Answer:abc defg hijk lmnop qrs tuv wx y and z
Explanation: now i know my abc's
On Earth, we experience lunar and solar eclipses. what types of eclipses (if any) would an inhabitant of the moon experience? Explain.
Answer:
However, those astronauts would experience a second spectacle: A solar eclipse caused by the Earth – the Sun disappearing behind the dark disc of the Earth. When Earth inhabitants witness a lunar eclipse, Moon inhabitants would, simultaneously be witnessing a solar eclipse.
Help me please!
On the earth, the gravitational field strength is 10 N/kg. On the Moon, the gravitational field strength is 1.6 N/kg.
If an object has a weight of 50 N on earth, what is its weight on the Moon?
A: 1.6 N
B: 5.0 N
C: 8.0 N
D: 80 N
Answer:
This is a way of measuring how much gravity there is. The formula is: weight/mass = gravitational field strength.
Gravitational field strength = Weight/mass unit is N/kg
Weight = mass x gravitational field strength unit is N
On Earth the gravitational field strength is 10 N/kg. Other planets have different gravitational field strengths. The Moon has a gravitational field strength of 1.6 N/kg. You might have seen films of astronauts leaping high on the moon.
Here on Earth, if I jump I am pulled back to ground by gravity. What is my weight? My mass is 80kg and if we multiply by gravitational field strength (10N/kg) - my weight is 800N. Now if I go to the moon, my mass will be the same, 80kg. We multiply that by the moon's gravitational field strength, which is 1.6 N/ kg. That means my weight on the moon is 128N. So I have different weights on the Earth and on the Moon. That's why astronauts can jump high into the air on the moon - they're lighter up there.
Jupiter is a very large planet with strong gravitational field strength of 25 N/ kg. My body is 80kg. If I go to Jupiter my weight is going to be 25 x 80 = 2,000 N. That means I wouldn't be able to get off the ground or stand up straight! I would probably be lying down all the time there. So weight varies depending on which planet you are on. You can find out more yourself by looking up tables of weight on different planets.
Lifting a stone block 146m to the top of the Great Pyramid required 146,000 J of work. How much work was done to lift the block halfway to the top?
A. 36,500 Joules
B. 73,000 Joules
C. 146,000 Joules
D. 292,000 Joules
Please help me.
Help!!! Need answer ASAP.
Answer:
a = 11.03 [m/s²]
Explanation:
To solve this problem we must use Newton's second law which tells us that the sum of forces is equal to the product of mass by acceleration.
ΣF = m*a
where:
F = force = 160000 [N]
m = mass = 14500 [kg]
a = acceleration [m/s²]
160000 = 14500*a
a = 11.03 [m/s²]
At 20 oC the densities of fresh water and ethyl alcohol are, respectively, 998 and 789 kg/m3. Find the ratio of the adiabatic bulk modulus of fresh water to the adiabatic bulk modulus of ethyl alcohol at 20 oC.
Answer:
The ratio is [tex]\frac{B_1}{B_2} = 1.265[/tex]
Explanation:
From the question we are told that
The density of fresh water is [tex]\rho__{f}} = 998 \ kg/m^3[/tex]
The density of ethanol is [tex]\rho_{e} = 789 \ kg /m^3[/tex]
Generally speed of a wave in a substance is mathematically represented as
[tex]v = \sqrt{\frac{B}{\rho} }[/tex]
Here B is the adiabatic bulk modulus of the substance while [tex]\rho[/tex] is the density of the substance
So at constant wave speed
[tex]\sqrt{\frac{B_1}{\rho_1} } = \sqrt{\frac{B_2}{\rho_2} }[/tex]
=> [tex]\frac{B_1}{\rho_1} = \frac{B_2}{\rho_2}[/tex]
=> [tex]B_1 \rho_2 = B_2\rho_1[/tex]
=> [tex]\frac{B_1}{B_2} = \frac{\rho_1}{\rho_2}[/tex]
Here [tex]\rho_1 =\rho__{f}} = 998 \ kg/m^3[/tex] and [tex]\rho_2 = \rho_{e} = 789 \ kg /m^3[/tex]
So
=> [tex]\frac{B_1}{B_2} = \frac{998}{789}[/tex]
=> [tex]\frac{B_1}{B_2} = 1.265[/tex]
What statement is TRUE about all the substances listed in the data table? A) All the substances conduct electricity. B) All the substances are strong electrolytes. All the substances have high dissociation constants. D) All the substances contain an equal amount of ions in solution. Eliminate
Answer:
A) all substance conduct electricity
In a warehouse, the workers sometimes slide boxes along the floor to move them. Two boxes were sliding toward each other and crashed. The crash caused both boxes to change speed. Based on the information in the diagram, which statement is correct? In your answer, explain what the forces were like and why the boxes changed speed.
Box 1 has more mass than Box 2.
Box 1 and Box 2 are the same mass.
Box 1 has less mass than Box 2.
**YOU MUST BE DESCRIPTIVE! Any short answers not explaining it wont get brainliest!**
Answer:
box 1 has larger mass than box 2 (which agrees with the first answer option given.
Explanation:
We need to consider the linear momentum of the boxes immediately before and after they crash.
Recall that momentum is defined as mass times velocity.
So for before the collision, the linear momentum of the system of two boxes is:
m1 * 4km/h - m2 * 8km/h
with m1 representing mass "1" on the left, and m2 representing mass 2 on the right.
Notice the sign of the linear momentum (one positive (moving towards the right) and the other one negative (moving towards the left)
For after the collision, we have or the linear momentum of the system:
- m1 * 2km/h - m2 * 1km/h
Then, since the linear momentum is conserved in the collision, we make the initial momentum equal the final and study the mass relationship between m1 and m2:
4 m1 - 8 m2 = - 2 m1 - m2
combining like terms for each mas on one side and another of the equal sign, we get;
4 m1 + 2 m1 = 8 m2 - m2
6 m1 = 7 m2
therefore m1 = (7/6) m2
which (since 7/6 is a number larger than one) tells us that m1 is larger than m2 by a factor of 7/6
Therefore, the first answer option is the correct answer.
When an aluminum bar is connected between a hot reservoir at 720 K and a cold reservoir at 358 K, 3.00 kJ of energy is transferred by heat from the hot reservoir to the cold reservoir. (a) In this irreversible process, calculate the change in entropy of the hot reservoir._______ J/K
(b) In this irreversible process, calculate the change in entropy of the cold reservoir.
_______ J/K
(c) In this irreversible process, calculate the change in entropy of the Universe, neglecting any change in entropy of the aluminum rod.
_______ J/K
(d) Mathematically, why did the result for the Universe in part (c) have to be positive?
Answer:
a. -4.166 J/K
b. 8.37 J/K
c. 4.21 J/K
d. entropy always increases.
Explanation:
Given :
Temperature at hot reservoir , [tex]$T_h$[/tex] = 720 K
Temperature at cold reservoir , [tex]$T_c$[/tex] = 358 K
Transfer of heat, dQ = 3.00 kJ = 3000 J
(a). In the hot reservoir, the change of entropy is given by:
[tex]$dS_h= -\frac{dQ}{t_h}$[/tex] (the negative sign shows the loss of heat)
[tex]$dS_h= -\frac{3000}{720}$[/tex]
= -4.166 J/K
(b) In the cold reservoir, the change of entropy is given by:
[tex]$dS_c= \frac{dQ}{t_c}$[/tex]
[tex]$dS_c= \frac{3000}{358}$[/tex]
= 8.37 J/K
(c). The entropy change in the universe is given by:
[tex]$dS=dS_h+dS_c$[/tex]
= -4.16+8.37
= 4.21 J/K
(d). According to the concept of entropy, the entropy of the universe is always increasing and never decreasing for an irreversible process. If the entropy of universe decreases, it violates the laws of thermodynamics. Hence, in part (c), the result have to be positive.
Imagine you are running PE class. You run the first 1000 meters in 3 minutes and then get tired and run the last 600 meters in 5 minutes. What was your *average* speed?
Answer:
200 meters per minute
Explanation:
You run the 1600 meters in a total of 8 minutes, so the average speed if 200 meters per minute.
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
A 4.00-kg particle moves along the x axis. Its position varies with time according to x= 5t +1 2.0t^3, where x is in meters and t is in seconds. Find:
a. the kinetic energy of the particle at any time t.
b. the acceleration of the particle and the force acting on it at time t.
c. the power being delivered to the particle at time t.
d. the work done on the particle in the interval t = 0 to t =5
Answer:
a) The kinetic energy of the particle at any time t is [tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex].
b) The acceleration of the particle at time t is [tex]a= 12\cdot t[/tex]. The force acting on the particle at time t is [tex]F = 48\cdot t[/tex].
c) The power being delivered to the particle at time t is [tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex].
d) The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.
Explanation:
a) The kinetic energy of the particle is entirely translational, whose formula is:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)
Where:
[tex]K[/tex] - Translational kinetic energy, measured in joules.
[tex]m[/tex] - Mass of the particle, measured in kilograms.
[tex]v[/tex] - Velocity of the particle, measured in meters per second.
The velocity of the particle is the rate of change of the position of the particle in time, that is:
[tex]v = 5+6\cdot t^{2}[/tex] (2)
Where [tex]t[/tex] is the time, measured in seconds.
By substituting on (1), we have the following expression: ([tex]m = 4\,kg[/tex])
[tex]K = 2\cdot (5+6\cdot t^{2})^{2}[/tex]
[tex]K = 2\cdot (25+60\cdot t^{2} +36\cdot t^{4})[/tex]
[tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex]
The kinetic energy of the particle at any time t is [tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex].
b) The acceleration of the particle is the rate of change of the velocity of the particle in time, that is:
[tex]a= 12\cdot t[/tex] (3)
Where [tex]a[/tex] is the acceleration of the particle, measured in meters per square second.
The acceleration of the particle at time t is [tex]a= 12\cdot t[/tex].
The force is obtained by multiplying (3) by the mass of the particle. That is to say: ([tex]m = 4\,kg[/tex])
[tex]F = m\cdot a[/tex] (4)
[tex]F = 48\cdot t[/tex]
The force acting on the particle at time t is [tex]F = 48\cdot t[/tex].
c) According to the Work-Energy Theorem, the change in kinetic energy of the particle equals the change in the net work done on the particle. In this case, the power is equal to the rate of change in kinetic energy.
[tex]\dot W = \dot K[/tex] (5)
[tex]\dot W = \frac{d}{dt}(50+120\cdot t^{2}+72\cdot t^{4})[/tex]
[tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex]
The power being delivered to the particle at time t is [tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex].
d) The work done on the particle ([tex]W[/tex]), measured in joules, is equal to the change of the kinetic energy of the particle:
[tex]W = K(5)-K(0)[/tex] (6)
[tex]W = [50+120\cdot (5)^{2}+72\cdot (5)^{4}]-[50+120\cdot (0)^{2}+72\cdot (0)^{4}][/tex]
[tex]W = 48000\,J[/tex]
The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.
what is the force of gravitational attraction between a 92 kg student and a 550 g slice of pizza that are 25 cm apart
Answer:
[tex]F = 5.4*10^{-8}\ N[/tex]
Explanation:
Given
Represent the mass of the student with M and the mass of the slice of pizza with m
[tex]M = 92kg[/tex]
[tex]m = 550g[/tex]
[tex]d = 25cm[/tex]
Required
Determine the force of attraction
This is calculated as:
[tex]F = \frac{GMm}{d^2}[/tex]
Where G = gravitational constant
[tex]G = 6.67408 * 10^{-11}\ m^3 kg^{-1} s^{-2}[/tex]
Convert both mass to kilogram and distance to metre
[tex]m = 550g[/tex]
[tex]m = 550kg/1000[/tex]
[tex]m = 0.55kg[/tex]
[tex]d = 25cm[/tex]
[tex]d = 25m/100[/tex]
[tex]d = 0.25m[/tex]
Substitute these values in [tex]F = \frac{GMm}{d^2}[/tex]
[tex]F = \frac{6.67408 * 10^{-11} * 92 * 0.55}{0.25^2}[/tex]
[tex]F = \frac{6.67408 * 92 * 0.55* 10^{-11} }{0.25^2}[/tex]
[tex]F = \frac{337.708448* 10^{-11} }{0.0625}[/tex]
[tex]F = 5403.335168* 10^{-11}[/tex]
[tex]F = 5.403335168* 10^3*10^{-11}[/tex]
[tex]F = 5.403335168*10^{3-11}[/tex]
[tex]F = 5.403335168*10^{-8}[/tex]
[tex]F = 5.4*10^{-8}\ N[/tex]
Calculate the distance covered by a bus moving at a rate of 11.5km/h with a time of 2060seconds
Answer:
6.56km
Explanation:
Given parameters:
Speed = 11.5km/hr
Time = 2060s
Unknown:
Distance covered = ?
Solution:
Speed is distance divided by the time taken.
Speed = [tex]\frac{distance}{time}[/tex]
Distance = Speed x time
Let us convert the seconds to hours;
3600s = 1hr
2060s = [tex]\frac{2060}{3600}[/tex] = 0.57hr
Now
Distance = 11.5km/hr x 0.57hr = 6.56km
Orb spiders make silk with a typical diameter of 0.15 mm. a. A typical large orb spider has a mass of 0.50 g. If this spider suspends itself from a single 12-cm-long strand of silk, by how much will the silk stretch?b. What is the maximum weight that a single thread of this silk could support?
Answer:
(a) the change in length of the silk is 0.001585 cm
(b) the maximum weight that a single thread can support is 17.67 N
Explanation:
Given;
mass of the spider, m = 0.50 g = 0.5 x 10⁻³ kg
length of the silk, L = 12 mm = 0.012 m
diameter of the silk, d = 0.15 mm
radius of the silk, r = d / 2 = 0.075 mm = 0.075 x 10⁻³ m
The cross sectional area of the silk;
A = πr² = π(0.075 x 10⁻³)²
A = 1.767 x 10⁻⁸ m²
The Young's modulus of elasticity of spider-silk is given by;
2.1 Gpa = 2.1 x 10⁹ N/m²
(a)
Apply Young's modulus of elasticity equation to determine the change in length of the silk;
[tex]E = \frac{FL}{Ax} = \frac{F_gL}{Ax}\\\\x = \frac{F_gL}{AE}\\\\x = \frac{(0.5*10^{-3}*9.8)(0.12)}{(2.1*10^9)(1.767*10^{-8})}\\\\x = 1.585*10^{-5} \ m\\\\x = 0.01585 \ mm[/tex]
[tex]x = 0.001585 \ cm[/tex]
(b)
the maximum weight that a single thread can support is given by;
[tex]T_{tensile \ strength} = \frac{F_{max}}{A}[/tex]
The tensile strength of spider-silk is given by 1 Gpa = 1 x 10⁹ N/m²
[tex]F_{max} = T_{tensile \ strength}*A\\\\F_{max} = (1*10^9)(1.767*10^{-8})\\\\F_{max} = 17.67 \ N[/tex]
What happens when the elevator is in free-fall, that is, what is the value for FN and what is the sensation experienced by the person in the elevator? Also, discuss the meaning of FN<0.
Answer:
Weightlessness
Explanation:
When the elevator is in free fall, this can only occur when the cord of the elevator breaks.
The acceleration of the elevator will be equal to the acceleration due to gravity. That is:
a = g
Then, the body will experience what we called WEIGHTLESSNESS
Where the normal reaction N of the person will tend to zero. That is
N = 0
The value of FN < 0 because the person inside the lift will experience weightlessness.
A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2?
A 0.31 s
B 0.56 s
C 4.3s
D 70s
Answer:
C. 4.3 seconds
Explanation:
B 0.56 s is the time period of a twirlers baton.
What is Centripetal Acceleration?Centripetal acceleration is defined as the property of the motion of an object which traversing a circular path.
Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration.
The centripetal acceleration is given by:
a = 4π²R/T²
Given values are:
a = 47.8 m/s²
D = 0.76 m so , R = 0.76/2 = 0.38m
Using this formula,
47.8*T² = 4π² x0.38
T² = [tex]\frac{4*3.14^2*0.38}{47.8}[/tex]
T = 0.56 s
Therefore,
A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2 which have time period of 0.56 s.
Learn more about Centripetal acceleration here:
https://brainly.com/question/14465119
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.
During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in the cannon,determine the average net force exerted on him in thebarrel of the cannon.
Answer:
2872.8 N
Explanation:
We have the following information
m =n72kg
Δy = 18m
t = 0.95s.
From here we use the equation
Δy=1/2at2 in order to solve for the acceleration.
So a
=( 2x 18m)/(0.95s²)
= 36/0.9025
= 39.9m/s2.
From there we use the equation
F = ma
F=(72kg) x (39.9)
= 2872.8N.
2872.8N is the average net force exerted on him in the barrel of the cannon.
Thank you!
A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocity 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.
Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
A parallel plate capacitor is made up of two metal squares with sides of length 8.8 cm, separated by a distance 5.0 mm. When a voltage 187 V is set up across the terminals of the capacitor, the charge stored on the positive plate is equal to __________ nC. g
Answer:
2.56 nC
Explanation:
By definition, the capacitance is expressed by the following relationship between the charge stored on one of the plates of the capacitor and the potential difference between them, as follows:[tex]C =\frac{Q}{V} (1)[/tex]
For a parallel-plate capacitor, assuming a uniform surface charge density σ, if the area of the plates is A, the charge on one of the plates can be written as follows:[tex]Q = \sigma * A (2)[/tex]
Assuming an uniform electric field E, the potential difference V can be expressed as follows:[tex]V = E*d (3)[/tex]
where d is the distance between plates.
Applying Gauss 'Law to a closed surface half within one plate, half outside it, we find that E can be written as follows:[tex]E =\frac{\sigma}{\epsilon_{0}} (4)[/tex]
Replacing (4) in (3), and (2) in (1), we can express the capacitance C as follows:[tex]C= \frac{\epsilon_{0}*A}{d} (5)[/tex]
Taking (1) and (5), as both left sides are equal each other, the right sides are also equal, so we can write the following equality:[tex]\frac{Q}{V} = \frac{\epsilon_{0}*A}{d} (6)[/tex]
Solving for Q, we get:[tex]Q = \frac{\epsilon_{0}*A*V}{d} = \frac{8.85e-12F/m*(0.088m)^{2}*187 V}{5.0e-3m} = 2.56 nC[/tex]
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
why does sound energy even exist
What ate the two safety precautions that should be taken before driving your car?
Answer:
When traveling behind other vehicles, there should be at least a four second space between your vehicles. When the car in front of you passes a stationary object, slowly count to yourself. If you pass the object before the allotted time, you should back off. When traveling at night or inclement weather, these times should be doubled.
Don't talk on a cell phone while driving. Phones detract from your ability to concentrate on the road and increase your chance of a collision by nearly 400%. If you must use the phone, pull over to a safe, well-lit parking lot and place your call there. After completing your call you may continue on your way.bey all speed limits and signs.
When does a magnet induce an electric current in a wire coil?
O A. When the wire is connected to the coil
O B. When the magnet is near the coil
O C. When the magnet is moving back and forth in the coil
D. When the magnet is very strong
Answer:
B I believe
Explanation:
A mass m at the end of a spring of spring constant k is undergoing simple harmonic oscillations with amplitude A.
Part (a) At what positive value of displacement x in terms of A is the potential energy 1/9 of the total mechanical energy?
Part (b) What fraction of the total mechanical energy is kinetic if the displacement is 1/2 the amplitude?
Part (c) By what factor does the maximum kinetic energy change if the amplitude is increased by a factor of 3?
Answer:
a) The potential energy of the system is 1/9 of the total mechanical energy, when [tex]x= \frac{1}{3}\cdot A[/tex].
b) The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.
c) The maximum kinetic energy is increased by a factor of 9.
Explanation:
a) From Mechanical Physics, we remember that the mechanical energy of mass-spring system ([tex]E[/tex]), measured in joules, is the sum of the translational kinetic energy ([tex]K[/tex]), measured in joules, and elastic potential energy ([tex]U[/tex]), measured in joules. That is:
[tex]E = K + U[/tex] (1)
By definitions of translational kinetic energy and elastic potential energy, we have the following expressions:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
[tex]U = \frac{1}{2}\cdot k\cdot x^{2}[/tex] (3)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Velocity of the mass, measured in meters per second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]x[/tex] - Elongation of the spring, measured in meters.
If we know that [tex]U = \frac{1}{9}\cdot E[/tex], [tex]k = k[/tex] and [tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex], then:
[tex]\frac{1}{18}\cdot k\cdot A^{2} = \frac{1}{2}\cdot k\cdot x^{2}[/tex]
[tex]\frac{1}{9}\cdot A^{2} = x^{2}[/tex]
[tex]x= \frac{1}{3}\cdot A[/tex]
The potential energy of the system is 1/9 of the total mechanical energy, when [tex]x= \frac{1}{3}\cdot A[/tex].
b) If we know that [tex]k = k[/tex], [tex]x = \frac{1}{2}\cdot A[/tex] and [tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex], then the equation of energy conservation associated with the system is:
[tex]\frac{1}{2}\cdot k\cdot A^{2} = \frac{1}{4}\cdot k\cdot A^{2}+K[/tex]
[tex]K = \frac{1}{4}\cdot k\cdot A^{2}[/tex]
The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.
c) From the Energy Conservation equation associated with the system, we know that increasing the amplitude by a factor of 3 represents an increase in the elastic potential energy by a factor of 9. Then, the maximum kinetic energy is increased by a factor of 9.
A person standing on top of a 30.0 m high building throws a ball with an initial velocity of 20. m/s at an angle of 20.0° below horizontal. How far from the base of the building will the ball land?
Answer:
Explanation:
horizontal component of velocity of throw = 20 cos20 = 18.8 m /s
vertical downwards component = 20 sin20 = 6.84 m /s
time to displace by height 30 m = t , initial velocity u = 6.84 m /s
h = ut + 1/2 gt²
30 = 6.84 t + .5 x 9.8 t²
4.9 t² + 6.84 t - 30 = 0
t = - 6.84 ±√( 6.84² + 4 x 4.9 x 30 ) / 2x 4.9
= - 6.84 ±√( 46.78 + 588 ) / 9.8
= - 6.84 ±√(634.78 ) / 9.8
= - 6.84 ±25.2 / 9.8
= 1.87 s
horizontal displacement in 1.87 s
= 18.8 x 1.87
= 35.15 m .
Discuss two ways to determine your muscular strength explain the advantages
Answer:
Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.
Explanation:
Answer:
Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.
Explanation:
Mr. Jones starts from rest and begins to accelerate straight to the bathroom at a rate of 0.5 m/s for 10 seconds. What kind of motion is this
A. linear
b centripetal
c.free fall
d projectile
Answer:
A. linear
Explanation:
because they are going in a straight line
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
Use the above picture to fill in the blanks for the following statement.
One of the element carbon combines with one of the element oxygen to form one of the compound carbon dioxide.
Answer:
C + 2O ------ CO2
Explanation:
"One" element of Carbon combines with "Two" elements of Oxygen, to form "One" compound of Carbon dioxide.
I didn't really get what you meant but this is my guess of what you meant
A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the horizontal. The ball rolls without slipping down the incline and at the bottom has a speed of 4.9 m/s. How many revolutions does the ball rotate through as it rolls down the incline
Answer:
Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!
The ball rotates 6.78 revolutions.
Explanation:
Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!
At the bottom the ball has the following angular speed:
[tex] \omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s [/tex]
Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:
[tex] sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m [/tex]
To find the revolutions we need the time, which can be found using the following equation:
[tex] v_{f} = v_{0} + at [/tex]
[tex] t = \frac{v_{f} - v_{0}}{a} [/tex] (1)
So first, we need to find the acceleration:
[tex] v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L} [/tex] (2)
By entering equation (2) into (1) we have:
[tex] t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}} [/tex]
Since it starts from rest (v₀ = 0):
[tex] t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s [/tex]
Finally, we can find the revolutions:
[tex] \theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev [/tex]
Therefore, the ball rotates 6.78 revolutions.
I hope it helps you!