Answer:
Height ball A will deflect the target is 2.65 m
Height ball A will deflectthe target is 2.2 m
Ball A, will deflect the target to greater height, thus i will throw ball A
Explanation:
Given;
mass of the target, m₁ = 5.00 kg
mass of the ball, m₂ = 1.5 kg
initial velocity of each throw, u = 12 m/s
Throwing ball A; apply the principle of conservation linear momentum for elastic collision;
m₁u₁ + m₂u₂ = v₁m₁ + v₂m₂
The initial velocity of the target, u₁ = 0
The ball bounced back at the same speed, v₂ = -12 m/s
the velocity of the target after collision, = v₁
0 + 1.5 x 12 = v₁(5) + (-12 x 1.5)
18 = 5v₁ - 18
18 + 18 = 5v₁
36 = 5v₁
v₁ = 36/5
v₁ = 7.2 m/s
The vertical height reached by the target is given by;
v₁² = u₁² + 2gh
v₁² = 0 + 2gh
v₁² = 2gh
h = v₁² / 2g
h = (7.2)² / (2 x 9.8)
h = 2.65 m
Throwing ball B; apply the principle of conservation energy for the inelastic collision;
the kinetic energy of the ball will be converted to the potential energy of the target.
¹/₂m₁u₂² = mgh
¹/₂(1.5)(12)² = (5 x 9.8)h
108 = 49h
h = 108 / 49
h = 2.2 m
Ball A will deflect the target to greater height, thus i will throw ball A.
A 2070 kg space station orbits Earth at an altitude of 5.37×105 m. Find the magnitude of the force with which the space station attracts Earth. The ma
Answer:
9.9 × 10^-16N
Explanation:
Given that a 2070 kg space station orbits Earth at an altitude of 5.37×105 m. Find the magnitude of the force with which the space station attracts Earth.
Calculating the gravitational field strength.
g = Gm/R^2
Where G = 6.67408 × 10-11 m3 kg-1 s-2
g =(6.67408 × 10-11 × 2070)/5.37 × 10^5
g = 1.38069×10^-7 / 2.88369^11
g = 4.788×10^-19 m/s
Force F = mg
Substitute g into the formula
F = 2070 × 4.79×10^-19
F = 9.9 × 10^-16N
Write the vectors labeled A, B and C with rectangular coordinates.
Focus on vector A. Start at the base of the vector which is at (1,1)
Then notice how we have to move 3 units to the right and 2 units up to get to the tip of the vector where the arrow is located. So this is at (4,3)
The fact that we moved 3 units to the right and 2 units up means that vector A is <3, 2>
So we'll type 3 in the first box and 2 in the second box.
======================================================
We use the same idea for vector B
Start at the base and move to the tip. Doing so has us move 1 unit left and 3 units down.
Vector B = <-1, -3>
We type -1 in the first box, and -3 in the second box.
======================================================
Finally onto vector C
Start at the base (0,0). To move to the tip, we need to go 3 units left and 3 units up
Vector C = <-3, 3>
-3 goes in the first box and 3 goes in the second box
======================================================
In general, vectors are useful to tell us a specific direction in which to go. They're useful in physics to determine how exactly a force is being applied, or how a particle is moving.
A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. She slows steadily, then continues on at 5.8 m/s . What is her acceleration on the rough ice?
Recall that
v² - u² = 2 a ∆x
where u and v are initial and final velocities, respectively; a is acceleration; and ∆x is the distance traveled (because the skater moves in only one direction).
So we have
(5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)
a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)
a = -3.65 m/s²
Centripetal acceleration must involve a change in
Answer:
Centripetal acceleration must involve a change in a. an object's tangential speed.
Explanation:
Centripetal acceleration must involve a change in direction of motion of the object.
What is centripetal acceleration?centripetal acceleration is the acceleration of a body traversing a circular path. Because velocity is a vector quantity (that is, it has both a magnitude, the speed, and a direction), when a body travels on a circular path, its direction constantly changes and thus its velocity changes, producing an acceleration. The acceleration is directed radially toward the center of the circle.
In your daily existence, you must have encountered numerous instances of centripetal acceleration. A centripetal acceleration occurs when you drive in a circle, and a centripetal acceleration also occurs when a satellite orbits the earth. Centripetal refers to being in the center.
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Which example illustrates Newton's first law?
Answer: The motion of a ball falling down through the atmosphere, or a model rocket being launched up into the atmosphere are both examples of Newton's first law. The motion of a kite when the wind changes can also be described by the first law.
Explanation:
A 10 [kg] object is dropped from rest. a. How far will it drop in 2 [s]?
Answer:
y=0.5 g t^2
=0.5*10*2^2
=20 m
In what climates would you expect soil to form fastest? Why?
Answer:
Soils develop faster in warm, moist climates and slowest in cold or arid ones. Rainfall is one of the most important climate factors in soil formation.
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Hope this helped God bless :)
Soil formation occurs more swiftly in warm, humid areas than it does in cold or arid ones.One of the most crucial climatic variables for the formation of soil is rainfall. The breakdown of large soil particles into smaller ones is accelerated by creatures such as bacteria, plants, and animals.
What climates soil to form fastest?The rate of chemical reactions is accelerated by rising temperature, which also accelerates soil formation. Faster plant and bacterial growth in warmer climates aids in the weathering of materials and the formation of soils.
The formation of soil is another indirect effect of climate on organisms. Weathering is accelerated by precipitation and warm temperatures, which increases the amount of soil development. Rainfall increases the amount of organic matter, lowers pH, accelerates the leaching of basic ions, moves mud, etc.
Therefore, In warm, wet regions, soil development happens more quickly than in cold or arid ones. One of the most crucial climatic variables for the formation of soil is rainfall.
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A net force of 275 N accelerates a 18.0 kg mass. What is the resulting acceleration? (5 points)
Answer:
The answer is 15.28 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
m is the mass
f is the force
We have
[tex]a = \frac{275}{18} \\ = 15.277777...[/tex]
We have the final answer as
15.28 m/s²Hope this helps you
A mountain biker is climbing a steep 20° slope at a constant speed. The cyclist and bike have a combined weight of 800 N. What can you say about the magnitude of the normal force of the ground on the bike?
Answer:
Normal force = 751.75 N
Explanation:
It is given that,
The combined weight of the cyclist and the bike is 800 N
We need to find the magnitude of the normal force of the ground on the bike. It can be given by the formula as follows :
[tex]N=mg\cos\theta[/tex]
mg = W, combined weight of cyclist and bike
[tex]N=W\cos\theta\\\\N=800\times \cos(20)\\\\N=751.75\ N[/tex]
So, the magnitude of the normal force of the ground on the bike is 751.75 N.
The magnitude of the normal force of the ground on the bike is 751.752 N
The formula for that will be used to calculate the normal force of the ground on the bike is expressed as:
[tex]N= Fcos \theta[/tex]
F is the combined weight
[tex]\theta[/tex] is the angle of inclination
Given the following paramters
F = 800 N
[tex]\theta[/tex] = 20 degrees
Substitute the given parameters into the formula:
[tex]N=800 \times cos20^0\\N=800 \times 0.9397\\N=751.752N[/tex]
Hence the magnitude of the normal force of the ground on the bike is 751.752 N
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Select the correct answer.
Which scientist discovered the relationship between electricity and magnetism?
OA.
Nikola Tesla
OB
Alessandro Volta
OC.
Hans Christian Oersted
OD
Isaac Newton
Answer:
Hans Christian Oersted
Explanation:
What is the momentum of a 76–kilogram horse running at a speed of
45 kilometers/hour?
A.
2.5 × 102 kilogram·meters/second
B.
3.4 × 104 kilogram·meters/second
C.
3.4 × 103 kilogram·meters/second
D.
9.5 × 102 kilogram·meters/second
Answer:
Option D, is closest to the answer = 950.
Answer: D. 9.5x102 kilogram.meters/second
Explanation:
correct for plato
Which statement describes S waves? A. They move side to side.B.They arrive betore P waves. C. They travel through liquids D. They occur at Earth's surtace.
Answer: they move side to side
Explanation:
i took the test
how is the process of finding the mass of a gas different from that of finding the mass of a liquid why is the difference necessary
This is about acceleration, velocity and all that mumbo jumbo
Which best depicts the data?
1, From 0 to 2 seconds, Anna experienced negative acceleration.
2, From 3 to 6 seconds, Elan accelerated while Anna did not.
3, From 6 to 8 seconds, Elan accelerated faster than Anna.
4, From 8 to 10 seconds, Elan experienced negative acceleration.
Answer:
[tex]3. \: From \: 6 \: to \: 8 \: seconds, \: Elan \: \\ accelerated \: \: faster \: than \: Anna. [/tex]
Someone please makeup a hypothesis for me rq make it make sense!
Will ice melt faster in salt, sugar or water?
Answer:
Water because the salt will just make it colder and the sugar won't do anything
What is the average velocity of a boy moving from initial position of 4 m to a final position of 8 min 20 s. Show steps of solving:
please please heeeeelp
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Answer:
0.2 m/s
Explanation:
The average velocity is the change in position divided by the change in time:
(8 m -4 m)/(20 s) = (4/20) m/s = 1/5 m/s = 0.2 m/s
The boy's average velocity is 0.2 m/s.
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3. A cat is moving at 18 m/s when it accelerates at 4 m/s? for 2 seconds. What is his new
velocity?
Answer:
26 m/s
Explanation:
v = u + at
Given that,
u(initial speed) = 18 m/sa (acceleration) = 4m/s^2t ( time) = 2sand v ( final velocity) = ?
v = 18 + ( 4 × 2)
v = 18 + 8
v = 26 m/s
Calculate the magnetic field at the center of a circular current loop of radius R divided by the magnetic field at a distance R away from a very long straight wire carrying the same current value I. (Note the loop and wire are not in electrical contact.)
Answer:
The magnetic field at the center of a circular current loop of radius R divided by the magnetic field at a distance R away from a very long straight wire carrying the same current value I is [tex]\pi[/tex] or 3.14
Explanation:
The magnetic field at center of a circular loop is given by
[tex]B = \frac{\mu _{o}I }{2R}[/tex]
Where B is the magnetic field
[tex]\mu _{o}[/tex] is the free space permeability constant ( [tex]\mu _{o}[/tex] = 4π × 10⁻⁷ N/A²)
[tex]I[/tex] is the current
and [tex]R[/tex] is the radius
For the magnetic field of a long straight wire, it is given by
[tex]B = \frac{\mu _{o}I }{2\pi R}[/tex]
Where B is the magnetic field
[tex]\mu _{o}[/tex] is the free space permeability constant ( [tex]\mu _{o}[/tex] = 4π × 10⁻⁷ N/A²)
[tex]I[/tex] is the current
and [tex]R[/tex] is the distance from the wire
Then, to calculate the magnetic field at the center of a circular current loop of radius R divided by the magnetic field at a distance R away from a very long straight wire carrying the same current value I, that will be
[tex]\frac{\mu _{o}I }{2R} \div \frac{\mu _{o}I }{2\pi R}[/tex]
= [tex]\frac{\mu _{o}I }{2R} \times \frac{2\pi R }{\mu _{o}I}[/tex]
= [tex]\pi[/tex]
(NOTE: [tex]\pi[/tex] = 3.14)
Hence, the magnetic field at the center of a circular current loop of radius R divided by the magnetic field at a distance R away from a very long straight wire carrying the same current value I is [tex]\pi[/tex] or 3.14.
1) A rigid tank contains 2 kg of 134a refrigerant at 800 kPa and 120°C. Determine tank volume and total internal energy.
Answer:
V = 0.07525 [m³]
U = 655.74 [kJ]
Explanation:
To solve this problem we must use the superheated refrigerant table for refrigerant 134a. The attached image shows the specific volume and specific internal energy values for a pressure of 800 [kPa] and a temperature of 120 [°C].
m = mass =2 [kg]
the specific volume = v = 0.037625 [m³/kg]
The volume of the tank can be found using the following expression:
V = v*m
V = 0.037625*2 = 0.07525 [m³]
The Specific internal energy = u = 327.87 [kJ/kg]
U = u*m
U = 327.87*2 = 655.74 [kJ]
Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a 100 W incandescent bulb uses only 23.0 W of power. The compact bulb lasts 10000 hours, on the average, and costs $ 12.00, whereas the incandescent bulb costs only $ 0.76, but lasts just 750 hours. The study assumed that electricity cost $ 0.090 per kilowatt-hour and that the bulbs were on for 4.0 h per day.
Required:
a. What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years?
b. How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb?
c. What is the resistance of a "100-W" fluorescent bulb?
Answer:
a
[tex]The \ price \ for \ Compact\ fluorescent\ bulbs \ is \ C_T = \$ 21.067\\The \ price \ for \ incandescent\ bulb \ is \ C_t = \$ 43.98[/tex]
b
[tex]The \ amount \ saved \ is \ C_S = \$ 22.913[/tex]
c
[tex]The \ resistance \ is \ R = 626.1 \Omega[/tex]
Explanation:
From the question we are told that
The power used by incandescent bulb [tex]P_r = 23.0 W[/tex]
The power used by Compact fluorescent bulb [tex]P_p = 100W[/tex]
The life span of Compact fluorescent bulb is [tex]t = 1000 \ hours[/tex]
The prices of Compact fluorescent bulb is [tex]C_c = \$ 12 [/tex]
The prices of incandescent bulb is [tex]C_i = \$ 0.76 [/tex]
The life span of incandescent bulb is [tex]t_k = 750 \ hours[/tex]
The cost of electricity is [tex]C_e = \$ 0.090/ kilowatt-hour = \frac{0.090}{1000} = \$0.00009 / W / h[/tex]
The duration of daily bulb usage is [tex]t_d = 4 \ hours[/tex]
Generally 3 years of bulb usage in hours is mathematically evaluated as
[tex]t_3 = 3 * 365 * t_d[/tex]
=> [tex]t_3 = 3 * 365 * 4[/tex]
=> [tex]t_3 = 4380 \ h [/tex]
Generally the cost to power a 23W bulb for the above duration is mathematically represented as
[tex]Z = 23 * t_3 * C_e[/tex]
=> [tex]Z = 23 * 4380 * 0.00009[/tex]
=> [tex]Z = \$ 9.0667[/tex]
The number of Compact fluorescent bulbs required to achieve this about of time is mathematically represented as
[tex]N = \frac{t_3}{ t}[/tex]
=> [tex]N = \frac{4380}{1000}[/tex]
=> [tex]N = 0.438 [/tex]
=> [tex]N \approx 1 [/tex]
The total cost of usage for the Compact fluorescent bulb is mathematically represented as
[tex]C_T = Z +C_c[/tex]
=> [tex]C_T = 9.0667 +12[/tex]
=> [tex]C_T = \$ 21.067 [/tex]
The number of incandescent bulbs required to achieve this about of time is mathematically represented as
[tex]M = \frac{t_3}{ t_k}[/tex]
=> [tex]M = \frac{4380}{750}[/tex]
=> [tex]M = 6 [/tex]
Generally the cost to power a 100W bulb for the above duration is mathematically represented as
[tex]Z = 100 * t_3 * C_e[/tex]
=> [tex]Z = 100 * 4380 * 0.00009[/tex]
=> [tex]Z = \$ 39.42[/tex]
The total cost of usage for the incandescent bulbs is mathematically represented as
[tex]C_t = C_e + [6 *C_i][/tex]
=> [tex]C_t = 39.42 + [0.76 * 6][/tex]
=> [tex]C_t = \$ 43.98[/tex]
Generally the total amount saved is mathematically represented as
[tex]C_S = C_t - C_T[/tex]
=> [tex]C_S = 43.98 - 21.067[/tex]
=> [tex]C_S = \$ 22.913[/tex]
Generally the resistance of the Compact fluorescent bulb is
[tex]R = \frac{V^2}{P}[/tex]
[tex]R = \frac{120^2}{23}[/tex]
[tex]R = 626.1 \Omega[/tex]
What is a factor in an experiment that can be changed or it can be kept the same?
Answer: If you are performing an experiment on the effect of spraying plants with different chemicals, you would try to maintain the same pressure and maybe the same volume when spraying your plants. The independent variable is the one factor that you are changing. It is one factor because usually in an experiment you try to change one thing at a time.
Explanation:
An airplane flying at 119 m/s is accelerated uniformly at the rate of 0.5 m/s2 for 10
seconds. What is its final speed?
Answer:
124m/s
Explanation:
v=u+at
v=119+(0.5×10)
v=119+5
v=124m/s
why cats cannot run on smoth table
Answer:
bc of their paw pads, they don't get the traction they need in order to be able to get a firm grip to push off of on the table for each bound
Explanation:
A 65.25 kg person holding a steel ball stands motionless on a frozen lake.
The person then throws the ball, which propels the person at 1.25 m/s to the
right and the ball 4.05 m/s to the left. If the initial momentum of the system is
zero, what is the mass of the steel ball?
Answer:
20.14 KG
Explanation:
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The mass is 20.14 kg
Object has a mass of 35g and a density of 5g/cm3.
What is the volume? *
Answer:
The answer is 7.0 mLExplanation:
The volume of a substance when given the density and mass can be found by using the formula
[tex]volume = \frac{mass}{density} \\[/tex]
From the question
mass = 35 g
volume = 5 g/cm³
We have
[tex]volume = \frac{35}{5} \\ [/tex]
We have the final answer as
7.0 mLHope this helps you
Which of the following is an example of homeostasis?
Answer:
Humans' internal body temperature is a great example of homeostasis. ...
Glucose is the most basic form of sugar, and the only type the body can use directly. ...
When bacteria or viruses that can make you ill get into your body, your lymphatic system kicks in to help maintain homeostasis.
Explanation:
what are Newton's 3 laws
Answer:
Newton 3 laws
Explanation:
1.object will not change its motion unless a force acts on it . 2.the force on an object is equal to it's time mass on acceleration. 3. When 2 o jecys interact they apply forces to each other of equal.
What would be the speed of an object just before hitting the ground if dropped 91.5 meters
Answer:
about 42.35 m/s
Explanation:
Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:
[tex]v_f^2=v_i^2+2\,a\,\Delta x[/tex]
which for our case would reduce to:
[tex]v_f^2=v_i^2+2\,a\,\Delta x\\v_f^2=0+2\,9.8\,(91.5)\\v_f^2= 1793.4\\v_f=\sqrt{1793.4} \\v_f \approx 42.35[/tex]
then the velocity just before hitting would be about 42.35 m/s
SOMEONE PLEASE PLEASE HELP QUICK PICTURE INCLUDED
Answer:
A
Explanation:
Switches are used to block electrical currents, thus needing to be made out of insulating material.