Answer:
Explanation:
a )
Time for first ball to reach top position
v = u - gt
0 = vi - gt
t = vi / g
Time to reach balcony while going downwards
= vi /g
Total time = 2 vi / g
Time to go down further to the ground = t₁
Total time = 2 vi / g + t₁
Time for the other ball to go to the ground = t₁
Time difference = ( 2 vi / g + t₁ ) - t₁
= 2vi / g .
( b )
v² = u² + 2gh
For both the throw ,
final displacement = h , initial velocity downwards = vi
( For the first ball also , when it go down while passing the balcony , it acquires the same velocity vi but its direction is downwards.)
vf² = vi² + 2gh
vf = √ ( vi² + 2gh )
(c )
displacement of first ball after time t
s₁ = - vi t + 1/2 g t² [ As initial velocity is upwards , vi is negative ]
displacement of second ball after time t
s₂ = vi t + 1/2 g t²
Difference = d = s₂ - s₁
= vi t + 1/2 g t² - ( - vi t + 1/2 g t² )
d = 2 vi t .
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
Problem 4.13: Sound waves travel through air at a speed of 330 m/s. A whistle blast at a frequency of about 1.0 kHz lasts for 2.0 s. (a) Over what distance in space does the "wave train" representing the sound extend? (b) What is the wavelength of the sound? (c) Estimate the precision with which an observer could measure the wavelength. (d) Estimate the precision with which an observer could measure the frequency.
Answer:
a) x = 660 m, b) λ = 0.330 m, c) Δλ = 0.1 cm, d) Δf = 104
Explanation:
a) the distance in which the train of waves extends can be obtained from the uniform movement
v = x / t
x = v t
x = 330 2
x = 660 m
b) the speed of sound is related to the wavelength and frequency
v = λ f
λ = v / f
λ = 330/1000
λ = 0.330 m
c) The precision in the measurement of the wavelength refers to the error or uncertainty in the measurement, if the measurement is direct with a tape measure the precision is the appreciation of the tape measure, in general it is 0.1 cm
d) the accuracy of the frequency in general the frequency is calculated from the measurements of period T
f = 1 / T
The precision of the period is data by the chronometer used, in general time a press of 0.01s, by the response time of the people
Δf = df / dT ΔT
Δf = 1 / T² ΔT
Δf = 1 / (0.001)² 0.01
Δf = 104
As we can see, a much more precise system is needed to reduce the error
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!
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 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 :)
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.
how is work and energy inter-related?? (in simple words)
Answer:
The work and energy is inter- related because if we are energetic then only we are able to work.
People who are weak and not energetic cannot work continuously.
Explanation:
hope this helps you....
what is permittivity
Answer:
Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and electric field intensity.
Which of the following describes the products of a chemical reaction?
A. The original materials
B. The substances that are changed
C. The chemicals on the left side of a chemical equation
Ο Ο
D. The chemicals on the right side of a chemical equation
Answer:
D The chemicals on the right side of a chamical equation
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.
A bicyclist accelerates from rest to a speed of
5.0 meters per second in 10 seconds. During the
same 10 seconds, a car accelerates from a speed
of 22 meters per second to a speed of 27 meters
per second. Compared to the acceleration of the
bicycle, the acceleration of the car is
Answer:
They have the same acceleration of 0.5m/s2 (please note m/s2 is the unit for acceleration and 2 is the power of s)
Explanation:
acceleration= velocity ÷ time
and the time is said to be 10seconds
velocity of car will be the new velocity- the initial velocity = 27-22= 5
acceleration= 5÷10
acceleration= 0.5
hope this helped
now hit that crown button :)
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:)
the distance between crest and the adjacent trough of water waves is 3m, they pass a given point at rate of 5m/s. what is the frequency and the speed of water waves?
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.
what type of reaction is being shown in this energy diagram?
X exothermic, because energy is absorbed from the surroundings
O exothermic, because energy is released into the surrounding
X endothermic, because energy is released into the surrounding
X endothermic, because energy is absorbed from the surroundings
best of luck nerds
Answer:
O exothermic, because energy is released into the surrounding
Explanation:
From the diagram the energy of the reactant is higher than the energy of the product, thereby making it exothermic. If you study diagram well, exothermic reaction means that the reactions releases energy into the surroundings.
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
4. What is the acceleration of the car in each section?
b
с
d
a
Answer:
0-4 acceleration comes at 12 m/s where (B) stagnates at 12 m/s and remains for 4 seconds (C) is breaks being activated slowing the car to 6 m/s in 2 seconds and (D) over the course of 4 seconds brings the car to 10 m/s.
Explanation:
Plzzz help me with this
I’ll give brainliest
Answer:
(A) By reducing friction
Answer:
A
Explanation:
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
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
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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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.
Two train whistles have identical frequencies of 180 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the train platform hears beats with a frequency of 2.00 beats/s when the whistles sound at the same time. What are the two possible speeds and directions that the moving train can have?
Actual answers :3.85 m/s away from the station and 3.77 m/s towards the station from the book. I just need to know how to get to the answers.
Answer:
-3.77 m/s
3.85 m/s
Explanation:
given that
Frequency at stationary = 180 Hz
Beat frequency = 2 Hz
Using Doppler effect, we know that
f' = f[(v ± v0) / (v ± vs)], where
v = speed of sound, 343 m/s
v0 = speed of the observer, 0
vs = speed of light, ?
f = stationary frequency, 180 Hz
f' = stationary ± beat frequency, 180 ± 2
Applying the formula, we have
f' = f[(v ± v0) / (v ± vs)]
182 = 180 [(343 + 0) / (343 + vs)]
182/180 = 343 / 343 + vs
343 + vs = 343 * 180/182
343 + vs = 339.23
vs = 339.23 - 343
vs = -3.77 m/s
Again, using
f' = f[(v ± v0) / (v ± vs)]
178 = 180 [(343 + 0) / (343 + vs)]
178/180 = 343 / 343 + vs
343 + vs = 343 * 180/178
343+ vs = 346.85
vs = 346.85 - 343
vs = 3.85 m/s
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
g Suppose that you seal an ordinary 60W lightbulb and a suitable battery inside a transparent enclosure and suspend the system from a very sensitive balance. (a) Compute the change in the mass of the system if the lamp is on continuously for one year at full power. (b) What difference, if any, would it make if the inner surface of the container were a perfect reflector
Answer:
kekemeeimdeiddnekem
Explanation:
mdjdjdiddmjd jjeneeiej
6th grade science I mark as brainliest
Answer:
8. organelle
Explanation:
9. Epithelial tissue
am i correct?
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.
Two students on ice skates stand one behind the other. Student 2 pushes student 1 in the back; student 1 has less mass than student 2; student 1 has a great acceleration than student 2. which law of motion is this (Newton's laws)
Answer:
law in acting
Explanation:
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
Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown to the right. The mass of the left block m1 = 1.4 kg and the mass of the right block m2 = 4.9 kg. The angle between the applied force and the horizontal is θ = 54°. The coefficient of kinetic friction between the blocks and the surface is μ = 0.38. Each block has an acceleration of a = 3.6 m/s2 to the right.
Answer:
Explanation:The Mass Of The Left Block M1 = 1.3 Kg And The Mass Of The Right Block M2 = 3.1 Kg. The Angle Between The String And The Horizontal Is ... (10%) Problem 8: Two blocks connected by a string are pulled across a horizontal surface by a ... m m, 50% Part (a) Write an equation for the magnitude of the force exerted by the ...