Momentum = (mass) x (speed)
If you work the problem in the same units as the given data, then you get the momentum in units of kilogram-meters per second, and your horse has 1,440 of them.
Answer:
A
Explanation:
1440 kg*m/s
PLS ANSWER FAST WILL GIVE BRAINLY!!!!!!!
explain at least one way that engineers and scientists work to prevent coastal erosion.
Answer:
the answer is the explaination
Explanation:
Present beach erosion prevention methods include sand dunes, vegetation, seawalls, sandbags, and sand fences. Based on the research conducted, it is evident that new ways to prevent erosion must be obtained. Each way that is currently used has extensive negative effects on beaches and their natural tendencies.
Answer:
They will construct strong sea walls with the intent to protect the coastline and infrastructure that could be affected by erosion.
Explanation:
Calculate the spring constant of the spring in a child's pogo stick if the child has a mass of 32 kg and bounces once every 2.7 seconds
Answer:
The spring constant is 173.12 N/m.
Explanation:
mass, m = 32 kg
Period, T = 2.7 s
let the spring constant is K.
Use the formula of period,
[tex]T = 2\pi\times \sqrt \frac{m}{K}\\\\2.7 =2\times 3.14\sqrt\frac{32}{K}\\\\K = 173.12 N/m[/tex]
A standing wave with 5 loops is created on a string that is 0.75 m long. If the
wave has a velocity of 300 m/s, what is the frequency of the wave?
1000 Hz
400 Hz
225 Hz
90 Hz
None of the above
wave velocity = frequency × wavelength
300 = frequency × 0.75
Frequency = 300÷0.75 = 400 HZ
Parallel rays from a distant object are traveling in air and then are incident on the concave end of a glass rod with a radius of curvature of 15.0 cm. The refractive index of the glass is 1.50. What is the distance between the vertex of the glass surface and the image formed by the refraction at the concave surface of the rod? Is the image in the air or in the glass?
Answer:
the distance of image from the vertex is 45 cm and the image formed is in the glass.
Explanation:
distance of object, u = - infinity
radius of curvature, R = - 15 cm
refractive index, n = 1.5
Let the distance of image is v.
Use the formula
[tex]-\frac{n1}{u}+\frac{n2}{v}=\frac{n2- n1}{R}\\\\-\frac{1}{\infty }+\frac{1.5}{v}=\frac{1.5-1}{-15}\\\\v=45 cm[/tex]
The image is in the glass.
Describe two 'beneficial' aspects from the last 10,000 years of humanity, as well as two 'dis-advantageous' aspects. Agriculture should be your guiding point here.
Answer:
Two beneficial aspects of the last 10,000 years of mankind have been the development of writing and the development of civilized communities.
Thus, writing has allowed the recording of the different historical events that have taken place during the development of humanity, setting precedents for subsequent generations, allowing documenting the progress that humanity has been achieving. On the other hand, the development of civilized communities around urban systems has allowed for greater economic and social development on the part of human groups, which through cooperation have been acquiring an increasingly higher quality of life.
On the other hand, two negative aspects of these years have been the world overpopulation and the pollution of the planet.
Thus, the amount of the world's population has been getting closer and closer to the limit of our planet's carrying capacity, which could lead humanity to a point where resources are definitely scarce, causing famines and various catastrophes. On the other hand, the pollution arising from the development of industrial societies has notably undermined the health of individuals and has damaged their standard of living, while causing irreparable damage to the natural environment.
HELP ME PLEASEEEEEEEEEEEEEE
Answer: The correct statements are:
The atoms are very attracted to one another.The atoms are held tightly together.Explanation:
Solid state: In this state, the molecules are closely packed and cannot move freely from one place to another that means no space between them and the intermolecular force of attraction between the molecules are strong.
In solid substance, the particles are very close to each other due to this the intermolecular forces of attraction are strongest.
The key point about solid are:
The atoms are very attracted to one another.The atoms are not moving freely.It will not spread out evenly to fill any container.The atoms are held tightly together.The forces of attraction are strong to bring molecules together.The atoms are close and in fixed positions.If a fisherman applies a horizontal force with magnitude 47.0 NN to the box and produces an acceleration of magnitude 3.20 m/s2m/s2, what is the mass of the box
Answer:
The correct solution is "14.6875 kg".
Explanation:
Given values:
Force,
F = 47.0 N
Acceleration,
a = 3.20 m/s²
Now,
⇒ [tex]Force=Mass\times Acceleration[/tex]
or,
⇒ [tex]F=ma[/tex]
⇒ [tex]47.0=m\times 3.20[/tex]
⇒ [tex]m=\frac{47.0}{3.20}[/tex]
⇒ [tex]=14.6875 \ kg[/tex]
100 POINTS AND BRAINLIEST!!! How does the egg sucked into a glass bottle trick work?
Answer: The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.
Explanation:
Quickly place the egg over the mouth of the bottle. The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.
With explanation pls
Answer:
[tex](x \times 1) = (400 \times 2.5) \\ x = 1000 \: newtons[/tex]
[tex]y = 0[/tex]
What is gravitational force?
-,-
Hello naina ji
Kaise h0 aap...
Hello yarr kha h0 sab k sab
Avvu
Mdhu sis
Naina sis
jonain
.....etc
A 6 kg ball experiences a 5 m/s^2 acceleration. What is the strength of the force felt by the ball?
a: 0.83kg
b: 30 newtons
c: 30 kg
d: 1.2 newtons
Answer:
30 newtons
explanations
data given
mass=6kg
acceleration=5
f=m×a
6×5=30
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8,32 meters per second take the speed of sound as 340 meters per second calculate frequency
Complete question:
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?
Answer:
The frequency reflected by the stationary wall to the bat is 41 kHz
Explanation:
Given;
frequency emitted by the bat, f = 39 kHz
velocity of the bat, [tex]v_b[/tex] = 8.32 m/s
speed of sound in air, v = 340 m/s
Apply the doppler-effect principle to solve this problem.
The apparent frequency of sound striking the wall is calculated as;
[tex]f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz[/tex]
The frequency reflected by the stationary wall to the bat is calculated as;
[tex]f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz[/tex]
[tex]f_s\approx 41 \ kHz[/tex]
g You decide to play fetch with your dog, who is sitting nextto you, so you throw a ball down a narrow hallway. The ballcomes to a stop 3.9 m down the hallway. The dog, startingfrom rest, runs after the ball with a constant acceleration of0.70m/s2until she reaches the ball. She grabs the ball whilestill running down the hallway uniformly accelerating(slowingdown) for 4.7 more seconds until she comes to a stop. What isthe total distance the dog travels to grab the ball and come toa final stop, starting from rest
You decide to play fetch with your dog, who is sitting nextto you, so you throw a ball down a narrow hallway. The ballcomes to a stop 3.9 m down the hallway. The dog, startingfrom rest, runs after the ball with a constant acceleration of0.70m/s2until she reaches the ball. She grabs the ball whilestill running down the hallway uniformly accelerating(slowingdown) for 4.7 more seconds until she comes to a stop. What isthe total distance the dog travels to grab the ball and come toa final stop, starting from rest
Some students are making a domino cascade for a fair project and want to use a ball rollingup and then down an incline in one part of it. The timing has to be very precise so theyneed to know exactly how long it will take the ball to go up and down. They have measuredthe acceleration of the ball on the incline to be 2.5 m/s2. If the ball starts part way up theramp, rolls up for 1.1 m, then down 0.7 m past its starting point, how long will it take
Answer:
total time taken is 2.1389 seconds
Explanation:
Given the data in the question and as illustrated in the diagram below;
They have measured the acceleration of the ball on the incline to be 2.5 m/s²
a = -2.5 m/s² { negative because its acting downward }
from the diagram, Motion from B to C will be;
v² = u² + 2as
given that distance s = 1.1 m
we substitute
v² - u² = 2as
(0)² - u² = 2 × (-2.5 m/s ) × 1.1 m
-u² = -5.5
u = √5.5
so Initial speed; u = 2.345 m/s
to get the time, we use the equation of motion;
v = u + at
we substitute
0 = 2.345 m/s + ( -2.5 m/s² × t₁ )
2.5t₁ = 2.345
t₁ = 2.345 / 2.5
t₁ = 0.938 s
Now, from C to B
we determine t₂
s = vt₂ + [tex]\frac{1}{2}[/tex]at₂²
1.1 m = 0×t₂ + [tex]\frac{1}{2}[/tex] × 2.5 m/s² × t₂²
1.1 m = [tex]\frac{1}{2}[/tex] × 2.5 m/s² × t₂²
1.1 m = 1.25 m/s² × t₂²
t₂² = 1.1 m / 1.25 m/s²
t₂² = 0.88 s²
t₂ = √0.88 s²
t₂ = 0.938 s
Next, from B to A; t₃ will be;
s = ut₃ + [tex]\frac{1}{2}[/tex]at₃²
we substitute
0.7 m = (2.345 m/s)t₃ + [tex]\frac{1}{2}[/tex] × 2.5 m/s × t₃²
0.7 m = (2.345 m/s)t₃ + ( 1.25 m/s )t₃²
1.25t₃² + 2.345t₃ - 0.7 = 0
using; x = [ -b ± √( b² - 4ac ) ] / 2a
we substitute
x = = [ -2.345 ± √( (2.345)² - (4 × 1.25 × -0.7) ) ] / 2(1.25)
x = [ -2.345 ± √( 5.499 + 3.5 ) ] / 2.5
x = [ -2.345 ± √( 8.999 ) ] / 2.5
x = [ -2.345 ± 2.9998 ] / 2.5
x = ( [ -2.345 - 2.9998 ] / 2.5 ) or ( [ -2.345 + 2.9998 ] / 2.5 )
x = -2.1379 or 0.2629
hence t₃ = 0.2629 s
∴ Total time = t₁ + t₂ + t₃
Total time = 0.938 s + 0.938 s + 0.2629 s
Total time = 2.1389 seconds
Therefore, total time taken is 2.1389 seconds
Harper suffers from nearsightedness equally severely in both eyes. The focal length of either of Harper's eyes is 16.7 mm in the relaxed state. What lens strength (a.k.a., lens power) of contact lenses should be prescribed to correct the nearsightedness in Harper's eyes
The lens strength or lens power of contact lenses that m should be prescribed to correct the nearsightedness is 59.9 diopters.
What is focal length of a lens?The focal length of a lens is the distance from a lens to the focal point of that lens.
It is a measure of how strongly a lens converses or diverges light.
The power of a lens is the inverse of the focal length when given in metres.
Power of a lens = 1/focal length (m)
The focal length = 16.7 mm = 0.0167 M
Power of lens required = 1/0.0167 = 59.9 diopters
Therefore, the lens strength or lens power of contact lenses that m should be prescribed to correct the nearsightedness is 59.9 diopters.
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The brake shoes of your car are made of a material that can tolerate very high temperatures without being damaged. Why is this so
they have rubber and meatal so its going to get hot of course cause it a engine pushing and the gasoline is just all coming together but damage wise I still it the protection on that part to keep it from damage and life risk
I hope this may help
What is the first quantum number of a 252 electron in phosphorus,
1322s22p3s23p3?
A. n=0
B. n= 3
O
C. n = 1
O D. n = 2
Answer:
the correct answer is B
Explanation:
The quantum numbers are the constants obtained when solving the Schrodinger equation, the first quantum number or principal quantum number (n), can take values from zero to infinity.
This quantum number is placed as a coefficient in the quantum distribution.
In this case for phosphorus, the number is n = 3
the correct answer is B
Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.How far must the faster car travel before it has a 15-min lead on the slower car
Answer:
The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Explanation:
Given;
speed of the faster car, v₁ = 60 mi/h
speed of the slower car, v₂ = 55 mi/h
Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles
[tex]\frac{x}{55} - \frac{x}{60} = \frac{15}{60}[/tex]
Note: divide 15 mins by 60 to convert to hours for consistency in the units.
[tex]\frac{x}{55} - \frac{x}{60} = \frac{15}{60}\\\\multiple \ through \ by \ 660\\\\12x - 11x = 165\\\\x = 165 \ miles[/tex]
Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
True or false, wrrect the false
statement:
• The magnetic field created by a flat coil is
uniform.
• Inside a solenoid, the lines of field are
oriented from the north face to the south
face.
• The magnetic field outside Helmholtz
coils is uniform.
• Le champ B à l'intérieur d'un solénoïde
est uniforme.
• The magnitude of B, created by a flat coil
of radius R, at any point in its plane is
B= 2m x 10-NI
R
• The designation of the faces of a wil
depend the sense of the current
traversing it.
Answer:
false
Explanation:
The volume of an ideal gas changes from 0.40 to 0.55 m3 although its pressure remains constant at 50,000 Pa. What work is done on the system by its environment
Answer:
w= p∆v 50000 ( 0.55-0.40) and calculate and you get it
The work done on the system by its environment is 7,500J.
To find the work done, we need to know about the work done on an ideal gas during isobaric process.
What is isobaric process?When the pressure is remained constant throughout a process, then the process is called isobaric process.
What is the work done on a system in the isobaric process?For a isobaric process, the work done is pressure × (final volume of the system - initial volume)Here, Final volume= 0.55 m³
Initial volume= 0.40 m³
Pressure= 50,000 Pa
Work done = 50,000 × (0.55 - 0.40)
= 7,500 J
Thus, we can conclude that 7,500 J of work is done on the system by its environment.
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what's the difference between mass and inertia in a tabular form
Answer:
to be honest i dont know
Explanation:
^^
A ship moves 330 m after 40,000 j of work is done on it. What force is used to do this work?
Explanation:
F=w/s
F=40,000j/330m
F=121.212N
state the two motion equations.
Answer:
Equations of motion relate the displacement of an object with its velocity, acceleration and time. s=vt where s is the displacement, v the (constant) speed and t the time over which the motion occurred. ...
Displacement with negative acceleration: s, equals, v, t, minus, one half, a, t, square...
Displacement with positive acceleration: s, equals, u, t, plus, one half, a, t, squared,s...
Velocity squared: v, squared, equals, u, squared, plus, 2, a, s,v2=u2+2as
Velocity: v, equals, u, plus, a, t,v=u+at
If the light energy passes through material in straight lines they are __?___ to light.
Answer:
transparent
Explanation:
[3 marks]
A particle of mass 100 g executes simple harmonic motion about x = 0 with angular frequency ω = 10 s-1. Its total mechanical energy is Etot = 0.45 J. Find the displacement of
the particle when its speed is 2 m/s. [3 marks]
Answer:
22.4 cm
Explanation:
The total mechanical energy of the particle Etot = K + U where K = kinetic energy = 1/2mv² where m = mass of particle = 100 g =0.1 kg and v = speed of particle = 2 m/s and U = potential energy = 1/2kx² where k = spring constant and x = displacement
Etot = K + U
U = Etot - K
U = Etot - 1/2mv²
Since Etot = 0.45 J, substituting the values of the other variables into the equation, we have
U = Etot - 1/2mv²
U = 0.45 J - 1/2 × 0.1 kg × (2 m/s)²
U = 0.45 J - 1/2 × 0.1 kg × 4 m²/s²
U = 0.45 J - 0.1 kg × 2 m²/s²
U = 0.45 J - 0.2 J
U = 0.25 J
Now U = 1/2kx² and the angular frequency ω = √(k/m) ⇒ ω²m = k
So, U = 1/2ω²mx²
x² = 2U/ω²m
x = (√2U/m)/ω
since ω = 10 rad/s, substituting the values of the variables into the equation, we have
x = (√2U/m)/ω
x = (√2 × 0.25 J/0.1 kg)/10 rad/s
x = (√0.5 J/0.1 kg)/10 rad/s
x = (√5 J/kg)/10 rad/s
x = 2.24/10 m
x = 0.224 m
x = 22.4 cm
You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/sm/s. Part A What is the speed of the putty just before it strikes the ceiling
Answer: [tex]4.65\ m/s[/tex]
Explanation:
Given
Distance putty has to travel is 3.5 m
The initial speed of putty is 9.50 m/s
Using equation of motion to determine the velocity of putty just before it hits ceiling
[tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s[/tex]
So, the velocity of putty just before hitting is [tex]4.65\ m/s[/tex]
3. Sally (45 kg) and Stan (90 kg) and arrive at Michigan Adventure and climb the stairs to ride the big
Shivering Timber coaster that is nearly 40 m tall and reaches speeds of 65 mi/hr.
a) Compare the potential energies of Sally and Stan as they wait in line at the top of the stairs.
Explain your reasoning.
Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 15 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 65 cm .
Required:
What is the wavelength of the sound?
Answer:
100 cm
Explanation:
It is given that there are two loudspeakers that produces [tex]$\text{sound waves }$[/tex] along x-axis.
The maximum intensity of the sound is [tex]$\text{when the speakers are}$[/tex] at a distance of = 15 cm apart.
The sound intensity becomes zero when the separation between the speakers are increased and becomes 65 cm.
Therefore, the sound waves are in the phase, [tex]$\Delta x_1=15 \ cm$[/tex]
The sound waves are out of phase when [tex]$\Delta x_2=65 \ cm$[/tex]
Therefore,
[tex]$\Delta x_2 - \Delta x_1 = \frac{\lambda}{2}$[/tex]
[tex]$\lambda= 2(\Delta x_2 - \Delta x_1)$[/tex]
= 2 (65 - 15)
= 2 x 50
= 100 cm
Hence the wavelength of the sound is 100 cm.
I need help with this assignment
Give 2 reasons for fitting heavy commercial vehicles with many tyres
As we know larger the area of contact lesser the pressure. So, in order to reduce the pressure heavy vehicles have broad tyres to increase the area of contact with the ground. Heavy vehicles have broad tyres because broad tyres have large area of contact and less pressure on the ground.
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