Answer:
1) v = 0.45 m/s
2) v = 0.65 m/s
3) v = 0.75 m/s
Explanation:
1) We can find the speed of the object by conservation of energy:
[tex] E_{i} = E_{f} [/tex]
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
Where:
k: is the spring constant = 280 N/m
v: is the speed of the object =?
m: is the mass of the object = 5.00 kg
x: is the displacement of the spring
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s [/tex]
2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s [/tex]
3) When the object is at the equilibrium position, the speed of the object is:
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s [/tex]
I hope it helps you!
(1) the speed of the object when compression of the spring is 8 cm is 0.449 m/s
(2) the speed of the object when compression of the spring is 5 cm is 0.648 m/s
(3) the speed of the object when the spring is at equilibrium is 0.748 m/s
Compression of spring and conservation of energy:
Given that the mass of the object, m = 5 kg
spring constant, k = 280 N/m
compression of the spring , x = 10 cm = 0.1m
(i) the spring compression is at d = 8cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.08)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.2016[/tex]
v = 0.449 m/s
(ii) the spring compression is at d = 5cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.05)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.42[/tex]
v = 0.648 m/s
(iii) the spring is at equilibrium so compression is at d = 0cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.56[/tex]
v = 0.748 m/s
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a pot of water at 20.0ºC is warmed to 100º when 40000cal is added. what is the mass of the water?
Answer:
500 g
Explanation:
m = Mass of water
c = Specific heat of water = [tex]1\ \text{cal/g}^{\circ}\text{C}[/tex]
[tex]\Delta T[/tex] = Temperature difference = [tex]100-20=80^{\circ}\text{C}[/tex]
Q = Heat added = 40000 cal
Heat is given by
[tex]Q=mc\Delta T\\\Rightarrow m=\dfrac{Q}{c\Delta T}\\\Rightarrow m=\dfrac{40000}{1\times 80}\\\Rightarrow m=500\ \text{g}[/tex]
The mass of water is 500 g.
A tall, open container is full of glycerine. At what depth h below the surface of the glycerine is the pressure 2370 Pa greater than atmospheric pressure? The density of glycerine is 1.26X10^3 kg/m^3
Answer:
So, at the depth of 24 cm below the surface of the glycerine the pressure is 2970 Pa. Hence, this is the required solution.
Explanation:
Given that,
Pressure exerted by the surface of glycerine, P = 2970 Pa and it is greater than atmospheric pressure.
The density of glycerine,
We need to find the depth h below the surface of the glycerine. The pressure due to some depth is given by :
h = 0.24 meters
or
h = 24 cm
The flywheel of an engine has moment of inertia 1.9 kg m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 537 rev/min in 6.78 s, starting from rest, in N-m
Answer:
τ = 15.76 N-m
Explanation:
For rigid bodies, the net external torque acting on a rotating body, is equal to the product of the moment of inertia about its rotation axis, times the angular acceleration of the body, as follows:[tex]\tau = I * \alpha (1)[/tex]
Since I is a given of the question, we need to find out the angular acceleration.By definition, the angular acceleration is the rate of change of the angular velocity with respect to time:α = Δω/Δt = (ωf-ω₀) / (tfi-t₀)Since ω₀ = 0, and choosing t₀ =0, as tfi is a given, we need just to find out ωf.We have this value in rev/min, so we need first to convert it to rad/sec, as follows:[tex]537 rev/min * (1min/60 sec)*(2*\pi *rad/rev) = 17.9 * \pi rad/sec (2)[/tex]
Replacing in the definition of α, we have:α = 17.9*π rad/sec / 6.78 s = 8.29 rad/sec²Replacing I and α in (1), we finally get:τ = 15.76 N-m.
A ball of mass 0.600 kg is carefully balanced on a shelf that is 2.20 m above the ground. What is its gravitational potential energy?
Answer:
Explanation:
Gravitational potential energy = mass x height of object x g
g is gravitational acceleration .
Gravitational potential energy of the ball = .600 x 2.20 x 9.8
= 12.936 J .
12.9 J .
What is the force used when a 7 kg bowling ball is rolled with an acceleration of 3 m /s2? Use the formula to find the force in Newtons (N): F = 7 kg × 3 m/s2
Answer:
21 Newtons
Explanation:
Force= mass * acceleration
Force= 7 kg * 3 m/s^2
Force= 21 Newtons
What is the wavelength of light falling on double slits separated by 2.00 μm if the third-order maximum is at an angle of 60.0∘?
Answer:
λ = 5.773 x 10⁻⁷ m = 577.3 nm
Explanation:
In order to solve this problem we will use the grating equation:
mλ = d Sin θ
where,
m = order = 3
λ = wavelength of light = ?
d = slit separation = 2 μm = 2 x 10⁻⁶ m
θ = angle = 60°
Therefore,
(3)λ = (2 x 10⁻⁶ m)Sin 60°
λ = 1.732 x 10⁻⁶ m/3
λ = 5.773 x 10⁻⁷ m = 577.3 nm
HELP ASAP!!!
Which graph shows the change in velocity of an object in free fall?
Answer:
the graph that show Change in velocity is no A
The graph that showing the velocity with respect to time for a free falling body is figure C where, the downward velocity will be higher due to the acceleration due to gravity.
What is velocity?Velocity of a moving body is the measure of the distance covered per unit time. Thus, it is the ratio of distance to the time. Velocity is expressed in the units of Km/h, m/s. miles/h, ft./s etc.
The rate of change of velocity is called acceleration. The acceleration by the force of a gravitational field is called acceleration due to gravity g having the value 9.8 m/s².
The velocity - time graph shows a diagonal relation for a free falling body. A free falling body is initially at rest and falls downward with the negative velocity with the acceleration due to gravity. Therefore, figure B shows the change in velocity of a free falling body.
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Which of the following requires the expenditure of more work?
a. Lifting a 110 newton [N] weight a height of 3 meters [m].
b. Exerting a force of 60 pounds-force [lbf] on a sofa to slide it 30 feet [ft] across a room.
Answer:
The correct answer is option B
Explanation:
Step one:
given data
a. force F= 110N
distance s= 3meters
we know that work= Force* distance
work= 110*3
Work= 330Joules
Step two:
data
Force= 60 pounds
distance= 30 ft
convert pounds to Newton
1 pound= 4.44822N
60 pounds= 60*4.44822
=266.9N
convert ft to meteres
1 ft = 0.3048meter
30ft= 0.3048*30
=9.144N
we know that work= Force* distance
work= 266.9N*9.144N
Work= 2440.53Joules
A block with a mass M = 4.85 kg is resting on a slide that has a curved surface. There is no friction. The speed of the block after it has slid along the slide sufficiently far for its vertical drop to be 19.6 m is:__________a. 19.6 m/s b. 384 m/s c. 93 m/s d. 43.2 m/s e. The problem cannot be solved because the shape of the curved slide is not given.
Answer:
The correct option is a
Explanation:
From the question we are told that
The mass of the block is [tex]m = 4.84 \ kg[/tex]
The height of the vertical drop is [tex]h = 19.6 \ m[/tex]
Generally from the law of energy conservation , the potential energy at the top of the slide is equal to the kinetic energy at the point after sliding this can be mathematically represented as
[tex]PE = KE[/tex]
i.e [tex]m * g * h = \frac{1}{2} * m * v^2[/tex]
=> [tex]gh = 0.5 v^2[/tex]
=> [tex]v = \sqrt{\frac{9.8 * 19.6}{0.5 } }[/tex]
=> [tex]v = 19.6 \ m/s[/tex]
A 45 kg object has a momentum of 225 kg-m/s northward. What is the object's velocity?
A. 180 m/s
B. 5.0 m/s
C. 10,125 m/s
D. 0.20 m/s
pls help quick. the number line shows the starting and ending velocities for ball 1 what's the change in velocity of ball 1 calculate the value mathematically and check using the number line
Answer:
The starting velocity for ball 1 is 1.00 meter/second. Its ending velocity is 0.25 meter/second.
The change in velocity for ball 1 is 0.25 – 1.00 = -0.75 meter/seconds
a car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achieve this speed?
The time taken by the car to achieve the final speed is 6.25 seconds.
What is the equation of motion?The equations of motion can be defined as the equation that represents the relationship between the time, velocity, acceleration, and displacement of a moving object.
The mathematical expressions for the equations of motions can be written as:
[tex]v= u+at\\S=ut+(1/2)at^2\\v^2-u^2=2aS[/tex]
Given, the initial speed of the car, u = 15 m/s
The final speed of the given car, v = 25m/s
The distance covered by car, S = 125 m
From the third equation of motion: v² = u²+ 2aS
(25)² = (15)² + 2×a× 125
a = 1.6 m/s²
From the first equation of motion we can find the time to achieve the final speed:
v = u+ at
25 = 15 + (1.6) × t
t = 6.25 sec
Therefore, 6.25 seconds will be taken by the car to catch the final speed.
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The particle accelerator at CERN can accelerate an electron through a potential
difference of 80 kilovolts. Calculate
(a) The kinetic energy (in keV) of the electron
Answer:
K.E = 1.28 × 10^-17 KeV
Explanation:
Given that a particle accelerator at CERN can accelerate an electron through a potentialdifference of 80 kilovolts.
To Calculate the kinetic energy (in keV) of the electron, let us first find the electron charge which is 1.60 × 10^-19C
The kinetic energy = work done
K.E = e × kV
Substitute e and the voltage into the formula
K.E = 1.60 × 10^-19 × 80
K.E = 1.28 × 10^-17 KeV
Therefore, the kinetic energy is approximately equal to 1.28 × 10^-17 KeV
which state can you infer that average home prices are the highest? Florida California Texas New York
Answer:
California
Explanation:
just took the test
The state you can infer where average home prices are the highest is California. Option B. This is further explained below.
What is an average home price?Generally, an average home price refers to a cumulative average to the house or home pricing in a state.
In conclusion, California has the highest average home prices highest in America.
Read more about House
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what are the formulae of momentun and their time of use
Answer:
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You are lifting a 10 kg block straight up at a constant speed of 10 m/s. How much force are you exerting on the block?
Answer:
The force exerted is [tex]F = 100 \ N[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m_b = 10 \ kg[/tex]
The speed is [tex]v = 10 \ m/s[/tex]
Generally the force exerted to lift the object at constant speed is equivalent to the wight of the ball, this is mathematically represented as
[tex]F = m * g[/tex] Here [tex]g = 10 \ m/s^2[/tex]
=> [tex]F = 10 * 10[/tex]
=> [tex]F = 100 \ N[/tex]
The force are you exerting on the block when the block is lifting straight up with constant speed is 98 N and this can be determined by using the given data.
Given :
You are lifting a 10 kg block straight up at a constant speed of 10 m/s.
The following steps can be used in order to determine the force are you exerting on the block:
Step 1 - According to the given data, the block is lifting straight up at a constant speed. So, the acceleration is zero.
Step 2 - So, the only force exerted on the block is the weight of the block.
Step 3 - So, the force are you exerting on the block is given by:
F = mg
F = 10 [tex]\times[/tex] 9.8
F = 98 N
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Which statement best describes energy and matter in a closed system? (2 po
O Energy and matter flow into and out of the system.
Energy can flow into and out of the system but matter cannot.
Energy and matter are contained within a closed system.
O There is no energy in a closed system; there is matter.
Answer:
Energy can flow into and out of the system but matter cannot.
Explanation:
In a closed system, energy can flow in and out of the system but matter cannot.
A closed system prevents double way flow of matter. A closed system conserves matter.For an isolated system, energy and matter cannot flow out of the system.
For open systems, energy and matter can flow out of the system.
Such systems are used for certain thermodynamics experiment.
In the Bohr model of the hydrogen atom, an electron in the 1st excited state moves at a speed of 2.19 106 m/s in a circular path having a radius of 5.29 10-11 m. What is the effective current associated with this orbiting electron?
Answer:
I = 1.05x10⁻³ A
Explanation:
By definition, an electric current is the rate of charge flow at a given time:
[tex] I = \frac{q}{t} [/tex]
Where:
q: is the electrons charge = 1.602x10⁻¹⁹ C
t: is the time
In a circular motion, the time is given by:
[tex] t = T = \frac{2\pi}{\omega} = \frac{2\pi}{v/r} = \frac{2\pi r}{v} [/tex]
Where:
ω: is the angular speed = v/r
v: is the speed = 2.19x10⁶ m/s
r: is the radius = 5.29x10⁻¹¹ m
[tex] t = \frac{2\pi r}{v} = \frac{2\pi 5.29 \cdot 10^{-11} m}{2.19 \cdot 10^{6} m/s} = 1.52 \cdot 10^{-16} s [/tex]
Now, the effective current is:
[tex] I = \frac{q}{t} = \frac{1.602 \cdot 10^{-19} C}{ 1.52 \cdot 10^{-16} s} = 1.05 \cdot 10^{-3} A [/tex]
Therefore, the effective current associated with this orbiting electron is 1.05x10⁻³ A.
I hope it helps you!
Mr Johnson launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, how much time does it take before the arrow hits the ground below (step 1 of a quesiton will need this answer for a future question)
a 2 seconds
b. 1 second
c.4 seconds
d 19.6
Answer:c
Explanation:
1. When asteroids collided some of the broken materials fall into Earth's orbit. What do
astronomers call the debris when it hits planet Earth?
(2 Points)
meteors
meteoroids
meteorites
metabots
Before they meet Earth -- meteoroids
While they're falling -- meteors
After they hit the ground -- meteorites
Which interaction contributes to the greenhouse effect?
A (Gases in the atmosphere absorb heat.
B (Dust particles scatter and reflect light from the Sun.
C (Green visible light is trapped in Earth’s atmosphere.
D (Light travels through a thick part of the atmosphere at sunrise and sunset.
Answer:
A
Explanation:
Answer:
A
Explanation:
Gases in the atmosphere absorb heat.
You know when you have a blanket around you and some heat gets trapped in but some still gets out. Thats basically what it is.
Plus I got it right in multiple questions including the test!
I hope that reassured you!
Have a good night!
Select all correct answers....Covalent compounds
The chemical equation below shows that hydrogen and oxygen chemically react to form water.
Which of these is true?
A.
Water is a pure substance and has the same properties as oxygen.
B.
Water is a pure substance and has the same properties as hydrogen.
C.
Hydrogen and oxygen are pure substances that have identical properties to each other.
D.
Hydrogen, oxygen, and water are pure substances that each have different properties.
Answer:
D
Explanation:
Hydrogen, oxygen, and water are pure substances that each have different properties.
Although the 3 of them are pure substances, but they nonetheless have different properties. These properties could either be their physical properties, or their chemical properties. This is particularly so because since Oxygen and Hydrogen the major components in water have different properties, it is only right if their product, water, has different properties also.
Answer:
D
Explanation:
got it right on study island
If a 46 kg person walks up two floors in the Physics building (about 10 meters up), this person's potential energy has increased by _____ J.a. 1400 J b. 10,000 J. c. 1000 J. d. 100 J
Answer:
B. 10000 J
Explanation:
The possible answers are not related to the statement at all. The correct statement is:
If a 100 kg person walks up two floors in the Physics building (about 10 meters up), this person's potential energy has increased by:
From definitions of work and gravitational potential, we get the following formula to calculate the change experimented in the person's potential energy after walking up two floors in the Physics building:
[tex]\Delta U_{g} = m\cdot g \cdot \Delta z[/tex] (1)
Where:
[tex]\Delta U_{g}[/tex] - Change in the gravitational potential energy, measured in joules.
[tex]m[/tex] - Mass of the person, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]\Delta z[/tex] - Change in height, measured in meters.
If we know that [tex]m = 46\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta z = 10\,m[/tex], then the change in the gravitational potential energy is:
[tex]\Delta U_{g} = (100\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (10\,m)[/tex]
[tex]\Delta U_{g} = 9807\,J[/tex]
The choice that best approximates this answer is B.
Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground. On his first try, Robin looses the arrow at v0=35.0 m/sv0=35.0 m/s at an angle of θ=30.0°θ=30.0° above the horizontal. The arrow has an initial height of y0=1.50 m,y0=1.50 m, and its tip is x=60.0 mx=60.0 m away from the target orange. Treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has reached the horizontal position xx of the orange? Use g=9.81 m/s2g=9.81 m/s2 for the acceleration due to gravity.
Answer:
h' = 55.3 m
Explanation:
First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:
s = vt
where,
s = horizontal distance between arrow and orange = 60 m
v = initial horizontal speed of the arrow = v₀ Cos θ
θ = launch angle = 30°
v₀ = launch speed = 35 m/s
Therefore,
60 m = (35 m/s)Cos 30° t
t = 60 m/30.31 m/s
t = 1.98 s
Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:
h = Vi t + (1/2)gt²
where,
Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°
Vi = 17.5 m/s
Therefore,
h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²
h = 34.6 m + 19.2 m
h = 53.8 m
since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:
h' = h + y
h' = 53.8 m + 1.5 m
h' = 55.3 m
Friction that occurs in gases and liquids is called
giving brainliest to first one who gets this right
a. Nitrogen, Oxygen,and Carbon Dioxide
hope im right but I dont really know
At an air show a jet flies at speed 1500 km/h on a day when the speed of sound is 342 m/s. What is the angle of the shock cone
Answer:
55 degrees
Explanation:
Given that an air show a jet flies at speed 1500 km/h on a day when the speed of sound is 342 m/s.
From the question above, we can get the below parameters
Object speed (V) = 1500 km/h
Sound speed ( v) = 342 m/s
Convert km/h to m/s
(1500 × 1000)/3600
Jet speed V = 416.67 m/s
Let's first calculate the mash number M.
M = V/v
M = 416.67 / 342
M = 1.2183
Formula for the angle of the shock cone is reciprocal of mash number. That is,
Sin Ø = 1 / M
Sin Ø = 1 / 1.2183
Sin Ø = 0.8208
Ø = sin^-1(0.8208)
Ø = 55 degree
Therefore, the angle of the shock cone is approximately 55 degrees
Why are the coral reefs suffering? (site 2) explain
Answer:
bcuz ov yo fat mamma
Explanation:
Answer:
Water pollution
Explanation:
What is the acceleration of gravity, in m/s2, on the surface (or outer limit) of Venus? The mass of Venus is 4.87 1024 kg and its radius is 6.05 106 m.
Answer:
8.9 m/[tex]s^{2}[/tex]
Explanation:
From Newton's law of universal gravitation,
F = [tex]\frac{GMm}{R^{2} }[/tex] .............. 1
and from Newton's second law of motion,
F = mg ........... 2
Equating the two expression,
mg = [tex]\frac{GMm}{R^{2} }[/tex]
g = [tex]\frac{GM}{R^{2} }[/tex]
Given that: mass of Venus = 4.87 x [tex]10^{24}[/tex] Kg, radius = 6.05 x [tex]10^{6}[/tex] and G = 6.67 x [tex]10^{-11}[/tex] N[tex]m^{2} Kg^{-2}[/tex]
Thus;
g = [tex]\frac{6.67*10^{-11}*4.87*10^{24} }{(6.05*10^{6} )^{2} }[/tex]
= [tex]\frac{3.24829*10^{14} }{3.66025*10^{13} }[/tex]
= 8.87450
g = 8.9 m/[tex]s^{2}[/tex]
the acceleration of gravity on the surface of Venus is 8.9 m/[tex]s^{2}[/tex].