Let both the balls have the same mass equals to m.
Let [tex]v_1[/tex] and [tex]v_2[/tex] be the speed of the ball1 and the ball2 respectively, such that
[tex]v_1>v_2\;\cdots(i)[/tex]
Assuming that both the balls are at the same level with respect to the ground, so let h be the height from the ground.
The total energy of ball1= Kinetic energy of ball1 + Potential energy of ball1. The Kinetic energy of any object moving with speed, [tex]v[/tex], is [tex]\frac 12 m v^2[/tex]
and the potential energy is due to the change in height is [tex]mgh[/tex] [where [tex]g[/tex] is the acceleration due to gravity]
So, the total energy of ball1,
[tex]=\frac 12 m v_1^2 + mgh\;\cdots(ii)[/tex]
and the total energy of ball1,
[tex]=\frac 12 m v_2^2 + mgh\;\cdots(iii)[/tex].
Here, the potential energy for both the balls are the same, but the kinetic energy of the ball1 is higher the ball2 as the ball1 have the higher speed, refer equation (i)
So, [tex]\frac 12 m v_1^2 >\frac 12 m v_2^2[/tex]
Now, from equations (ii) and (iii)
The total energy of ball1 hi higher than the total energy of ball2.
True or false: points that lie on the same plane are Collinear
r = (2+2+1) i - (t+1)] + t3 k
what is the direction of initial velocity
Answer:
In the - j direction, that is negative of the y-axis
Explanation:
As typed in the question, the position of the object is given by the expression in three component ( i, j, k) form:
r (t) = 5 i - (t + 1 ) j + t^3 k
and since the velocity is the derivative of position with respect to time, by doing the derivative of this expression we get:
v(t) = 0 i - 1 j +3 t^2 k
which for the initial velocity requested (that is at time zero) we have:
v(t) = 0 i - 1 j +3 (0)^2 k = = 1 j
Then the direction of the initial velocity is entirely in the direction of the j versor, that is pointing to the negative of the y-axis.
what type of organism does not use photosynthesis
(a) plants
(b) bacteria
(c) algae
(d) humans
Answer:
D) Humans
Explanation:
Photosynthesis is a useless ability without some way of exposing yourself to as much of the Sun's energy as possible. That requires a large surface area, relative to their volume. Plants achieve that with large, horizontal, light-capturing surfaces – leaves.
The group that receives treatment is called the?
Tested Group
Control Group
Placebo Group
Experimental Group
Answer:
experimental group
Explanation:
please mark me as a brainlist..
Answer:
Experemtial group
Explanation:
The placebo is just sugar pills tested is ones already done and control is a mix of the two
What is Bill's average running speed?
Answer:
Hello!
Sorry you haven't put up an image of your question! Without it we can't answer your question!
Explanation:
Maybe put up another one and it'll be answered!
:D
How did oxygen get into the earths atmosphere in its early days? plz explain
Answer:
The answer is tiny organisms known as cyanobacteria, or blue-green algae. These microbes conduct photosynthesis: using sunshine, water and carbon dioxide to produce carbohydrates and, yes, oxygen. In fact, all the plants on Earth incorporate symbiotic cyanobacteria (known as chloroplasts) to do their photosynthesis for them down to this day.
Explanation:
A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of the air in the classroom?
Answer:
the mass of the air in the classroom = 2322 kg
Explanation:
given:
A classroom is about 3 meters high, 20 meters wide and 30 meters long.
If the density of air is 1.29 kg/m3
find:
what is the mass of the air in the classroom?
density = mass / volume
where mass (m) = 1.29 kg/m³
volume = 3m x 20m x 30m = 1800 m³
plugin values into the formula
1.29 kg/m³ = mass
1800 m³
mass = 1.29 kg/m³ ( 1800 m³ )
mass = 2322 kg
therefore,
the mass of the air in the classroom = 2322 kg
If A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, then the mass of the air in the classroom is 2322Kg.
What is density??Density is the ratio of mass to volume. it tells how much mass a body is having for its unit volume. for example egg yolk has 1027kg/m³ of density, means if we collect numbers of egg yolk and keep it in a container having volume 1 m³ then total amount of mass it is having will be 1027kg. Density is a scalar quantity. when we add egg yolk into the water, egg yolk has greater density than water( 997 kg/m³), because of higher density of egg yolk it contains higher mass in same volume as water. hence due to higher mass higher gravitational force is acting on the egg yolk therefore it goes down on the inside the water. water will float upon the egg yolk. same situation we have seen when we spread oil in the water. ( in that case water has higher density than oil. thats why oil floats on the water).
Given,
Height = 3 m
Width = 20 m
length= 30 m
Density of air = 1.29kg/m³
The volume of the room = 3×20×30 m³
Volume V = 1800m³
By formula,
Density = Mass/Volume
1.29kg/m³ = Mass/1800m³
Mass of the air = 1.29×1800 = 2322 Kg
The mass of the air is classroom is 2322Kg.
To know more about density :
https://brainly.com/question/29775886
#SPJ2.
Ricardo finds an online site about the gas laws. The site shows the equation below for Charles’s law. What change would correct the error on the site? “Charles’s law” should read “Gay-Lussac’s law,” which explains changes in volume and temperature. The symbol for T2 should be smaller than for T1 because if volume increases, then temperature should decrease. Each T should be replaced by a P in the equation because Charles’s law describes changes in volume and pressure. The volume should be divided by temperature on each side of the equation.
Answer:
D. The volume should be divided by temperature on each side of the equation.
Explanation:
Charles's Law is [tex]\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }[/tex].
Answer:
The volume should be divided by temperature on each side of the equation.
Explanation:
1. Looking at the planet vs. eccentricity table, which two planets have the greatest eccentricity?
Answer:
Pluto & Mercury
Explanation:
Pluto's eccentricity is 0.248
Mercury's eccentricity is 0.206
blutions math practice
Maddie Hicks: Attempt 1
07.5 ml
Question 4 (1 point)
You have a stock bottle of 3mM tetrodotoxin solution, but your protocol states that
you need 6ml of 1mM tetrodotoxin (MW=319.27 g/mol). How much of your 3mM
stock solution (liquid) should you use to make the 1mM solution?
Answer:
[tex]V_1=2mL[/tex]
Explanation:
Hello,
In this case, considering this as a dilution problem, the first step here is to consider that the moles of tetrodotoxin remains the same and just the volumes and concentrations are modified from the initial stock (1) and the required dilution (2):
[tex]V_1M_1=V_2M_2[/tex]
Whereas V is referred to volume and M to molar concentration, in this case in mM. In such a way, solving the V1 as the volume of the 3-mM solution we obtain:
[tex]V_1=\frac{V_2M_2}{M_1} \\\\V_1=\frac{6mL*1mM}{3mM} \\\\V_1=2mL[/tex]
It means you need 2 mL of the 3-mM solution.
Best regards.
A cyclist accelerates from rest to 8 m/s in 3 seconds. How far did the cycles travel in 3 seconds?
Answer:
[tex] \boxed{\sf Distance \ travelled = 12 \ m} [/tex]
Given:
Initial speed (u) = 0 m/s (Accelerates from rest)
Final speed (v) = 8 m/s
Time taken (t) = 3 seconds
To Find:
Distance travelled by cyclist (s)
Explanation:
From equation of motion of object moving with uniform acceleration in straight line we have:
[tex] \boxed{ \bold{s = (\frac{v + u}{2} )t}}[/tex]
By substituting value of v, u & t in the equation we get:
[tex] \sf \implies s = ( \frac{8 + 0}{2} ) \times 3 \\ \\ \sf \implies s = \frac{8}{2} \times 3 \\ \\ \sf \implies s = 4 \times 3 \\ \\ \sf \implies s = 12 \: m[/tex]
[tex] \therefore[/tex]
Distance travelled by cyclist (s) = 12 m
Answer:
s(distance) =36m
Explanation:
u(initial velocity) =0 m/s
a =8 m/s^2
t=3s
s=ut+1/2at^2
s=1/2(8)(3)^2
s=1/2(8)(3)(3)
s=4(9)
s=36m
seagull is flying at a rate of 20 miles per hour south, it encounters wind blowing 20 miles per hour north. What is the resultants
Answer:
the winds will make the bird stop
Explanation:
is basically 20 - 20
Which has the fastest wave speed, a high frequency sound or a low frequency sound?
Answer:
high frequent sound
Explanation:
because if its low than its slower.
A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A that splits at point B and attaches to the ship at points C and D. The two rope segments BC and BD angle away from the center of the ship at angles of ϕ = 26.0 ∘ and θ = 21.0 ∘, respectively. The tugboat pulls with a force of 1200 lb . What are the tensions TBC and TBD in the rope segments BC and BD?
Answer:
The tensions in [tex]T_{BC}[/tex] is approximately 4,934.2 lb and the tension in [tex]T_{BD}[/tex] is approximately 6,035.7 lb
Explanation:
The given information are;
The angle formed by the two rope segments are;
The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°
The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°
Therefore, we have;
The tension in rope segment BC = [tex]T_{BC}[/tex]
The tension in rope segment BD = [tex]T_{BD}[/tex]
The tension in rope segment AB = [tex]T_{AB}[/tex] = Pulling force of tugboat = 1200 lb
By resolution of forces acting along the line A_F gives;
[tex]T_{BC}[/tex] × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = [tex]T_{AB}[/tex] = 1200 lb
[tex]T_{BC}[/tex] × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = 1200 lb............(1)
Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;
[tex]T_{BC}[/tex] × sin(26.0°) + [tex]T_{BD}[/tex] × sin(21.0°) = 0...........................(2)
Which gives;
[tex]T_{BC}[/tex] × sin(26.0°) = - [tex]T_{BD}[/tex] × sin(21.0°)
[tex]T_{BC}[/tex] = - [tex]T_{BD}[/tex] × sin(21.0°)/(sin(26.0°)) ≈ - [tex]T_{BD}[/tex] × 0.8175
Substituting the value of, [tex]T_{BC}[/tex], in equation (1), gives;
- [tex]T_{BD}[/tex] × 0.8175 × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = 1200 lb
- [tex]T_{BD}[/tex] × 0.7348 + [tex]T_{BD}[/tex] ×0.9336 = 1200 lb
[tex]T_{BD}[/tex] ×0.1988 = 1200 lb
[tex]T_{BD}[/tex] ≈ 1200 lb/0.1988 = 6,035.6938 lb
[tex]T_{BD}[/tex] ≈ 6,035.6938 lb
[tex]T_{BC}[/tex] ≈ - [tex]T_{BD}[/tex] × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb
[tex]T_{BC}[/tex] ≈ -4934.1733 lb
From which we have;
The tensions in [tex]T_{BC}[/tex] ≈ -4934.2 lb and [tex]T_{BD}[/tex] ≈ 6,035.7 lb.
You are trying to get to class on time using the UCF Shuttle. You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. In 40.9 m you will reach a barrier and you must catch the shuttle before that point. The shuttle has a constant acceleration of 4.5 m/s2. What is the minimum velocity you have to run at to catch the bus before it reaches the barrier
Answer:
20.1 m/s
Explanation:
Since You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. And In 40.9 m you will reach a barrier and you must catch the shuttle before that point.
Given that the shuttle has a constant acceleration of 4.5 m/s2.
The total distance to cover is:
Total distance = 40.9 + 3.9 = 44.8 m
Assuming you are starting from rest. Then initial velocity U = 0
Using the 3rd equation of motion to calculate the minimum velocity.
V^2 = U^2 + 2as
V^2 = 0 + 2 × 4.5 × 44.8
V^2 = 403.2
V = sqrt (403.2)
V = 20.1 m/s
Therefore, the minimum velocity you have to run at to catch the bus before it reaches the barrier is 20.1 m/s
- When a mixture contains substances that are not evenly mixed, it is called?
Answer: Heterogenous mixture
Explanation:
A heterogeneous mixture in which the components of the mixture are not evenly mixed or uniform, allowing one to identify the different constituents of each components and enable the mixture to be separated physically.
Examples of heterogeneous mixtures includes
-- sand and nails
---rice and beans
--- water and oil
Another type of mixture is the homogeneous mixture in which all the components are evenly mixed causing that each components cannot be visible with the eye and therefore separated chemically.
How many meters are in 5.0 cm?
500
0.050
0.0005
0.5
Explanation:
100cm=1m
5 cm= x
cross multiple
x=5/100
=0.05m
There are 0.05 meters in 5 cm. The correct option among the following is option (B).
In the metric system, a centimeter (cm) is a unit of length. It is one centimeter (1 cm = 0.01 m) in size. Small distances or measurements, such as the length, height, or width of items, are frequently measured in centimeters. Due to its practical size for many routine measurements, it is also frequently employed in scientific and mathematical computations.
In the International System of Units (SI), the meter serves as the base unit and is frequently used to measure lengths, heights, and other measurements. It serves as the basic building block for other metric units like centimeters (one meter equals one hundred centimeters) and kilometers (one kilometer equals one thousand meters).
So, to convert 5.0 cm to meters:
5.0 cm ÷ 100 = 0.05 m
Hence, 5.0 cm is equal to 0.05 meters. The correct option is (B).
To learn more about meters, here:
https://brainly.com/question/13187866
#SPJ4
What is the net Force needed to get a 16 kg box moving 4 m/s^2?
Answer:
The net force should be of a magnitude of 64 N
Explanation:
We use Newton's second Law for this:
[tex]F_{net} = m\,*\,a[/tex]
which for our case gives:
[tex]F_{net} = m\,*\,a=16\,(4)\,N= 64\,\,N[/tex]
A hockey puck initially travelling to the right at 34 m/s. It moves for 7 before
coming to a stop. How far did it move in 7 seconds?
You can use kinematic equations
Answer:
[tex]x=119m[/tex]
Explanation:
Hello,
In this case, since the hockey puck was moving at 34 m/s and suddenly stopped (final velocity is zero) in 7 seconds, we can first compute the acceleration via:
[tex]a=\frac{v_f-v_o}{t}=\frac{0m/s-34m/s}{7s}\\ \\a=-4.86m/s^2[/tex]
In such a way, we can compute the displacement via:
[tex]x=\frac{v_f^2-v_o^2}{2a}\\ \\x=\frac{0^2-(34m/s)^2}{2*-4.86m/s^2}\\ \\x=119m[/tex]
Best regards.
what role do control groups play
Answer:
Control groups let the one who is expermenting compare the effect of the varibles in the expermental group.
Explanation:
An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C. (a) What is the speed of the electron 1.3 ns after entering this region? (b) How far does the electron travel during the 1.3 ns interval?
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
A box measures 3.12 ft in length, 0.0455 yd in width and 7.87 inches in height. Since volume can be found using length x width x height, find the volume in cubic centimeters. Now find the volume in gallons.
Explanation:
This problem is on units conversion
given the dimensions
3.12 ft in length,
0.0455 yd in width and
7.87 inches in height
we have to convert to cm first
3.12 ft to cm=
1 ft-------30.48cm
3.12--------x
x=95.09cm
0.0455 yd to cm
0.0455yd ft to cm=
1 yd-------91.44cm
0.0455--------x
x=4.16cm
7.87 in to cm
1 in-------2.54cm
7.87--------x
x=19.98cm
the volume is =95.09cm*4.16cm*19.98cm
volume= 7907.45cm^3
the volume in gallon is
1 cm^3------0.000264172gal
7907.45cm^3-------x
x=7907.45*0.000264172
x=2.088gallons
A brick is lying on a table in a state of static equilibrium. If the mass of the brick is 7.52 kilograms, what is the normal force exerted by the table on the brick? A. -73.7 newtons B. 73.7 newtons C. 80.7 newtons D. 7.52 newtons E. 8.07 newtons
The answer is OPTION B : 73.7 N
Answer:
B
Explanation:
during which stage do you have sleepwalking and sleeptalking
A. stage 2
B. stage 4
C. REM
D. stage 1
What potential difference is required to produce a current of 0.5 A in a bulb of power 60 W
Answer:
ans is 120
Explanation:
I= 0.5 A
p= 60w
p=Iv
60/0.5
v= 120
Answer:
V=P/I
V=60/0.5
V=120
Explanaton:
the formula P=IV is converted to V=P/I to make V (voltage) the subject. then sub in the values to get your answer
note that potential difference is same as voltage
4. What is the velocity of an object that doesn't move?
It depends on the object b. it depends on the speed c. it depends on the height
O mis
help
Answer:
Acceleration /Speed
Explanation:
An objects Velocity can be determined by acceleration,
Please pay attention in your middle school class, speed and velocity quiz.
Aldis is swinging a ball tied to the end of a string over his head. Suddenly, the string breaks and the ball flies away. Arrow best represents the path the ball follows after the string breaks.
Answer:
Straight line in the direction of the tangential velocity the ball had at the moment the string broke
Explanation:
After the string breaks, the ball now disconnected from the centripetal force that was exerted via the string, continues its travel in a straight line in the direction of the tangential velocity it had at the moment the string broke.
Answer:
B
Explanation:
just took the test :)
.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattened. Suppose the maxi-mum depth of the dent is on the order of 1 cm. Find the order of magnitude of the maximum acceleration of the ball while it is in contact with the pavement. State your assumptions, the quantities you estimate, and the values you estimate for them.
Answer:
a = 1.1 10⁵ m / s²
Explanation:
This is a momentum exercise, where we use the relationship between momentum and momentum
I = ∫ F dt = Δp
= p_f - p₀
as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction
F t = 2 (m v)
therefore the average force is
F = 2 m v / t
where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy
starting point. Done the ball is released with zero initial velocity
Em₀ = U = mgh
final point. Upon reaching the ground, just before the deformation begins
Em_f = K = ½ m v²
energy is conserved in this system
Em₀ = Em_f
m g h = ½ m v²
v = √ (2gh)
This is the velocity of the body when it reaches the ground, so the force remains
F = 2m √(2gh) /t
where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds
knowing this force let's use Newton's second law
F = m a
a = F / m
a = 2 √(2gh) / t
We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s
a = 2 √ (2 9.8 1.5) / 10⁻³
a = 1.1 10⁵ m / s²
The chilled water system for a 27-story building has a pump located at ground level. The lost head in a vertical riser from the pump to an equipment room on the twenty-seventhfloor is 40ftof water, and the pump produces 270ft of head. What is the pressure on the suction side of the pump for a pressure of 8 psig to exist in the riser on the twenty-fifth floor
This question is incomplete, the complete question is;
The chilled water system for a 27-story building has a pump located at ground level. The lost head in a vertical riser from the pump to an equipment room on the twenty-seventh floor is 40ft of water, and the pump produces 270ft of head. What is the pressure on the suction side of the pump for a pressure of 8 psig to exist in the riser on the twenty-fifth floor
Assume 12ft of elevation per floor
Answer: 48.68 psig
Explanation:
First we calculate the elevation of the building
hb = 27 story * 12ft per floor/story
hb = 324 ft
given that the head lost in the vertical riser hL = 40 ft
now the delivery head required in the riser on he 27th floor;
hd = 8 psig * (2.31 ft / 1 psig)
hd = 18.46 ft
Now calculate the suction head required by balancing the energy per unit weight of water, considering pump as the control volume
hp = (hb + hL + hd) - hs
hs = hb + hL + hd - hp
where hp is the head developed by the pump (270 ft)
hb is the elevation of the 27th floor of the building ( 324 ft)
hL is the head lost in the vertical riser ( 40 ft)
hd is the head required to exist in the riser on the 27th floor (18.46 ft)
so we substitute
hs = 324 ft + 40 ft + 18.46 ft - 270 ft
hs = 112.46
so 112.46ft * (1 psig / 2.31 ft)
= 48.68 psig
The overall length of a piccolo is 30.0 cm. The resonating air column is open at both ends. (a) Find the frequency of the lowest note a piccolo can sound. (Assume that the speed of sound in air is 343 m/s.) Hz (b) Opening holes in the side of a piccolo effectively shortens the length of the resonant column. Assume the highest note a piccolo can sound is 3 000 Hz. Find the distance between adjacent antinodes for this mode of vibration.
Answer:
(a) the frequency of the lowest note the piccolo can sound is 571.7 Hz
(b) the distance between adjacent antinodes is 5.72 cm
Explanation:
(a)
Given;
length of piccolo, L = 30 cm = 0.3 m
the speed of sound in air is 343 m/s
The wavelength of a pipe open at both ends, for the first harmonic is given;
L = A → N + N → A
L = λ / 4 + λ / 4
L = λ / 2
λ = 2L
λ = 2 x 0.3 = 0.6 m
The fundamental frequency (lowest frequency) is given by;
f₀ = v / λ
f₀ = (343 / 0.6)
f₀ = 571.7 Hz
(b)
Given;
highest note, f = 3000 Hz
the distance between adjacent antinodes is given by;
[tex]d = \frac{v}{2f}\\\\ d = \frac{343}{2*3000}\\\\ d = 0.0572 \ m\\\\d = 5.72 \ cm[/tex]