Answer:
7.43m/s
Explanation:
Given parameters:
Initial velocity = 1.7m/s
Height = 2.5m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we apply one of the motion equation. The most fitting one given the parameters is;
V² = U² + 2gh
where V is the final velocity
U is the initial velocity
g is the acceleration due to gravity = 9.8m/s
h is the height
Input the parameters and solve;
V² = 2.5² + 2(9.8)(2.5)
V = 7.43m/s
Describe and give an example of a normal force.
Answer:
Normal force is the force exerted on an object that is in contact with another non-moving object. Force is push and pull
Ex: When a book rests on a surface (like a table), the table is exerting a force on the book in order to support the weight of it.
Answer:
Normal Force is a contact force. When 2 objects are touching each other, they are in contact. Like when a cup is sitting on a table they are in contact cause the table has to hold the force of the cup up on the table.
A example of normal force is when a book is sitting on a table, and the table is holding the force of the book up.
Explanation:
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elastic limit is equal to:(a) stress (b)strain (c) young's modulus (d)modulus of rigidity
Answer:
A. stress
Explanation:
Hope this helped. Have an amazing day/night!!
We have that for the Question "elastic limit is equal to" it can be said that
Young's modulusThis speaks to a body retaining it shape after changes occur
Option C
From the question we are told
elastic limit is equal to
(a) stress
(b)strain
(c) young's modulus
(d)modulus of rigidity
Where
Where
Stress is the force acting on a material
and
strain
The effect of force acting on a material
Young's modulus
This speaks to a body retaining it shape after changes occur
Modulus of rigidity
m=sheer stress/shear strain
Now having that
Elastic limit is a point where an elastic body does not return to its original shape
The option chosen is'
Young's modulus
This speaks to a body retaining it shape after changes occur
Option C
Generally the equation for the is mathematically given as
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A spherical balloon is inflated with gas at a rate of 800 cubic centimeters per minute. (a) How fast is the radius of the balloon changing at the instant the radius is 70 centimeters
Answer:
0.013 cm/minute
Explanation:
We are given;
Volume rate; dV/dt = 800 cm³/min
Now, volume of a sphere is given by the formula;
V = (4/3)πr³
We want to find the rate at which the radius of the balloon is changing. This is represented by dr/dt.
Now, dr/dt will be gotten from the relation;
dr/dt = dV/dt ÷ dV/dr
Now, dV/dr = 3 × (4/3)πr²
dV/dr = 4πr²
Thus;
dr/dt = 800/(4πr²)
At r = 70 cm
dr/dt = 800/(4π × 70²)
dr/dt = 0.013 cm/minute
40 m north west
50m Northwest
120 m southeast
40 m south east
Which answer
Answer:
120 north west
Explanation:
did he wash the shoe.change it to passive voice
Answer:
Did he wash the shoes?-Were the shoes washed by him?
What has the least potential energy in this world
Answer:
Anything that has the most kinetic has the least potential energy.
Explanation:
NEED BRAINLYEST PLZZ!!!!!!!!!!!!!!!
Two equal masses travel in opposite directions with equal speeds. They collide in a collision that is between elastic and inelastic. Just after the collision, their velocities are
Answer:
less in magnitude and opposite in direction to their original velocities
Explanation:
The nature of collisions can be determined by the coefficient of restitution.
Coefficient of restitution is given by
[tex]e=\dfrac{\text{Kinetic energy after collision}}{\text{Kinetic energy before collision}}[/tex]
In the case described here the value of [tex]e[/tex] is in between 0 and 1.
This means that the kinetic energy after collision is less than the kinetic energy before collision.
If the kinetic energy after collision is less this means that the velocity is not balanced with the velocity before the collision however the direction of the masses are opposite Some amount of energy is dissipated as heat or through other forms of energy.
So, just after the collision, their velocities are less in magnitude and opposite in direction to their original velocities.
A 227−lb fullback runs the 40−yd dash at a speed of 17.25 pm 0.10 mi/h. Enter your answers in scientific notation. (a) What is his de Broglie wavelength (in meters)? × 10 (select) m (b) What is the uncertainty in his position based on the de Broglie wavelength?
Answer:
His de Broglie wavelength is 8.35×10⁻³⁷ m
The uncertainty in his position is 1.15 × 10⁻³⁵ m
Explanation:
First, Convert 227-lb to kg and convert the unit of the speed from mi/h to m/s.
To convert 227-lb to kg,
1-lb = 0.453592 kg
∴ 227-lb = 227 × 0.453592 kg
227-lb = 102.97 kg
To convert 17.25 ± 0.10 mi/h to m/s
1 mi = 1609.34 m
and 1 h = 3600 s
Therefore,
17.25 mi/h = (17.25 ×1609.34)/3600 m/s = 7.71 m/s
and 0.10 mi/h = (0.10 ×1609.34)/3600 m/s = 0.044704 m/s
Hence, the speed 17.25 ± 0.10 mi/h = 7.71 ± 0.044704 m/s
Now
(a) To determine the de Broglie wavelength,
De Broglie wavelength is given by
λ = h/mv
Where λ is the de Broglie wavelength
h is Planck's constant (h = 6.626×10⁻³⁴ kgm²/s)
m is the mass
and v is the speed (velocity)
From the question
m = 102.97 kg
v = 7.71 m/s
Therefore,
λ = 6.626×10⁻³⁴ / (102.97×7.71)
λ = 8.35×10⁻³⁷ m
Hence, his de Broglie wavelength is 8.35×10⁻³⁷ m
(b) To calculate the uncertainty in his position
From
Δx = h/(4πmΔv)
Where Δx is the uncertainty in the position
h is Planck's constant (h = 6.626×10⁻³⁴ kgm²/s)
π is a constant ( π = 3.14)
m is the mass
Δv is the uncertainty in speed
From the question
m = 102.97 kg
Δv = 0.044704 m/s
Hence,
Δx = 6.626×10⁻³⁴ / (4×3.14×102.97×0.044704)
Δx = 1.15 × 10⁻³⁵ m
Hence, the uncertainty in his position is 1.15 × 10⁻³⁵ m.
An 80 N box initially at rest is pulled by a horizontal rope on a horizontal table. The coefficients of kinetic and static friction between the box and the table are 0.25 and 0.5 respectively. What is the friction force on this box if the pull is 25N?
A. 20 N
B. 40 N
C. 25 N
D. 0 N
Answer:
C. 25 N
Explanation:
w = Weight of box = 80 N
[tex]\mu_k[/tex] = Coefficient of kinetic friction = 0.25
[tex]\mu_s[/tex] = Coefficient of static friction = 0.5
F = Pulling force = 25 N
Static friction
[tex]f_s=\mu_s w\\\Rightarrow f_s=0.5\times 80=40\ \text{N}[/tex]
[tex]F<f_s[/tex]
The force of pulling is less than the static frictional force so this makes the friction force equal to the force of pulling.
So, the friction force on the box is 25 N.
Jennifer hits a stationary 0.04-kg ball, and it leaves her racket at 41.3 m/s. Time-lapse photography shows that the ball was in contact with the racket for 26.5 ms. What average force, in N, did the ball exert on the racket?
Answer:
62.33NExplanation:
We know that the relationship for the impulse and momentum is given as
Ft=mv
given data
mass m=0.04kg
t= 26.5ms= 26.5/1000= 0.0265seconds
velocity v= 41.3m/s
substituting in the expression we have
Ft=mv
make F subject of formula we have
F=mv/t
F=(0.04*41.3)/0.0265
F=1.652/0.0265
F=62.33N
Which statement is true about an object that is traveling in uniform circular motion?
Part A What electron current is driven through a 2.3-mm-diameter iron wire by a 0.062 V/m electric field?
We know, Current is given by :
[tex]I = \dfrac{EA}{\rho}[/tex]
Here, [tex]\rho[/tex] is resistivity and for iron.
[tex]\rho=9.71\times 10^{-8}\ \Omega\ m[/tex]
Putting [tex]\rho[/tex] in above equation, we get :
[tex]I = \dfrac{0.062\times \pi\times (2.3\times 10^{-3})^2}{4\times 9.71\times 10^{-8}}[/tex]
[tex]I=2.65\ A[/tex]
Therefore, the electron current is 2.65 A.
Hence, this is the required solution.
A ball is thrown up in the air with an initial velocity of 40m/s what is the balls velocity after 2 seconds?
Answer:
20 m/s
Explanation:
velocity after 2 sec.
v=40+(−10)t
=40−10∗2=20m/s
We have that for the Question "A ball is thrown up in the air with an initial velocity of 40m/s what is the balls velocity after 2 seconds?" it can be said that the balls velocity after 2 seconds
v=67.44m/s
From the question we are told
A ball is thrown up in the air with an initial velocity of 40m/s what is the balls velocity after 2 seconds?
Generally the Newtons equation for the Motion is mathematically given as
[tex]v=u+at\\\\Therefore\\\\v=40+9.8*2.8\\\\v=67.44m/s\\\\[/tex]
Therefore
the balls velocity after 2 seconds
v=67.44m/s
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i. What is the gravitational potential energy added to the steel beam when the beam reaches a height of 18.4 m? (1 point)
Answer:
59,566.32JoulesExplanation:
The question is incomplete. Here is the complete question.
At a construction site, a crane lifts a steel beam with a mass of 330 kg.
i. What is the gravitational potential energy added to the steel beam when the beam reaches a height of 18.4 m? (1 point)
Gravitational potential energy GPE = mgh
m is the mass of the steel beam = 330kg
g is the acceleration due to gravity = 9.81m/s²
h is the height the beam reaches = 18.4m
Substitute
GPE = 330 * 9.81 * 18.4
GPE = 3237.3 * 18.4
GPE = 59,566.32Joules
Hence the gravitational potential energy is 59,566.32Joules
Answer:
The increase in gravitational potential energy is approximately 60,000 J.
Explanation:
GPE=mgh
m=330
g=9.8^2
h=18.4
GPE=330*9.81*18.4
GPE=59,566.32 J
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Object A has a mass of 12 g and a density of 3 g/cm3. Object B has a mass of 12 g and a density of 5 g/cm3. Which object has a greater volume and by how much? (4 points)
Object A, by 2 6 cubic centimeters
Object A, by 1.6 cubic centimeters
Object B, by 2 6 cubic centimeters
Object B, by 1.6 cubic centimeters
Answer:
object B has a greater volume and by2 g/cm3
A car driver traveling on a linear segment of a level road at 60 miles per hour (mph) slows down to 40 miles per hour in a road construction zone. If the car traveled a distance of 360 ft from the original speed to the final speed then what is the car deceleration rate in ft per sec2
Answer:
a=5.98 ft/secExplanation:
given
initial velocity u=60mph to ft per sec= 88ft/s
final velocity v=40 mph to ft per sec= 58.6 ft/s
distance s=360ft
convert feet to mile 360ft is 0.06miles
we can apply the expression
v^2=u^2-2as
58.6^2=88^2-2*a*360
3433.96=7744-720a
3433.96-7744=-720a
-4310.04=-720a
a=4310.04/720
a=5.98 ft/sec
Consider the situation a book a sliding to a stop while moving across the classroom floor. Of the forces listed identify which act upon the book
The forces which act upon the book are Normal, Gravity, Air resistance, and Friction. Therefore, the (A), (B), (D), and (E) options are correct.
What are Normal force, Gravity, Air resistance, and Friction?The normal force is a contact force that surfaces exert to prevent objects from passing through each other. If two surfaces are not in direct contact, they are able to exert a normal force on each other. The book and the floor are exerting normal on each other.
Gravity is the force by which the earth draws objects toward its center. The force of gravity keeps everything on the surface of the earth. Therefore, the sliding book on the floor experiences the force of gravity.
Air resistance is a force that arises due to air. The force acts in the opposite direction to the motion of an object through the air. As the book slides across the floor, it experiences resistance due to air.
Friction is the resistance caused by the surfaces that are in contact when they slide over each other. Friction arises due to the irregularities on the two surfaces. The book and floor both have irregularities on their surfaces which is caused friction between them.
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Your question is incomplete, but most probably the complete question was,
Consider the situation a book a sliding to a stop while moving across the classroom floor. Of the forces listed identify which act upon the book:
A)Normal,
B) Gravity
C)Applied
D) Air resistance,
E) Friction
F) Tension
Two masses, m_1m ​1 ​​ = 3.75 kg and m_2m ​2 ​​ = 6.34 kg, are connect by a string of negligible mass. The string passes over a frictionless pulley so that m_1m ​1 ​​ and m_2m ​2 ​​ hang down on opposite sides of the pulley. The whole system is released from rest. Calculate the tension in the string as the masses are moving.
Answer:
Explanation:
For the mass m1;
The sum of forces acting on the body is expressed according to Newton's second law as;
\sum Fx = m1ax
T - Ff = m1a .... 1
T is the tension
Ff is the frictional force acting on m1
For the mass m2:
\sumFy = m2a
W - T = m2a
m2g - T = m2a.... 2
W is the weight
g is the acceleration due to gravity
If the acceleration of the system is 0, the equation becomes;
From 2:
m2g - T = m2a.... 2
a = m2g-T/m2
From 1:
a = T-Ff/m1
Equate both accelerations
m2g-T/m2 = T-Ff/m1
Cross multiply
m1m2g - Tm1 = m2T-m2Ff
m1m2g+m2Ff = m2T+m1T
m1m2g+m2Ff = T(m2+m1)
T = m1m2g+m2Ff/m2+m1
Hence the tension in strings is expressed as
T = m1m2g+m2Ff/m2+m1
How does the temperature affect the amount of weathering
The fastest recorded pitch in major league baseball, thrown by Aroldis Chapman in 2010, was clocked at 169.1km/hr. If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically (in meters) by the time it reached the home plate, 18.3m away?
Answer:
The ball would fall vertically for 112.45m
Explanation:
The vertical height of a horizontal throw is affected only by the speed of the throw and the acceleration due to gravity.
At this point, we may use this formula to determine the vertical height of the throw
vertical height = [tex]-v^2/2g[/tex]
The initial throwing speed has to be converted to m/s to ensure uniformity during the calculations. To do this we multiply by 1000 and divide by 3600
169.1km/hr = 46.972m/s
Maximum height = [tex]46.972^2 /2 \times 9.81=112.45[/tex] m
The pitch would make the ball fall vertically for 112.45m by the time it reached the home plate 18.2 m away
what is hubbles law? please put in simple terms
Answer:
the law that the velocity of recession of distant galaxies from our own is proportional to their distance from us.
Explanation:
Hubble's law, also known as the Hubble–Lemaître law, is the observation in physical cosmology that galaxies are moving away from the Earth at speeds proportional to their distance. In other words, the farther they are the faster they are moving away from Earth.
A hungry rabbit sprints... what is the average velocity of the rabbit between the times t = 2 s and t = 12 s
Answer: -0.10m/s
Explanation:
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A soccer player kicks a 0.5 kg ball and accelerates it at 50 m/s2. What force does the player apply to the ball?
Answer:
The answer is 25 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × accelerationFrom the question we have
force = 0.5 × 50
We have the final answer as
25 NHope this helps you
Large igneous provinces are formed by ____________________, which can flow tens to hundreds of kilometers.
Answer:
Large igneous provinces are formed by flood basalt, which can flow tens to hundreds of kilometers.
Explanation:
What is the difference between fast-twitch and slow-twitch muscle fibers?
A. They are almost identical.
B. Fast-twitch fibers can deliver a quick burst of power. Slow-twitch
fibers can maintain a contraction for a longer time.
C. Fast-twitch fibers are what keep us going on a long bike ride or
hike. Slow-twitch fibers are what we use while dashing for the bus
or jumping up for a rebound.
D. Fast-twitch fibers take a long time to warm up, but are stronger
than slow-twitch fibers once they get going.
Answer:
B. fast-twitch fibers can deliver a quick burst of power. Slow-twitch fibers can maintain a contraction for a longer time.
Answer:
the person above is right, just came for answers lol :')
Explanation:
When a velocity vs time graph is slowed downwards, is the object still moving forward? Explain why or why not.
Answer:
Since velocity is considered a vector quantity and vectors show you in what direction it goes the object wont be moving forward but backwards because its a negative and negatives go to the left and that would be considered backwards.
Explanation:
Hope this helps
In the amusement park ride known as Magic Mountain Superman, powerful magnets accelerate a car and its riders from rest to 45 m/s (around 100 mi/h) in a time of 6.8 s. The mass of the car and riders is 5100 kg. Find the average net force exerted on the car and riders by the magnets.
Answer:
The average net force exerted on the car and riders by the magnets is 33751.8 newtons.
Explanation:
Let assume that car and its riders accelerate at constant rate, such that acceleration ([tex]a[/tex]), measured in meters per square second, can be found by using the following kinematic equation:
[tex]a = \frac{v_{f}-v_{o}}{t}[/tex] (Eq. 1)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the car and its riders, measured in meters per second.
[tex]t[/tex] - Acceleration time, measured in seconds.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v_{f} = 45\,\frac{m}{s}[/tex] and [tex]t = 6.8\,s[/tex], the average acceleration of the car is:
[tex]a = \frac{45\,\frac{m}{s}-0\,\frac{m}{s}}{6.8\,s}[/tex]
[tex]a = 6.618\,\frac{m}{s^{2}}[/tex]
By the Second Newton's Law, we find that average force exerted on the car and riders by the magnets ([tex]F[/tex]), measured in newtons, is:
[tex]F = m\cdot a[/tex] (Eq. 2)
Where [tex]m[/tex] is the mass of the car-riders system, measured in kilograms.
If we know that [tex]m = 5100\,kg[/tex] and [tex]a = 6.618\,\frac{m}{s^{2}}[/tex], the net force exerted on the car and riders is:
[tex]F = (5100\,kg)\cdot \left(6.618\,\frac{m}{s^{2}} \right)[/tex]
[tex]F = 33751.8\,N[/tex]
The average net force exerted on the car and riders by the magnets is 33751.8 newtons.
A playground merry-go-round has a radius R and a rotational inertia I. When the merry-go-round is at rest, a child with mass m runs with speed v along a line tangent to the rim and jumps on. The angular velocity of the merry-go-round is then:
Answer:
The angular velocity of the merry-go-round is [tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex].
Explanation:
After a careful reading of the statement, we notice that merry-go-round-child system is a system with no external force exerting on it, such that the Principle of Angular Momentum Conservation can applied to analyze the system:
[tex]m_{C}\cdot \left(\frac{v}{R} \right) = (I+m_{C}\cdot R^{2})\cdot \omega[/tex] (Eq. 1)
Where:
[tex]m_{C}[/tex] - Mass of the child, measured in kilograms.
[tex]v[/tex] - Initial speed of the child, measured in meters per second.
[tex]R[/tex] - Radius of the playground merry-go-round, measured in meters.
[tex]I[/tex] - Moment of inertia, measured in kilograms per square meter.
[tex]\omega[/tex] - Angular velocity of the merry-go-round-child system, measured in radians per second.
Then, we clear the final angular speed:
[tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex]
The angular velocity of the merry-go-round is [tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex].
A car traveling 14 m/s accelerates at a rate of 0.95 m/s2 for an interval of 8 s. What is the final
velocity of the car? Remember: Diagram, Knowns, Equation, Rearrange, Solve.
Suppose a small compact car with a mass of 1000 kg is traveling north on Morewood Avenue at a speed of 15 m/s. At the intersection of Morewood and Fifth Avenues, it collides with a truck with a mass of 2000 kg that is traveling east on Fifth Avenue at 10 m/s. Treating each vehicle as a particle, find the total momentum (magnitude and direction) just before the collision.
Explanation:
We are expected to solve for the total momentum before collision
the expression for momentum is given as
p = mv
p =momentum
m=mass
v=velocity
For small compact truck
m1=1000kg
v1=15m/s
For small truck
m1=2000kg
v1=10m/s
the total momentum can also be expressed as
Ptotal=m1v1+m2v2
Ptotal=1000*15+2000*10
Ptotal=15000+20000
Ptotal=35000kg•m/s
The direction is east because the momentum of the truck is more than the car
