⚠️I need help with the last question!⚠️
Answer:
I can't do your work for you but I can explain the last question;
The want you to tell them (In at least 3 sentences) Why you think your answers are correct or how your answer's match your hypothesis.
A hypothesis (for a little more help) is an idea or explanation that you then test through study and experimentation.
Outside science, a theory or guess can also be called a hypothesis. A hypothesis is something more than a wild guess but less than a well-established theory. Anyone who uses the word hypothesis is making a guess.
Explanation:
Sorry I didnt give you the exact answer but I hope this helps :)
Help Please! Been stuck for some time
Answer:
pretty sure A is correct, as they are pushing in opposite directions, hence canceling each other
Explanation:
Orbital speed of a satellite is dependent of its mass
I'm assuming the question is supposed to be like this:
How does the Orbital velocity of a satellite depends on the mass of the satellite?
Answer
It is independent of mass of satellite.
A ball is thrown horizontally from the top of a 24.7 m building with a speed of 13.2m/s. Assuming level ground, the ball lands ___ m from the base of the building.
Answer:
The horizontal distance traveled by the ball is 29.7 m.
Explanation:
Given;
height of the building, h = 24.7 m
initial horizontal velocity, Vₓ = 13.2 m/s
The time taken for the ball to fall from the vertical height is calculated as;
[tex]h = V_yt + \frac{1}{2} gt^2[/tex]
where;
[tex]V_y[/tex] is initial vertical velocity = 0
[tex]h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 24.7}{9.8} } \\\\t = 2.25 \ s[/tex]
The horizontal distance traveled by the ball is calculated as;
[tex]X = V_x t\\\\X = 13.2 \times2.25\\\\X = 29.7 m[/tex]
Therefore, the horizontal distance traveled by the ball is 29.7 m.
A cat is sleeping on the floor in the middle of a 2.8-m-wide room when a barking dog enters with a speed of 1.40 m/s. As the dog enters, the cat (as only cats can do) immediately accelerates at 0.85 m/s2 toward an open window on the opposite side of the room. The dog (all bark and no bite) is a bit startled by the cat and begins to slow down at 0.10 m/s^2 as soon as it enters the room.
Required:
How far is the cat in front of the dog as it leaps through the window?
Answer:
the cat is 0.4238 m in front of the dog as it leaps through the window
Explanation:
Given that;
acceleration a = 0.85 m/s²
speed v = 1.40 m/s
the cat is at rest, so initial velocity u = 0
we know that, since the cat is sleeping on the floor in the middle of a 2.8-m-wide room, it needs to cover (2.8 m / 2 ) distance to get to the window;
using the second equation equation of motion;
s = ut + 1/2 at²
we substitute
2.8/2 = 0×t + 1/2 × 0.85 × t²
1.4 = 0.425t²
t = √( 1.4 / 0.425 )
t = 1.81497 sec
now, at acceleration 0.10 m/s²
the dog has to cover the distance;
s = ut + 1/2 at²
s = ( 1.4 × 1.81497) - 1/2 × 0.10 × 1.81497²
s = 2.540958 - 0.1647
s = 2.3762 m
The cant in front of the dog as it leaps through the window;
distance = 2.8 m - 2.3762 m
distance = 0.4238 m
Therefore, the cat is 0.4238 m in front of the dog as it leaps through the window
The temperature of an object is
A. The average kinetic energy of its atoms and molecules
B. The total potential and kinetic energy of its atoms and molecules
C. The transfer of kinetic energy from one place or object to another through collisions between atoms and molecules
3% of earth's water is?
Is earth’s freshwater
Only about 3 percent of Earth’s water is fresh water.
1. If a radioactive sample has an initial count rate of 600 Bq. What is its count
rate after
i) l half-life?
ii) 2 half-lives
iii) 3 half-lives
iv) 4 half-lives?
Answer:
See explanation
Explanation:
Since the original count rate is 600 Bq,
i) after 1 half life, the count rate decreases to 1/2 of 600 Bq = 300 Bq
ii) after 2 half lives, the count rate decreases to 1/4 of 600 = 150 Bq
iii) after 3 half lives, the count rate decreases to 1/6 of 600 = 100 Bq
iv) after 4 half lives, the count rate decreases to 1/8 of 600 = 75 Bq
550 nm light passes through a diffraction grating with 3000 lines per centimeter. The screen is 115 cm away from the grating. What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?
Answer:
69.7 cm
Explanation:
What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?
For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m, m = order of fringe and λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.
Also, tanθ = x/D where x = distance of nth order fringe from central maximum and D = distance of screen from grating = 115 cm = 1.15 m
Now sinθ = d/mλ, Since θ is small, sinθ ≅ tanθ
So, d/mλ = x/D for a second order bright fringe, m = 2.
So, d/2λ = x/D
x = dD/2λ
So, x =
For a dark fringe, we have
d/(m + 1/2)λ = x'/D where x' is the distance of the fringe from the central maximum.
For a second-order dark fringe, m = 2. So,
d/(2 + 1/2)λ = x'/D
d/(5/2)λ = x'/D
2d/5λ = x'/D
x' = 2dD/5λ
So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is x" = dD/2λ - 2dD/5λ
x" = dD/10λ
Substituting the values of the variables into the equation, we have
x"= 1/3 × 10⁻⁵ m × 1.15 m/(10 × 550 × 10⁻⁹ m)
x" = 1.15/165 × 10² m
x" = 0.00697 × 10² m
x" = 0.697 m
x" = 69.7 cm
The distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe 69.7 cm
What is difraction?Diffraction of light occurs when a light wave passes by a corner or through an opening or slit that is physically the approximate size of, or even smaller than that light's wavelength
For a diffraction grating, dsinθ = mλ
where,
d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m,
m = order of fringe
λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.
Also, [tex]tan\theta=\dfrac{x}{D}[/tex]t where
x = distance of nth order fringe from central maximum
D = distance of screen from grating = 115 cm = 1.15 m
Now [tex]Sin\theta =\dfrac{d}{m\lambda}[/tex] , Since θ is small, sinθ ≅ tanθ
So, [tex]\dfrac{d}{m\lambda}=\dfrac{x}{D}[/tex]
for a second order bright fringe, m = 2.
So, [tex]\dfrac{d}{2\lambda}=\dfrac{x}{D}[/tex]
[tex]x=\dfrac{dD}{2\lambda}[/tex]
For a dark fringe, we have
[tex]\dfrac{d}{(m+\dfrac{1}{2})\lambda}=\dfrac{X'}{D}[/tex]
where x' is the distance of the fringe from the central maximum.
For a second-order dark fringe, m = 2. So,
[tex]\dfrac{d}{(m+\dfrac{1}{2})\lambda}=\dfrac{X'}{D}[/tex]
[tex]\dfrac{d}{ \dfrac{5}{2}\lambda}=\dfrac{X'}{D}[/tex]
[tex]X'=\dfrac{2dD}{5\lambda}[/tex]
So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is
[tex]X''=\dfrac{dD}{2\lambda}-\dfrac{2dD}{5\lambda}[/tex]
[tex]X''=\dfrac{dD}{10\lambda}[/tex]
Substituting the values of the variables into the equation, we have
[tex]x''=\dfrac{1}{3\times10^{-5}}\times \dfrac{1.15}{10\times 550\times 10^{-9}}[/tex]x
x" = 0.00697 × 10² m
x" = 0.697 m
x" = 69.7 cm
Hence the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe 69.7 cm
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Static Friction
Now let’s examine the static case. Remain on the “Force graphs” tab at the top of the window. Make sure the box labeled “Ffriction” is checked at the left of the screen, this will allow us to measure to force of friction experienced by an object as it slides down the ramp.
Draw a free body diagram for an object sitting on the incline at rest, assuming the incline is at the maximum angle BEFORE the object starts to move. Be sure to include friction and stipulate whether it is kinetic or static.
A bucket is filled partly with water such that its combined mass is 2.17 kg. It is tied to a rope and whirled in a circle with a radius of 1.13 m. The speed at the top of the circle is 4.42 m/s and the speed at the bottom of the circle is 7.1 m/s. (a) Determine the acceleration, net force and tension force at the top of the circle.
Answer:
1) a = 17.3 m/s²
2) Fnet = 37.5 N
3) T = 16.2 N
Explanation:
1)
When the bucket is at the top of the circle, there are two forces acting on it: the tension force (T) which pulls from the bucket, so it is directed downward, and the force due to gravity, that also points downward, so both forces add:[tex]F_{net} = T + m*g (1)[/tex]According to Newton's 2nd Law, this net force must be equal to the mass of the bucket, times the acceleration.Now, due to the bucket is moving around a circle, there must be a force that keeps the bucket following a circular trajectory, that is the centripetal force, and always aims toward the center of the circle.This force is not a new type of force, it's always the net force that aims toward the center.At the top of the circle, because as the tension force as gravity point downward, the centripetal force, is just this net force.It can be showed that the centripetal force can be written as follows:[tex]F_{c} = m*a_{c} = m*\frac{v^{2}}{r} (2)[/tex]
Since we have already said that a = ac (At the top of the circle), we can solve (1) for a, simplifying and replacing v and r by their values, as follows:[tex]a = a_{c} = \frac{v^{2} }{r} = \frac{(4.42m/s)^{2} }{1.13m} = 17.3 m/s2 (3)[/tex]
2)
Once we got the value of a, applying Newton's 2nd law, we can find easily the net force on the bucket at the top of the circle, as follows:[tex]F_{net} = m*a = 2.17 kg * 17.3 m/s2 = 37.5 N (4)[/tex]
3)
We have already said, that at the top of the circle, the net force is just the sum of the tension T and the force of gravity, as follows:[tex]F_{net} = T + m*g = 37.5 N (5)[/tex]
Replacing m and g by their values, we can solve (4) for T:[tex]T = 37.5 N - m*g = 37.5 N - (2.17kg*98m/s2) \\ = 37.5 N - 21.3 N = 16.2 N (6)[/tex]
What would we need to do to make an electromagnet strong enough to move cars and trains
Answer:
The combined magnetic force of the magnetized wire coil and iron bar makes an electromagnet very strong. In fact, electromagnets are the strongest magnets made. An electromagnet is stronger if there are more turns in the coil of wire or there is more current flowing through it.
A freight train has a mass of [02] kg. The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 km/h?
Answer:
t = 300.3 seconds
Explanation:
Given that,
The mass of a freight train, [tex]m=1.01\times 10^7\ kg[/tex]
Force applied on the tracks, [tex]F=7.5\times 10^5\ N[/tex]
Initial speed, u = 0
Final speed, v = 80 km/h = 22.3 m/s
We need to find the time taken by it to increase the speed of the train from rest.
The force acting on it is given by :
F = ma
or
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s[/tex]
So, the required time is 300.3 seconds.
A soccer ball was kicked over the edge of a wall and traveled 35 m horizontally at a speed of 5.6m/s. Calculate the vertical height of the wall.
Answer:
Are you sure it was soccer ball? Or meine hearts
Explanation:
A ball is thrown straight up into the air. Which of the following best describes the energy present at various stages?
There is more energy at the top of the ball's path than there is at the bottom.
The total amount of energy varies, with more energy at the bottom and less at the top of the path.
At the very top, most of the energy is potential and just before it hits the ground, most of the energy is kinetic.
At the very top, most of the energy is kinetic and just before it hits the ground, most of the energy is potential.
Answer:
Uhh 2 one
Explanation
Captain Jack Sparrow has been marooned on an island in the Atlantic by his crew, and decides to builda raft to escape. The wind seems quite steady, and first blows him due east for 11km, and then 6km ina direction 6degrees north of east. Confident that he will eventually find himself in safety, he fallsasleep. When he wakes up, he notices the wind is now blowing him gently 11degrees south of east -but after traveling for 21km, he finds himself back on the island.
Variable Name Min Max Step Sample Value
thetab 5 10 1 6
a 10 20 11 1
b 5 15 1 6
c 20 30 1 21
thetac 10 15 11 1
Required:
How far (in km) did the wind blow him while he was sleeping?
Answer:
d₃ = 37,729 km, θ= 5.1º North of West
Explanation:
This is a velocity addition problem, the easiest way to solve it is to decompose the velocities in a Cartesian system, the x-axis coincides with the West-East direction and the y-axis with the South-North direction
* first displacement is
d₁ₓ = 11 km
* second offset is
cos 6 = d₂ₓ / d₂
sin 6 = d_{2y} / d₂
d₂ₓ = d₂ cos 6
d_{2y} = d₂ sin 6
d₂ₓ = 6 cos 6 = 5.967 km
d_{2y} = 6 sin 6 = 0.6272 km
* third displacement is unknown
* fourth and last displacement
cos (-11) = d₄ₓ / d₄
sin (-11) = d_{4y} / d₄
d₄ₓ = d₄ cos (-11)
d_{4y} = d₄ sin (-11)
d₄ₓ = 21 cos (-11) = 20.61 km
d_{4y} = 21 sin (-11) = -4.007 km
They tell us that at the end of the tour you are back on the island, so the displacement must be zero
X axis
x = d₁ₓ + d₂ₓ + d₃ₓ + d₄ₓ
0 = 11 +5.967 + d₃ₓ + 20.61
d₃ₓ = -11 - 5.967 - 20.61
d₃ₓ = -37.577 km
Y axis
y = d_{1y} + d_{2y} + d_{3y} + d_{4y}
0 = 0 + 0.6272 + d_{3y} -4.007
d_{3y} = 4.007 - 0.6272
d_{3y} = 3.3798 km
This distance can be given in the form of module and angle
Let's use the Pythagorean theorem for the module
d₃ = [tex]\sqrt{d_{3x}^2 + d_{3y}^2}[/tex]
d₃ = [tex]\sqrt{37.577^2 + 3.3798^2}[/tex]
d₃ = 37,729 km
Let's use trigonometry for the angle
tan θ = d_{3y} / d₃ₓ
θ = tan⁻¹ [tex]\frac{d_{3y}}{d_{3x}}[/tex]
θ = tan-1 (-3.3798 / 37.577)
θ = 5.1º
Since the y coordinate is positive and the x coordinate is negative, this angle is in the second quadrant, so the direction given in the form of cardinal coordinates is
θ= 5.1º North of West
3. Sodium-24 has a half-life of 15 hours. If a sample of sodium-24 has an
original activity of 800 Bq, what will
its activity be after:
i) 15 hours?
ii) 30 hours?
iii) 45 hours?
iv) 60 hours?
Answer:
See explanation
Explanation:
From the formula;
0.693/t1/2 = 2.303/t log (Ao/A)
t1/2 = half life of Sodium-24
Ao = initial activity of Sodium-24
A= activity of Sodium-24 at time = t
So,
0.693/15 = 2.303/15 log (800/A)
0.0462 = 0.1535 log (800/A)
0.0462/0.1535 = log (800/A)
0.3 = log (800/A)
Antilog(0.3) = (800/A)
1.995 = (800/A)
A = 800/1.995
A = 401 Bq
ii) 0.693/15 = 2.303/30 log (800/A)
0.0462 = 0.0768 log (800/A)
0.0462/0.0768 = log (800/A)
0.6 = log (800/A)
Antilog (0.6) = (800/A)
3.98 = (800/A)
A = 800/3.98
A = 201 Bq
iii)
0.693/15 = 2.303/45 log (800/A)
0.0462 = 0.0512 log (800/A)
0.0462/0.0512 = log (800/A)
0.9 = log (800/A)
Antilog (0.9) = (800/A)
7.94 = (800/A)
A = 800/7.94
A= 100.8 Bq
iv)
0.693/15 = 2.303/60 log (800/A)
0.0462 = 0.038 log (800/A)
0.0462/0.038 = log (800/A)
1.216 = log (800/A)
Antilog(1.216) = (800/A)
16.44 = (800/A)
A = 800/16.44
A = 48.66 Bq
You use a 160 cm plank to lift a large rock. If the rock is 20cm from the fulcrum, what is the plank’s IMA?
Answer:
IMA = 8
Explanation:
Given the following data:
Length of plank = 160cm
Length of resistance = 20cm
To find the ideal mechanical advantage (IMA);
[tex] IMA= \frac {length \; of \; effort}{length \; of \; resistance} [/tex]
[tex] IMA= \frac {160}{20} [/tex]
IMA = 8
Clouds
Clouds have a big influence on weather. They are a necessary precursor of precipitation, although not all of them
produce precipitation. Clouds also prevent some solar radiation from reaching the ground and absorb some of the
heat that is re-radiated from the surface. As a result, cloudy days are likely to be cooler and cloudy nights warmer
than clear days and nights.
Water vapor condenses out of the air when the temperature reaches the dew point. Air may reach its dew point
when humidity increases or air temperature decreases. The latter commonly happens when warm, moist air rises.
For clouds to form, water vapor must condense around tiny particles called nuclei (singular, nucleus). A nucleus
might be a speck of dust or smoke, or it might be a salt crystal. The condensation of many water molecules around
a nucleus forms a tiny droplet of liquid water. If billions of these water droplets come together, they make a cloud.
Clouds are classified in several ways. The most common classification used today divides clouds into groups based on altitude
• Middle clouds form at middle altitudes and consist of ice crystals, water droplets, or both. Examples of middle
clouds include altocumulus and altostratus clouds.
• Low clouds form at low altitudes and consist entirely or mainly of water droplets. Examples of low clouds
include stratus, stratocumulus, and nimbostratus clouds.
• Vertical clouds grow upward and have their bases at low altitude and their tops at middle or high altitude.
They form when strong air currents carry warm air upward. Examples of vertical clouds include cumulus and
nimbocumulus clouds.
Questions
1. How do clouds influence weather?
2. Explain how clouds form.
3. Outline how clouds are classified by altitude
Answer:
clouds from when moist, warm rising air cools and expands in the atmosphere. The water vapor in the air condenses to form tiny water droplets which are the basis of clouds.
Explanation:
what is electric potential
Answer:
The electric potential is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field with negligible acceleration of the test charge to avoid producing kinetic energy or radiation by test charge.
SI unit: volt
Other units: statvolt
In SI base units: V = kg⋅m^2⋅A^−1⋅s^−3
Dimension: M L^2 T^−3 I^−1
Explanation:
Hope it is helpful....
Explanation:
common symbol - V
SI unit - volt
other unit - statvolt
Assume all of the resistors in the following circuits have a value of 5 ohms. Which ammeter will have the highest
reading?
Answer:
Circuit B
Explanation:
To know the correct answer to the question, we shall determine the reading of the ammeter i.e the current in each circuit. This can be obtained as follow:
For circuit A:
Resistance (R) = 5 ohms
Voltage (V) = 1.5 V
Current (I) =?
V = IR
1.5 = I × 5
Divide both side by 5
I = 1.5 / 5
I = 0.3 A
For circuit B:
We'll begin by calculating the total resistance in the circuit. This can be obtained as follow:
Resistance 1 (R₁) = 5 ohms
Resistance 2 (R₂) = 5 ohms
Total resistance (Rₜ) =?
Since they are in parallel connections, the total resistance can be obtained as illustrated below:
Rₜ = R₁ × R₂ / R₁ + R₂
Rₜ = 5 × 5 / 5 + 5
Rₜ = 25 / 10
Rₜ = 2.5 ohms
Finally, we shall determine the current in the circuit.
Resistance (R) = 2.5 ohms
Voltage (V) = 1.5 V
Current =?
V = IR
1.5 = I × 2.5
Divide both side by 2.5
I = 1.5 / 2.5
I = 0.6 A
For circuit C:
We'll begin by calculating the total resistance in the circuit. This can be obtained as follow:
Resistance 1 (R₁) = 5 ohms
Resistance 2 (R₂) = 5 ohms
Total resistance (Rₜ) =?
Since they are in series connections, the total resistance can be obtained as illustrated below:
Rₜ = R₁ + R₂
Rₜ = 5 + 5
Rₜ = 10 ohms
Finally, we shall determine the current in the circuit.
Resistance (R) = 10 ohms
Voltage (V) = 1.5 V
Current =?
V = IR
1.5 = I × 10
Divide both side by 10
I = 1.5 / 10
I = 0.15 A
SUMMARY:
Circuit >>>>>>> Current
A >>>>>>>>>>>> 0.3 A
B >>>>>>>>>>>> 0.6 A
C >>>>>>>>>>>> 0.15 A
from the above calculations, circuit B has the highest ammeter reading.
Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C. If the particles are free to move, what are their speeds (in m/s) after 47.6 ns?
Explanation:
Given that,
Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C.
We need to find their speeds after 47.6 ns.
For electron,
The electric force is given by :
[tex]F=qE\\\\F=1.6\times 10^{-19}\times 570\\\\=9.12\times 10^{-17}\ N[/tex]
Let a be the acceleration of the electron. So,
F = ma
m is mass of electron
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{9.12\times 10^{-17}}{9.1\times 10^{-31}}\\\\a=10^{14}\ m/s^2[/tex]
Let v be the final velocity of the electron. So,
v = u +at
u = 0 (at rest)
So,
[tex]v=10^{14}\times 47.6\times 10^{-9}\\\\v=4.76\times 10^6\ m/s[/tex]
For proton,
Acceleration,
[tex]a=\dfrac{9.12\times 10^{-17}}{1.67\times 10^{-27}}\\\\=5.46\times 10^{10}\ m/s^2[/tex]
Now final velocity of the proton is given by :
[tex]v=5.46\times 10^{10}\times 47.6\times 10^{-9}\\\\v=2598.96\ m/s[/tex]
Hence, this is the required solution.
Derase
An electric heater Consumes 1.8 MJ When connected to a 250V supply for 30 minutes. Find the power rating of the heater and the current taken from the supply
Answer:
a. Power = 1000 Watts or 1 Kilowatts.
b. Current = 4 Amperes.
Explanation:
Given the following data;
Energy consumed = 1.8MJ = 1.8 × 10^6 = 1800000 Joules
Voltage = 250V
Time = 30 minutes to seconds = 30 * 60 = 1800 seconds
To find the power rating;
Power = energy/time
Substituting into the equation, we have;
Power = 1800000/1800
Power = 1000 Watts or 1 Kilowatts.
b. To find the current taken from the supply;
Power = current * voltage
1000 = current * 250
Current = 1000/250
Current = 4 Amperes.
Can any one help pls
Answer:
A and D are correct as they reduce GHG emissions while maintaining people's standard of living.
Answer:
Ig A and D
Explanation:
as :
a : Solar powered cars- redices emissions, relies on renewable sources and maintains current living.
d: a : WIND TURBINES - redices emissions, relies on renewable sources and is the closest to maintaining current living.
I hope im right !!!
What type of bond is CO2?||
Answer:
Lol
Explanation:
how many pennies can 4 folds of a paper hold?
Albert Bandura emphasized the idea of __________, which is the belief one has in one’s own ability to succeed. A. operant conditioning B. determinism C. self-efficacy D. self-worth
Answer:
C
Explanation:
Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
What is Self efficacy?
A person's self-efficacy relates to their confidence in their ability to carry out the behaviors required to achieve particular performance goals (Bandura, 1977, 1986, 1997).
The belief in one's capacity to exercise control over one's own motivation, behavior, and social environment is known as self-efficacy. The goals for which people strive, the amount of effort put out to obtain goals, and the possibility of achieving particular levels of behavioral performance are all influenced by these cognitive self-evaluations.
Self-efficacy beliefs, unlike conventional psychological notions, are anticipated to change according to the operating domain and the environment in which an action occurs.
Therefore, Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
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A boy of mass 60 kg is sledding down a 70 m slope starting from rest. The slope is angled at 15° below the horizontal. After going 20 m along the slope he passes his friend of mass 50 kg, who jumps on the sled. They now move together to the bottom of the slope. The coefficient of kinetic friction between the sled and the snow is 0.12. Ignoring the mass of the sled, find their speed at the bottom.
39. What is the change in momentum for a 5,000 kg ship in
outer space that experiences no net force over a 1 hr
period?
Answer:
Change in momentum is zero.
Explanation:
The following data were obtained from the question:
Mass (m) = 5000 kg
Time (t) = 1 h
Net force (F) = 0
Change in momentum =?
Force = Rate of change of momentum
0 = change in momentum
Change in momentum = 0
We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.
Two bodies, A and B, have equal kinetic energies. The mass of A is nine times that of B. The ratio of the momentum of A to that of B is:_______.A. 1:9B. 1:3C. 1:1D.3:1E. 9:1
Answer:
D. 3 : 1
Explanation:
Let suppose that A and B are particles. From statement we know that [tex]K_{A} = K_{B }[/tex] (same kinetic energy) and [tex]m_{A} = 9\cdot m_{B}[/tex]. Then,
[tex]K_{A} = K_{B}[/tex]
[tex]\frac{1}{2}\cdot m_{A}\cdot v_{A}^{2} = \frac{1}{2}\cdot m_{B}\cdot v_{B}^{2}[/tex]
[tex]m_{A} \cdot v_{A}^{2} = m_{B}\cdot v_{B}^{2}[/tex]
[tex]\frac{v_{B}}{v_{A}} = \sqrt{\frac{m_{A}}{m_{B}} }[/tex]
[tex]\frac{v_{B}}{v_{A}} = 3[/tex]
And the ratio of the momentum of A to the momentum of B is:
[tex]r = \frac{m_{A}\cdot v_{A}}{m_{B}\cdot v_{B}}[/tex]
[tex]r = 9\times \frac{1}{3}[/tex]
[tex]r = 3[/tex]
Hence, the correct answer is D.
