The energy stored in a spring is elastic potential energy. The formula for calculating the elastic potential energy is expressed as
elastic potential energy = 1/2kx^2
where
k represents spring constant
x represents the distance through which the spring was compressed
From the information given,
k = 150 N/m
x = 2cm
we would convert 2cm to m
recall, 100cm = 1 m
2 cm = 2/100 = 0.02 m
By substituting these values into the formula,
elastic potential energy = 1/2 x 150 x 0.02^2
elastic potential energy = 0.03 J
A crane used 600,000 Joules of work to move a beam to the top of a building in 30 seconds. How much power did the crane use?
The power crane used is2× 10⁵ W.
What is power, exactly?The amount of work finished in a specific length of time is a definition of power. The SI unit of power is the watt (W), which is derived from joules per second (J/s). Horsepower (hp) is a unit of measurement sometimes used to express the power of motor vehicles and other equipment. It is roughly comparable to 745.7 watts.
How to Calculate?Given parameters:
Work done = 600,000J
Time taken = 30s
Unknown:
Power of the crane = ?
Mathematically,
Power = Work done /time taken
= 600,000J/30
Power= 200000 W
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What is the weight of a girl with a mass of 30 kg on Earth?
N
Use g= 9.81 m/s² and do not include units in your answer.
The weight of a girl would be 294 N on Earth with a mass of 30 Kg.
Arithmetic operations can also be specified by the addition, subtract, divide, and multiply built-in functions.
+ Addition operation: Adds values on either side of the operator.
For example 4 + 2 = 6
- Subtraction operation: Subtracts the right-hand operand from the left-hand operand.
for example 4 -2 = 2
* Multiplication operation: Multiplies values on either side of the operator
For example 4*2 = 8
The weight on Earth of a girl would be the product of the mass and acceleration due to gravity on Earth
So the required solution would be:
⇒ weight = mass × acceleration due to gravity on Earth
⇒ weight = 30 × 9.8
⇒ weight = 294 N.
Hence, the weight of a girl would be 294 N on Earth with a mass of 30 Kg.
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at what stage of solar system formation is the sun extremely active and gas is blown out to the outer solar system by the suns radiation pressure and strong solar wind
Using the theories of solar system, we got that at stage-2 of solar system formation is the sun extremely active and gas is blown out to the outer solar system by the suns radiation pressure and strong solar wind.
Protostar is created as the gas collapsed over the time, it became denser and hotter. Over time, the spherical inner core started forming, which was hotter than rest of the star.
Most of the collapsing mass which is collected in the center, forming the Sun, while the rest of the mass flattened into a protoplanetary disk out of which the planets, moons.
Hence, the stage of solar system formation in which the sun extremely active and gas is blown out to the outer solar system by the suns radiation pressure and strong solar wind is stage-2.
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a hoopof moment of inertia released from the top of an inclined plane 20m high and length 32m rolls drawn without slipping determine the speed and acceleration when it reaches the foot of the inclined plane
The speed and acceleration when it reaches the foot of the inclined plane is 14√2 m/s
Torricelli's theorem states that the speed of a liquid flowing through an orifice is equal to the speed it would attain if falling freely a distance equal to the height of the liquid's free surface above the orifice.
Given that
A hoop of the moment of inertia released from the top of an inclined plane 20m high and length 32m rolls drawn without slipping
W have to determine the speed and acceleration when it reaches the foot of the inclined plane
By using Torricelli's theorem
V= √2gh
Here g= 9.8 and h = 20m
V= √2x9.8x20
V= √4x9.8
V= 14√2 m/s
Therefore the speed and acceleration when it reaches the foot of the inclined plane is 14√2 m/s
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the ebola virus can often have symptoms similar to bacteria pneumonia, an infection of these symptome include chest pain, shortness of breath and fever. which statement best summarize the differen es betweens these two diseas?
The statement that best summarizes the difference between the two diseases is viruses are not living organisms, but bacteria are living. The correct option is b.
What are viruses and bacteria?Bacteria are living organisms, and they are both bad and good, they come under archaea. They are present everywhere on earth. Viruses are a non-living entity, and it only gets alive when it gets a living body or host.
Since bacteria can also proliferate, not just viruses, there is no difference. There is something known as "good bacteria," but the virus is frequently the one that is more hazardous.
Therefore, the correct option is b. Viruses are not living organisms, but bacteria are living.
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The question is incomplete. Your most probably complete question is given below:
a. Viruses can replicate because they are living organisms, while bacteria cannot.
b. Viruses are not living organisms, but bacteria are living.
c. Viruses can be beneficial to living organisms, but bacteria are often infectious.
d. Viruses are living organisms, while bacteria are not.
A rocket is launched with horizontal velocity of 42.3 m and a vertical velocity of 78%. what is the rocket's hang time, max vertical displacement, and horizontal displaceuent?
The rocket's hang time if it is launched with horizontal velocity of 42.3 m / s and a vertical velocity of 78% is 30.53 s
If vertical velocity is 78 % then the horizontal velocity must be 22 %
So 1 % = 42.3 / 22 = 1.92 m / s
Vertical velocity = 78 * 1.92
Vertical velocity = 149.97 m / s
Initial velocity, u = √ uy² + ux²
u = √ ( 149.97 )² + ( 42.3 )²
u = √ 22491.82 + 1789.29
u = 155.82 m / s
tan θ = uy / ux
tan θ = 149.97 / 42.3
tan θ = 3.55
θ = 74.3°
a ) Hang time,
T = 2 u sin θ / g
T = 2 * 155.82 * sin 74.3° / 9.8
T = 30.53 s
b ) Maximum vertical displacement,
H = u² sin²θ / 2 g
H = 155.82² sin²( 74.3 ) / 2 * 9.8
H = 644.16 m
c ) Maximum horizontal displacement,
R = u² sin 2θ / g
R = 155.82² sin² ( 74.3 ) / 9.8
R = 2283.3 m
Therefore,
a ) Hang time = 30.53 s
b ) Maximum vertical displacement = 644.16 m
c ) Maximum horizontal displacement = 2283.3 m
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A bowling ball is dropped from the top of a building that has a height of h = 159 m.a) Assuming air resistance is negligible, what is the ball's speed, vf in m/s, when it strikes the ground? b) If drag produces an opposing force of 10 N while the ball's falling, what is the ball's speed, vf in m/s, when it strikes the ground? Assume the ball has a mass of 5 kg.c) A different ball was dropped and the final speed was measured to be vf = 6.5 m/s while the drag force was measured to be 50 Newtons. What is the mass of this ball in kg?
ANSWER:
a) 55.82 m/s
b) 49.8 m/s
c) 5.17 kg
STEP-BY-STEP EXPLANATION:
Given:
Height (h) = 159 m
Initial speed (vi) = 0 m/s
a)
We calculate the final speed using the following formula:
[tex]\begin{gathered} (v_f)^2=(v_i)^2+2gh \\ \\ \text{ we replacing:} \\ \\ (v_f)^2=(0)^2+2(9.8)(159) \\ \\ v_f=\sqrt{3116.4} \\ \\ v_f=55.82\text{ m/s} \end{gathered}[/tex][tex]\begin{gathered} (v_f)^2=(v_i)^2+2gh \\ \\ \text{ we replacing:} \\ \\ (v_f)^2=(0)^2+2(9.8)(159) \\ \\ v_f=\sqrt{3116.4} \\ \\ v_f=55.82\text{ m/s} \end{gathered}[/tex]b)
To determine the final velocity with the opposing force, we have to determine the net force and then calculate the acceleration that would be used to calculate the velocity, like this:
[tex]\begin{gathered} F_n=mg-10 \\ \\ \text{ we replacing} \\ \\ F_n=5\cdot9.8-10 \\ \\ F_n=49-10=39\text{ N} \\ \\ F=m\cdot a \\ \\ a=\frac{F}{m}=\frac{39}{5} \\ \\ a=7.8\text{ m/s^^b2} \\ \\ \text{ Therefore:} \\ \\ (v_f)^2=(v_i)^2+2ah \\ \\ \text{ we replacing:} \\ \\ (v_f)^2=(0)^2+2(7.8)(159) \\ \\ v_f=\sqrt{2480.4} \\ \\ v_f=49.8\text{ m/s} \end{gathered}[/tex]c)
We can calculate the mass in the following way, first, we must calculate the acceleration to be able to determine the mass:
[tex]\begin{gathered} (v_{f})^{2}=(v_{i})^{2}+2ah \\ \\ \text{ we replacing:} \\ \\ 6.5^2=0^2+2\cdot a\cdot(159) \\ \\ 318a=42.25 \\ \\ a=\frac{42.25}{318} \\ \\ a=0.133\text{ m/s^^b2} \\ \\ \text{ Now we can calculate the mass using the force balance:} \\ \\ ma=mg-50 \\ \\ a=g-\frac{50}{m} \\ \\ \text{ we replacing:} \\ \\ 0.133=9.8-\frac{50}{m} \\ \\ 1.333-9.8=-\frac{50}{m} \\ \\ -9.667m=-50 \\ \\ m=\frac{-50}{-9.667} \\ \\ m=5.17\text{ kg} \end{gathered}[/tex]Describe where infrared is found on the ems compared to the other six forms of radiation.
Between visible light and microwave is where infrared falls. Infrared has a higher energy than radio but a lower energy than gamma rays.
Where, in relation to the other six types of radiation, is infrared found on the EMS?The term "infrared" refers to electromagnetic waves that exist within the electromagnetic spectrum at frequencies slightly below those of red visible light and just above those of microwaves. According to the California Institute of Technology, the wavelengths of infrared radiation are longer than those of visible light (Caltech).
Where can I find radio waves on the EMS?The majority of the radio portion of the electromagnetic spectrum (EM spectrum) is found between 1 cm and 1 km, or 30 GHz and 300 kHz in frequency. The EM spectrum's radio region is relatively vast. Astronomers usually employ wavelength in the infrared and optical spectrum.
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Spiral galaxy rotation curves are generally fairly flat out to large distances. Suppose that spiral galaxies did not contain dark matter. How would their rotation curves be different?.
Spiral galaxy rotation curves are generally fairly flat out to large distances. Suppose that spiral galaxies did not contain dark matter. Their rotational curves be different as the orbital speeds would fall off sharply with increasing distance from the galactic center.
Plotting the orbital speeds vs radius graph gives us the rotational curves. According to the theoretical calculations, the curve increases as the radial distance decrease and then decreases as the radial distance increases. The theoretical graph does not take the dark matter into account.
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The dock workerbpushes on a 1200 N crate but it doesnt move. How much work is performed by him.
Work is performed by him is zero.
What is work?Work in physics is the energy that is transferred to or from an item when a force is applied along a displacement. In its simplest form, it equals the product of the force's magnitude and the distance traveled for a constant force directed in the direction of motion.
Given in the question a force of 1200 N crate but it doesn't move.
Work = force x distance. In units, Joules = Newtons x meters.
So: Work = 1200 Newtons x 0 meters
Work = 0 joules.
Work is performed by him is zero.
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The pressure of a gas goes from 1 atm to 0.5 atm. based on the relationship between pressure and volume, what will happen to the volume of the gas?
Based on the relationship between pressure and volume, the volume of the gas will increase if the pressure of a gas goes from 1 atm to 0.5 atm.
According to ideal gas equation,
PV = nRT
P = Pressure
V = Volume
n = Number of moles
R = Gas constant
T = Temperature
At constant Temperature,
P ∝ 1 / V → ( 1 )
From ( 1 ), it can be clearly seen that pressure is inversely proportional to volume. So as pressure increases the volume will decrease and if the pressure decreases, the volume will increase. In the given situation, the pressure is decreased. So the volume will increase.
Therefore, the volume of the gas will increase if the pressure of a gas goes from 1 atm to 0.5 atm.
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6. Suppose that two chrome steel balls attract each other with a gravitational
force of 16 units. If the mass of both objects was doubled, and if the distance
between the objects was doubled, then what would be the new force of attraction
between the two objects?
4 units
8 units
16 units
32 units
64 units
Answer:
Explanation:
Given:
m₁
m₂
R₁
F₁ = 16 units
m₁' = 2·m₁
m₂' = 2·m₂
R₂ = 2·R₁
_________
F₂ - ?
The force of gravity:
F₁ = G·m₁·m₂ / (R₁)²
F₂ = G·m₁'·m₂' / (R₂)² = G·(2·m₁)·(2·m₂) / (2·R₁)² =
= 4·m₁·m₂ / (4·R₁²) = m₁·m₂ / (R₁)² = F₁
F₂ = F₁ = 16 units
The image I've attached holds the question.What is the satellite’s speed?Answer in units of m/s.What is the period of the satellite’s orbit?Answer in units of h.
Given:
The gravitational constant, G=6.67259×10⁻¹¹ N·m²/kg⁻²
The mass of the moon, M=7.36×10²² kg
The radius of the orbit of the satellite, R=955.2 km=955.2×10³ m
To find:
1. Satellite's speed.
2. The period of satellite.
3. The acceleration of the satellite.
Explanation:
1.
The gravitational force applied by the moon on the satellite provides the satellite with the centripetal force that is neccessary for the satellite to orbit the moon.
Thus,
[tex]\begin{gathered} F_c=F_g \\ \frac{mv^2}{R}=\frac{GMm}{R^2} \\ v=\sqrt{\frac{GM}{R}} \end{gathered}[/tex]Where F_c is the centripetal force, F_g is the gravitational force, and v is the orbital velocity of the satellite.
On substituting the known values,
[tex]\begin{gathered} v=\sqrt{\frac{6.67259\times10^{-11}\times7.36\times10^{22}}{955.2\times10^3}} \\ =2267.46\text{ m/s} \end{gathered}[/tex]2.
The orbital period of the satellite in seconds is given by,
[tex]T=\frac{2\pi R}{v}[/tex]Thus the period of the satellite in hours is given by,
[tex]T=\frac{2\pi R}{v\times3600}[/tex]On substituting the known values,
[tex]\begin{gathered} T=\frac{2\pi\times955.2\times10^3}{2267.46\times3600} \\ =0.74\text{ hr} \end{gathered}[/tex]3.
The acceleration of the satellite is given by,
[tex]a=\frac{v^2}{R}[/tex]On substituting the known values,
[tex]\begin{gathered} a=\frac{2267.46^2}{955.2\times10^3} \\ =5.38\text{ m/s}^2 \end{gathered}[/tex]Final answer:
1. The orbital velocity of the satellite is 23267.46 m/s
2. The period of the satellite is 0.74 hr
3. The acceleration of the satellite is 5.38 m/s²
A car travelling with an initial velocity of
15.0 m/s, accelerates at 2.40 m/s² over a
distance of 180 m. What is the final
velocity of the car (m/s)?
?] m/s
The final velocity of the car 120.93 m/sec.
What is velocity?When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.
Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.
Given in the question, a car travelling with an initial velocity of 15.0 m/s, accelerates at 2.40 m/s² over a distance of 180 m
Using equation of motion,
v² = u² + 2as
v² = 15² + 2*40*180
v = 120.93 m/sec
The final velocity of the car 120.93 m/sec.
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light ray passing through the focus is undeviated true or false
Answer:true
Explanation:
A 250 kg object is lifted to a height of 4 m. What is the work done? How much
potential energy does the object have? What is the maximum kinetic energy on impact
if the object is released?
Answer:
9800J
Explanation:
It appears that potential energy is given to the object, so to calculate total energy in the system, we can use the potential energy formula
Potential energy = 250 kg • 9.8 m/s/s • 4 m
Potential energy = 9800J
You'll notice that the equation for the potential energy and work in this scenario is the same due to the fact that force would be equal to mass multiplied by gravity anyway.
Assuming the system has 100% efficiency, the total kinetic energy will be equivalent to the potential in the system currently
a mouse runs a distance of 2.0 meters in 16 seconds. calculate the speed of the mouse
S=d/t
S=2/16
S=0.125m/s
dentify the objective of this lab. select all that apply: group of answer choices to submurged different objects in the fluid. to determine density of unknown materials using archimedes' principle to discover the relationship between aarchimedes' principal and hydrostatic pressure in object submerged in a fluid. to test archimedes' principle for objects of different densities
The objective of the experiment of submerging different objects in fluid is to determine the density of the unknown material using Archimedes' principle.
Archimedes' principle- The amount of the upward force felt by the object when submerged in fluid is equal tot he weight of the mass of the fluid displaced.
So, in this experiment,
The objects of unknown densities are submerged in liquid, they displaces some amount of liquid. The amount of liquid is related to the buoyancy of the unknown object as,
Vₐpg = Vσg
where,
p is the density of the unknown object,
σ is the density of the fluid and Vₐ is the volume of the body submerged in fluid and V is the volume of the water displaced.
So, by this relation, we can find the density of the unknown material.
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On Saturday, Ashley rode her bicycle to visit Maria. Maria’s house is directly east of Ashley‘s. The graph shows how far Ashley was from her house after each minute of her trip.
1. Ashley Road at a constant speed for the first four minutes of her trip. What was her constant speed?
2. what was Ashley‘s average speed for the entire trip?
3. what was her average velocity for the entire trip?
4. Ashley stop to talk with another friend during her trip. How far is she from her house when she stopped?
5. what is the slope of the line after Ashley stopped to talk with her friend?
6. how is the slope of the line related to her speed?
1. The constant speed of Ashley for first four minute of her trip is calculated here
For first four minutes Ashley travelled 800 m distance (d) , substituting this value in the formula of speed
Speed(s) = distance (d)/ time (t)
Speed(s) = 800/4
Speed(s)= 200m/s
2. Ashley's average speed for the entire trip is
The total distance (d) traveled by Ashley is 1600 m and time (t) taken for that is 10 minutes.
Speed (s)= distance(d)/time(t)
Speed (s)= 1600/10
Speed(s) = 160m/s
3. The average velocity of Ashley is 160 m/s in east direction. As velocity is nothing but speed with direction.
4. In graph there is linear slop when Ashley was 800 m away from her house where net displacement is zero means she stopped there to talk with friend.
5. Slope of line after Ashley stopped to talk to her friend is
Slope = [tex]X_{2}[/tex] - [tex]X_{1}[/tex]/ [tex]Y_{2}[/tex]-[tex]Y_{1}[/tex]
Slope= 8-0/800-800
Slope= 0 as it is a linear line and slope of it is zero.
6. The slope of a line is related to her speed as speed increases the slope increases.
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A ray diagram without the produced image is shown.Which describes the image produced by the lens?smaller than the object and uprightsmaller than the object and invertedreal and uprightreal and inverted
smaller than the object and upright.
We have a negative lens, the image is between the object and the lens.
The top of the virtual image must be below and touching the ray that connects the top of the car with the lens (green)
So, the image will be smaller than the car and it won't be inverted.
1. If a cart with a mass of 5 kg is accelerated at a rate of 12 m/s/s, then what is theNUMERICAL value for the force?
*The mass of the cart is
m=5 kg
*The acceleration is
[tex]undefined[/tex]Help with explaining please!
As the elevator moves upward, the cables are under 1,960 N of tension.
The given parameters are,
The mass of the elevator (m) = 200 kg
According to Newton's Second Law of Motion, an object's force is directly inversely proportional to its mass times its acceleration.
The tension in the cables as the elevator travel upwards is calculated by applying the 2nd law of Newton,
T = ma +mg
where a is the acceleration of the elevator and g is the acceleration due to gravity,
At constant velocity, acceleration is zero. (a =0)
T = m(0) +mg
T = mg
T = 200*9.8
T = 1,960 N
Thus, As the elevator moves upward, the cables are under 1,960 N of tension.
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What is a helicapterHi
A helicopter is a type of aircraft which uses spinning blades ( wings) to fly .
one type of wagon wheel consists of a 2.0-kgkg hoop fitted with four 0.80-kgkg thin rods placed along diameters of the hoop so as to make eight evenly spaced spokes. part a for a hoop of radius 0.40 mm , what is the rotational inertia of the wheel about an axis perpendicular to the plane of the wheel and through the center?
The rotational inertia of the wheel is 1.344 × 10⁻⁴ kgm²
What is rotational inertia?Rotational inertia of an object is the measure of the ability for the object to rotate about an axis which may or may not pass thorugh the object.
Hoiw to find the rotational inertia of the wheel?Since in one type of wagon wheel consists of a 2.0-kg hoop fitted with four 0.80-kg thin rods placed along diameters of the hoop so as to make eight evenly spaced spokes and for a hoop of radius 0.40 mm, we need to find the rotational inertia of the wheel about an axis perpendicular to the plane of the wheel and through the center?
The rotational inertia of the wheel is given by I = I₁ + I₂ where I₁ = rotational inertia of hoop = and I₂ = rotational inertia of thin rods
Now the rotational inertia of the hoop is, I₁ = m₁r² where
m₁ = mass of hoop = 2.0 kg and r = radius of hoop = 0.40 mmAlso, the rotational inertia of each thin rod is I₃ = m₂L²/2 where
m₂ = mass of thin rod = 0.80 kg and L = length of rod = 2r where r = radius of hoop = 0.4 mm = 0.004 mSince there are four thin rods, I₂ = 4I₃
= 4m₂L²/2
= 2m₂L²
= 2m₂(2r)²
= 2m₂(4)r²
= 8m₂r²
So, I = I₁ + I₂
I = m₁r² + 8m₂r²
I = (m₁ + 8m₂)r²
So, substituting the values of the variables into the equation, we have
I = (m₁ + 8m₂)r²
I = (2.0 kg + 8 × 0.80 kg)(0.004 m)²
I = (2.0 kg + 6.4 kg)(0.000004 m²)
I = (8.4 kg)(0.000016 m²)
I = 0.0001344 kgm²
I = 1.344 × 10⁻⁴ kgm²
The rotational inertia of the wheel is 1.344 × 10⁻⁴ kgm²
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A cannonball
of mass 25kg is
fired straight up at an initial
velocity of 8m/s. The
gravitational field strength is
9.8m/s. Calculate the
maximum height it can reach?
If a cannonball of mass 25 kg is fired straight up at an initial velocity of 8 m / s, the maximum height it can reach is 3.27 m
v² = u² + 2 a s
v = Final velocity
u = Initial velocity
a = Acceleration
s = Distance
u = 8 m / s
a = - 9.8 m / s²
At maximum height,
v = 0
Using the equation from equations of motion,
0 = 8² + ( 2 * - 9.8 * s )
19.6 s = 64
s = 3.27 m
Therefore, the maximum height it can reach is 3.27 m
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how do you find your body composition
Answer (with explanation):
BMI = body composition
formula to find BMI = weight/height^2
so for example:
weight = 50kg and height = 150cm = 1.5m (make sure convert cm to m)
BMI = 50/(1.5)^2 = 50/2.25 ≈ 22.22
so the BMI would be about 22.22
Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the angular acceleration when it has an angle of 10. 0 degrees from the vertical? a typical pencil has an average length of 15. 0 cm and an average mass of 10. 0 g. Assume the tip of the pencil does not slip as it falls.
Answer: 0.4299s.
Explanation: To find the angular acceleration ,Newton's second angular law will be used.
τ = I α
Wr = I α
The weight is applied in the middle of the pencil , sin 10 = r / (L/2).
r = L/2 sin 10
the moment of inertia will be: I = 1/3 M L²
mg L / 2 sin 10 = (1/3 m L²) α
αf = 3/2 g / L sin 10
α = [3/2 × 9.8] / [0.150 sin10]
α = 17 rad/s²
angle from the vertical to the ground θ = π / 2
t = √ (2 π / (2 α ))
t = √(π / α )
t = √(π / 17)
t = √3.142 / 17
t = 0.4299s
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A ball is fired at 5 m/s to the right and bounces off a second ball, initially at rest, and comes back to the left at 1m/s. What is the velocity of the second ball? Assume both balls have a mass of 3.86kg and make to the right positive.
ANSWER:
6 m/s
STEP-BY-STEP EXPLANATION:
Given:
Mass of balls (m) = 3.86 kg
Initial speed of ball 1 (u1) = 5 m/s
Final speed of ball 1 (v1) = - 1 m/s
Initial speed of ball 2 (u2) = 0 m/s
We make the balance for the moment and we are left with the following:
[tex]\begin{gathered} m_1\cdot u_1+m_2\cdot u_2=m_1\cdot v_1+m_2\cdot v_2 \\ \\ \text{ We repalcing:} \\ \\ 3.86\cdot5+3.86\cdot0=3.86\cdot-1+3.86\cdot v_2 \\ \\ 19.3=-3.86+3.86v_2 \\ \\ 3.86v_2=19.3+3.86 \\ \\ v_2=\frac{23.16}{3.86} \\ \\ v_2=6\text{ m/s} \end{gathered}[/tex]Therefore, the speed of the second ball is equal to 6 m/s
A 29-g rifle bullet traveling 220 m/s embeds itself in a 3.9-kg pendulum hanging on a 3.0-m-long string, which makes the pendulum swing upward in an arc.
The vertical component (height) of the pendulum's maximum displacement is 0.146 meters.
What is maximum displacement?In physics, the maximum displacement or range traveled by a point on a vibrating body or wave calculated from its equilibrium position is known as amplitude. It is one-half of the length of the vibration path.
We have a collision when the bullet hits the pendulum, and momentum is preserved in any collision. Let the pendulum's beginning momentum and the bullet's initial momentum be, p₁ and p₂, respectively.
Following the impact, the pendulum and the bullet travel as one object with the same velocity, v. Let p₃ be the momentum of the pendulum and as well as the bullet following the impact.
Hence,
First momentum = Last Momentum
p₁ + p₂ = p₃
[3.4kg x 0m/s] +([2.90 x 10⁻²] x 200m/s) = 3.48kg x v
Simplified, we have
5.8kg · m/s = 3.49kg · v
Making v the subject of the formula, we have
v = 1.69m/s
As a result, the combined velocity of the pendulum and bullet after the contact is v = 1.69 m/s.
We are now using the conservation of energy equation. The potential energy at the maximum height will be equal to the kinetic energy of the pendulum plus the bullet.
mgh = 1/2mv²
Where
m is the mass in kilograms
g is the acceleration due to gravity
h is height in meters; and
v is velocity
If we divide both sides with m and make h the subject of the formula, we have:
h = v²/2g
= (1.69m/s)²/2 (9.81)m/s²
= 0.14557084607m
Hence,
The vertical component of the pendulum's maximum displacement (h) [tex]\approx[/tex] 0.146m
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Full Question:
A 29-g rifle bullet traveling 220 m/s embeds itself in a 3.9-kg pendulum hanging on a 3.0-m-long string, which makes the pendulum swing upward in an arc.
Determine the vertical component of the pendulum's maximum displacement. Express your answer in two significant figures and include the appropriate units.
. a 1.20 kg box is firmly attached to a spring with a spring constant of 32.0 n/m (the other end of the spring is attached to a wall); the surface the box is on is frictionless surface. you stretch the spring by 0.500 m and hold the box steady for a moment. you then begin pushing it with a constant 4.00 n force directed parallel to the surface (the figure shows the point where the spring is neither stretched nor compressed and the direction of the force you apply). hot compress will the spring will the spring be at the next instant the box is at rest?
If the spring is pushed with a constant force of 4 N force directed parallel to the surface, the spring will be compressed by 0.5 m at the next instant the box is at rest.
Us = 1 / 2 k x²
W = F x
Us = Potential energy of the spring
k = Spring constant
x = Displacement
F = Force
W = Work done
k = 32 N / m
F = 4 N
xi = 0.5 m
According to law of conservation of energy,
Change in potential energy = Total Work done by the force
1 /2 k xf² - 1 / 2 k xi² = F xf + F xi
1 / 2 * 32 ( xf² - 0.5² ) = 4 ( xf + 0.5 )
4 ( xf² - 0.25 ) = xf + 0.5
4 xf² - xf - 1.5 = 0
This is in the form of a quadratic equation,
x = - b ± √ ( b² - 4 a c ) / 2 a
a = 4
b = - 1
c = - 1.5
xf = - ( - 1 ) ± √ [ ( - 1 )² - ( 4 * 4 * -1.5 ) ] / ( 2 * 4 )
xf = 1 ± √ 25 / 8
xf = 1 ± 5 / 8
xf = 0.75 or - 0.5
Since it is compressed,
xf = - 0.5 m
Therefore, the spring will be compressed by 0.5 m at the next instant the box is at rest.
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