v_s^2} - ( \frac{2t_o}{v_s} + \frac{2}{g} ) h + t_o^2 = 0[/tex]h2 / vs2 - (2nd / vs + 2 / g) h + to2 - = 0
[tex]h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0[/tex]
To know the height, you must solve the second degree equation, it is much easier with numerical values.
The correct result is that of a positive height
How can you tell whether an object is neutral
or charged? What would you have to do to test
that object?
Answer:
The number of electrons that surround the nucleus will determine whether or not it is electrically charged or electrically neutral
Explanation:
Drag the tiles to the correct boxes to complete the pairs
Match the particles with their characteristics.
subatomic particles with a positive charge
subatomic particles with a negative charge
subatomic particles with no charge
made of atoms
neutrons
electrons
protons
malaria
Answer:
1. Protons.
2. Electrons.
3. Neutrons.
4. Molecules.
Explanation:
1. Protons: subatomic particles with a positive charge. They are bound together in the nucleus of an atom due to strong nuclear forces.
2. Electrons: subatomic particles with a negative charge. Electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.
3. Neutrons: subatomic particles with no charge. The negative charge of the electrons cancels the positive charge of the protons.
4. Molecules: they are made of atoms.
Generally, molecules attach on the inside of a mineral to give it shape. Therefore, the molecule of a mineral is a crystal three-dimensional regular structure (arrangement) of chemical particles that are bonded together and determines its shape.
Due to the fact that these molecules are structurally arranged or ordered and are repeated by different symmetrical and translational operations they determine the shape of minerals.
1. Clara stops for 10 minutes to catch up with a friend.
Answer:
Clara has speed of 80m/min
Explanation:
Clara was jogging at 600 m in 5 minutes. She stopped suddenly which reduced her velocity and then she waited for 10 minutes so that her friends comes near her. She stopped to catch her friend. During this 10 minutes the velocity of Clara is zero. She started to walk again at a slower speed of 80m/min.
A child holds a sled at rest on frictionless snow covered hill. if the sled weighs 77N,find the force T exerted by the rope on the sled and the force n exerted by the hill on the sled
Answer:62
Explanation:
The weight of the sled is 77 N. The force by the hill on the sled is equal to its weight that is 77 N. Then the tension force exerted by the rope on the sled is being 77N sin θ, where θ be the angle of inclination.
What is force?Force is an external agent acting on an object to change its motion or to deform it. There are various kinds of force like magnetic force, tension force, frictional force, gravitational force etc.
The weight that an object experience on earth is due to the gravitational force. The force that is exerted by a rope on an object is tension force since it is pulling from a side.
The normal force by the hill on the sled is equal to its weight that is 77 N. The tension force on the sled by the rope is dependent on the angle of inclination θ. If know the angle we can find T by the equation:
T = 77 sin θ.
Find more on tension force:
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If all pairs of adjacent sides of a quadrilateral are congruent then it is called _________.
(A) rectangle (B) parallelogram (C) trapezium, (D) rhombus
Answer:
D
Explanation:
If you need an explanation feel free to ask.
Question 15 of 25
What is the period of a wave that has a frequency of 30 Hz?
Answer:
0.033 seconds
Explanation:
Period = 1/30 = 0.033 seconds
Answer:
The answer is 0.03 s
Explanation:
A.P.E.X.
Which statement is true of a glass lens that diverges light in air?
A.
It is thick near the center and thin at the edges.
B.
It is thin near the center and thick at the edges.
C.
It is uniformly thick.
D. It is uniformly thin.

Answer: it is thin near the center and thick at the edges
Explanation: took the test on Plato :)
An ordinary ruler is used to measure the area and its error of a rectangle. It is found that their sides are 5.0 cm long and 2.0 cm width. The error in area (in cm) is
Answer:
You need to know the accuracy to which you can read the ruler:
Suppose that you can read the read the ruler to the nearest milimeter
A = L * W your calculated area of the rectangle
A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA
Or ΔA = L ΔW + W ΔL
Where we have subtracted A = L * W and the term ΔL * ΔA is very small
So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2
Then you report A = 10 cm^2 +- .7 cm^2 including the - sign for completeness
In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 2.15 kg. They then hang the object on a pivot located 0.163 m from the object's center of mass and set it swinging at a small amplitude. As two of the team members carefully count 113 cycles of oscillation, the third member measures a duration of 241 s. What is the moment of inertia of the object with respect to its center of mass about an axis parallel to the pivot axis
Answer:
0.339 kgm²
Explanation:
We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.
Since T = 2π√(I/mgh), making I subject of the formula, we have
I = mghT²/4π²
Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.
So, T = 241 s/113 = 2.133 s
So, Substituting the values of the variables into I, we have
I = mghT²/4π²
I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²
I = 15.63/4π² kgm²
I = 0.396 kgm²
Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis
I' = I - mh²
I' = 0.396 kgm² - 2.15 kg × (0.163 m)²
I' = 0.396 kgm² - 0.057 kgm²
I' = 0.339 kgm²
Galileo _____.
did not believe friction existed
believed that friction stopped objects in motion
believed that friction kept objects in motion
assumed that in a frictionless environment objects would never move
Answer:
friction help to slow motion in other word it oppose motion, but in a frictionless environment object would move with difficult stopping point.
The force of gravity on a person or object on the surface of a planet is called
A. gravity
ОВ.
B. free fall
OC
c. terminal velocity
D. weight
Answer:
D. Weight
Explanation:
Hope that helps:)
What are two things that happen to the sugars that are made by the plant during photosynthesis?
I
Answer:
The sugars produced by photosynthesis can be stored, transported throughout the tree, and converted into energy which is used to power all cellular processes. Respiration occurs when glucose (sugar produced during photosynthesis) combines with oxygen to produce useable cellular energy.
Explanation:
I think this is correct lol.
Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end of a rope is attached to the bus and the other is tied around Magnus's waist. If a force gauge placed halfway down the rope reads out a constant 2100 Newtons while Magnus pulls the bus a distance of 1.30 meters, how much work does the tension force do on Magnus
Answer:
Workdone = -2730 J
Explanation:
Formula for workdone is;
W = Force × Displacement
Now, according to Newton's 3rd law of motion, to every action, there is an equal and opposite reaction.
In the question given, we are told that a force gauge placed halfway down the rope reads out a constant 2100 Newtons while Magnus pulls the bus. This means that the force exerted by the rope on Magnus acts in an opposite direction to that which Magnus does to the rope.
Therefore, the force will be in the negative direction.
So;
Workdone = -2100 N × 1.3 m
Workdone = -2730 J
A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 46.1 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding
Answer:
The minimum coefficient of friction required is 0.35.
Explanation:
The minimum coefficient of friction required to keep the crate from sliding can be found as follows:
[tex] -F_{f} + F = 0 [/tex]
[tex] -F_{f} + ma = 0 [/tex]
[tex] \mu mg = ma [/tex]
[tex] \mu = \frac{a}{g} [/tex]
Where:
μ: is the coefficient of friction
m: is the mass of the crate
g: is the gravity
a: is the acceleration of the truck
The acceleration of the truck can be found by using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} [/tex]
Where:
d: is the distance traveled = 46.1 m
[tex] v_{f}[/tex]: is the final speed of the truck = 0 (it stops)
[tex]v_{0}[/tex]: is the initial speed of the truck = 17.9 m/s
[tex] a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2} [/tex]
If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.
[tex] \mu = \frac{a}{g} [/tex]
[tex] \mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}} [/tex]
[tex] \mu = 0.35 [/tex]
Therefore, the minimum coefficient of friction required is 0.35.
I hope it helps you!