Answer: the earth
Explanation: Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth. Over a very long time, the moon's rotations created fiction with the Earth's tugging back, until the moon's orbit and rotational locked with Earth.
and that's why the earth pulls the moon
What do mammoths and tigers need energy for
What is the key for a successful relationship? and Why?
Answer:
communication, if you don't talk you'll never know what's going on.
Explanation:
A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping distance of the car?
v² - u² = 2 a ∆x
where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.
So
0² - (27 m/s)² = 2 (-8 m/s²) ∆x
∆x = (27 m/s)² / (16 m/s²)
∆x ≈ 45.6 m
The stopping distance of car achieved during the braking is of 45.56 m.
Given data:
The initial speed of car is, u = 27 m/s.
The final speed of car is, v = 0 m/s. (Because car comes to stop finally)
The magnitude of deacceleration is, [tex]a = 8\;\rm m/s^{2}[/tex].
In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.
Therefore,
[tex]v^{2}=u^{2}+2(-a)s[/tex]
Here, s is the stopping distance.
Solving as,
[tex]0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m[/tex]
Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.
Learn more about the kinematic equation of motion here:
https://brainly.com/question/11298125
The boys are finally old enough to compete in the box car derby race at the local fair. They have been working on their cars since the conclusion of the race last year. One boy's car raced down the track and placed 2nd in his race. However, the other boy's car started well but half-way through the race a wheel came off and his car came to a complete stop. The boy was very disappointed and the other boy felt horrible for his friend. Which of the following graphs best represents the motion of boy's car that stopped?
A
6. All other changeable factors that must
be kept the same to ensure a fair test
(what you keep the same).
Answer:
a constant variable?
Explanation:
A constant variable is any aspect of an experiment that a researcher intentionally keeps unchanged throughout an experiment.
Experiments are always testing for measurable change, which is the dependent variable. You can also think of a dependent variable as the result obtained from an experiment. It is dependent on the change that occurs
A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration?
Answer:68.15m/s
Explanation:
Given:
v₁=15m/s
a=6.5m/s²
v₁=?
x=340m
Formula:
v₁²=v₁²+2a (x)
Set up:
=[tex]\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)[/tex]
Solution:68.15m/s
Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birmingham—1963. Explain how they are similar or different in a few sentences.
Answer:
they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta
Explanation:
use in your own words teachers know when your not trust me.
Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The expression is [tex]W_c = P_o V_o ln (R_v)[/tex]
Explanation:
Generally smallest workdone done by a gas is mathematically represented as
[tex]dW = PdV[/tex]
Generally for an isothermal process
[tex]PV = nRT = constant [/tex]
=> [tex]P = \frac{nRT}{V}[/tex]
Generally the total workdone is mathematically represented as
[tex]W_c = \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]
=> [tex]W_c = nRT \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]
=> [tex]nRT [lnV] | \left \ {V_f}} \atop {V_o}} \right.[/tex]
=> [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]
=> [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]
From the question [tex]\frac{V_f}{V_o } = R_v[/tex]
=> [tex]W_c = P Vln (R_v)[/tex]
at initial state
[tex]W_c = P_o V_o ln (R_v)[/tex]
A car stops in 130 m. If it has an acceleration of -5 m/s2 what was the cars starting velocity?
after
Variables:
Equation and Solve:
Answer:
We are given:
displacement (s) = 130 m
acceleration (a) = -5 m/s²
final velocity (v) = 0 m/s [the cars 'stops' in 130 m]
initial velocity (u) = u m/s
Solving for initial velocity:
From the third equation of motion:
v² - u² = 2as
replacing the variables
(0)² - (u)² = 2(-5)(130)
-u² = -1300
u² = 1300
u = √1300
u = 36 m/s
Why do you feel that you are being thrown upward out of your seat when going over an arced hump on a roller coaster
Answer: The options are not given.
Here are the options.
a) There is an additional force lifting up on you.
(b) At the top you continue going straight and the seat moves out from under you.
(c) You press on the seat less than when the coaster is at rest.Thus the seat presses less on you. (
d) Both b and c are correct.
(e) a, b, and c are correct.
The correct option Is D.
B.At the top you continue going straight and the seat moves out from under you. C.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less.
Explanation:
At the top you continue going straight and the seat moves out from under you.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less because it is as a result of a phenomenon called Weightlessness. This occur when there is no force or little force is acting on your body. At the top you continue going straight and the seat moves out from under you because there is no force acting on your body and when the body is in free fall i.e acceleration due to gravity , the person is not supported by any thing at.
That is the scenarion that occur...
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
This question is incomplete, the complete question is;
The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answer:
the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Explanation:
Given that;
P₁ = 1.00 atm
P₂ = ?
V₁ = 1 L
V₂ = 1.60 L
the temperature of the gas is kept constant
we know that;
P₁V₁ = P₂V₂
so we substitute
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Write a haiku
poem
explaining
why graphing
is useful.
If you are
able, share
your poem
with others.
Answer:
Explanation:
graphing is helpful
helps visualize the line
of your equation
A 1870 kg car traveling at 13.5 m/s collides with a 2970 kg car that is initally at rest at a stoplight. The cars stick together and move 1.93 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.
Answer:
The value is [tex] \mu = 0.72 [/tex]
Explanation:
From the question we are told that
The mass of the first car is [tex]m_1 = 1870\ kg[/tex]
the initial speed of the car is [tex]u = 13.5 \ m/s[/tex]
The mass of the second car is [tex]m_2 = 2970\ kg[/tex]
The distance move by both cars is s = 1.93 m
Generally from the law of momentum conservation
[tex]m_1 * u_1 + m_2 * u_2 = (m_1 + m_2 ) * v_f[/tex]
Here [tex]u_2 = 0[/tex] because the second car is at rest
and [tex]v_f[/tex] is the final velocity of the the two car
So
[tex]1870* 13.5+ 0= ( 1870 + 2970 ) * v_f[/tex]
=> [tex]v_f = 5.22\ m/s[/tex]
Generally from kinematic equation
[tex]v_f^2 = u_2^2 + 2as[/tex]
here a is the deceleration
So
[tex]5.22^2 = 0 + 2 *a * 1.93[/tex]
=> [tex]a = 7.06 \ m/s^2 [/tex]
Generally the frictional force is equal to the force propelling the car , this can be mathematically represented as
[tex]F_f = F[/tex]
Here F is mathematically represented as
[tex]F = (m_1 + m_2) * a[/tex]
[tex]F = (1870 + 2970) * 7.06 [/tex]
[tex]F =34170.4 \ N[/tex]
and
[tex]F_f = \mu * (m_1 + m_2 ) * g[/tex]
[tex]F_f = 47432 * \mu [/tex]
So
[tex] 47432 * \mu = 34170.4 [/tex]
=> [tex] 47432 * \mu = 34170.4 [/tex]
=> [tex] \mu = 0.72 [/tex]
The Earth's magnetic field is modeled as that of a bar magnet with the geographic poles being Magnetic poles of the bar magnet, Based on our definitions of Magnetic Poles, if you were to go to the Earth's Geographic North Pole, you would be at a Magnetic _______________ of the bar magnet.
Answer:
South pole
Explanation:
In a bar magnet, field lines go from the North Pole to the South Pole (outside the magnet).
As the earth magnetic field lines go from South Pole (geographic) to the North one, this means that the North pole (geographic) really behaves as a South Pole (magnetic).
Your TV has a resistance of 10 ohms and a wall voltage of 120 V. How much current and power does it use
Answer:
Current used is 12 ampere.
Power is 1440 watts.
Explanation:
To find the current used by the TV.
Current (I) = voltage/resistance
Current= 120/10
Current is 12Ampere.
To get power used by the TV,
Power = voltage × current.
= 120× 12
Power = 1440 watts.
If Mary runs 5 miles in 50 minutes, what is her speed with the correct
label?
A vector of components (−23, −22) is multiplied by the scalar value of −6. What is the magnitude and direction of the resultant vector?
Answer:
(1,)
Explanation:
Answer:
magnitude: 21.6; direction: 33.7°
Explanation:
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop. (a) What is the kinetic energy of the ball just bef
Answer:
(a) The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the bowling ball is 113.272 joules.
Explanation:
The statement is incomplete. The complete question is:
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.
(a) What is the kinetic energy of the ball just before it hits the mattress?
(b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)?
(c) How much work does the gravitational force do on the ball while it is compressing the mattress?
(d) How much work does the mattress do on the ball? (You’ll need to use the results of parts (a) and (c)）
(a) Based on the Principle of Energy Conservation, we know that ball-earth system is conservative, so that kinetic energy is increased at the expense of gravitational potential energy as ball falls:
[tex]K_{1}+U_{g,1} = K_{2}+U_{g,2}[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Kinetic energies at top and bottom, measured in joules.
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Gravitational potential energies at top and bottom, measured in joules.
Now we expand the expression by definition of gravitational potential energy:
[tex]U_{g,1}-U_{g,2} = K_{2}-K_{1}[/tex]
[tex]K_{2}= m\cdot g \cdot (z_{1}-z_{2})+K_{1}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the bowling ball, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the bowling ball, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex], [tex]z_{2} = 0\,m[/tex] and [tex]K_{1} = 0\,J[/tex], the kinetic energy of the ball just before it hits the matress:
[tex]K_{2} = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1.5\,m-0\,m)+0\,m[/tex]
[tex]K_{2} = 102.974\,J[/tex]
The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The gravitational work done by the gravitational force of Earth ([tex]\Delta W[/tex]), measured in joules, is obtained by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,1}-U_{g,2}[/tex]
[tex]\Delta W = m\cdot g\cdot (z_{1}-z_{2})[/tex] (Eq. 2)
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex] and [tex]z_{2} = 0\,m[/tex], then the gravitational work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.5\,m-0\,m)[/tex]
[tex]\Delta W = 102.974\,J[/tex]
The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) The work done by the gravitational force of Earth while the bowling when mattress is compressed is determined by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,2}-U_{g,3}[/tex]
Where [tex]U_{g,3}[/tex] is the gravitational potential energy of the bowling ball when mattress in compressed, measured in joules.
[tex]\Delta W = m\cdot g \cdot (z_{2}-z_{3})[/tex]
Where [tex]z_{3}[/tex] is the height of the ball when mattress is compressed, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2}= 0\,m[/tex] and [tex]z_{3} = -0.15\,m[/tex], the work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0\,m-(-0.15\,m)][/tex]
[tex]\Delta W = 10.298\,J[/tex]
Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the ball equals the sum of kinetic energy just before mattress compression and the work done by the gravitational force when mattress is compressed:
[tex]\Delta W' = K_{2}+\Delta W[/tex]
([tex]K_{2} = 102.974\,J[/tex], [tex]\Delta W = 10.298\,W[/tex])
[tex]\Delta W' = 113.272\,J[/tex]
The work done by the mattress on the bowling ball is 113.272 joules.
A 10-ohm resistor has a constant current. If 1200 C of charge flow through it in 4 minutes what
is the value of the current?
A. 3.0 A
B 5.0 A
C. 11 A
D. 15 A
E. 20A
Answer:
B 5.0 A .
Explanation:
Hello.
In this case, since we know the charge (1200 C), time (4 min =240 s) and resistance (10Ω) which is actually not needed here, we compute the current as follows:
[tex]I=\frac{Q}{t}[/tex]
Then, for the given data, we obtain:
[tex]I=\frac{1200C}{4min}*\frac{1min}{60s}\\\\I=5A[/tex]
Therefore, answer is B 5.0 A .
Best regards!
The Jamaican Bobsled Team is sliding down a hill in a toboggan at a rate of 5 m/s when he reaches an even steeper slope. If he accelerates at 2 m/s2 for the 5 m slope, how fast is he traveling when he reaches the bottom of the 5 m slope?
Answer:
6.7 m/s
Explanation:
Given:
Δx = 5 m
v₀ = 5 m/s
a = 2 m/s²
Find: v
v² = v₀² + 2aΔx
v² = (5 m/s)² + 2 (2 m/s²) (5 m)
v = 6.7 m/s
A student uses a microwave oven to heat a meal. The wavelength of the radiation is 8.97 cm. What is the energy of one photon of this microwave radiation? Multiply the answer you get by 1025 to be able to input a number more easily into canvas. Enter to 2 decimal places.
Answer:
The energy of one photon is 2.21x10⁻²⁴ J. Multiplied by 10²⁵ is 22.10 J.
Explanation:
The energy (E) of a photon is:
[tex] E = h\frac{c}{\lambda} [/tex]
Where:
h: is the Planck's constant = 6.62x10⁻³⁴ J.s
λ: is the wavelength of the radiation = 8.97 cm
c: is the speed of light = 3.00x10⁸ m/s
[tex] E = h\frac{c}{\lambda} = 6.62 \cdot 10^{-34} J.s\frac{3.00\cdot 10^{8} m/s}{8.97 \cdot 10^{-2} m} = 2.21 \cdot 10^{-24} J [/tex]
Hence, the energy of one photon is 2.21x10⁻²⁴ J.
Now, if we multiply the answer by 10²⁵ we have:
[tex] E = 2.21 \cdot 10^{-24} J \cdot 10^{25} = 22.10 J [/tex]
I hope it helps you!
The energy of one photon is 2.21x10⁻²⁴ J. Multiplied by 10²⁵ is 22.10 J.
Calculation of energy:We know that
[tex]E = h\frac{c}{\lambda}[/tex]
Here
h be the Planck's constant = 6.62x10⁻³⁴ J.s
λ be the wavelength of the radiation = 8.97 cm
c be the speed of light = 3.00x10⁸ m/s
Now
Here we need to multiply the answer 10^25 so that the correct answer could come.
[tex]E = 6.62.10^{-34} \frac{3.00.10^{8}}{8.97.10^{-2}}[/tex]
= 2.21x10⁻²⁴ J.
= 22.10 J.
Hence, the energy of one photon is 2.21x10⁻²⁴ J. Multiplied by 10²⁵ is 22.10 J.
learn more about energy here; https://brainly.com/question/24719731
If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be Imax
The complete question is;
A person with body resistance between his hands of 10 kΩ accidentally grasps the terminals of a 16-kV power supply. What is the power dissipated in his body?
A) If the internal resistance of the power supply is 1600 Ω , what is the current through the person's body?
B) What is the power dissipated in his body?
C) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be I_max = 1.00mA or less?
Answer:
A) I = 1.379 A
B) P = 19016.41 W
C) r = 15990000 Ω
Explanation:
A) We are given;
Internal resistance of the power supply; r = 1600 Ω
Body resistance between hands; R = 10kΩ = 10000 Ω
Power supply voltage; E =16 kV = 16000 V
Formula for the current through the person's body with internal resistance is given by;
I = E/(R + r)
Thus;
I = 16000/(10000 + 1600)
I = 1.379 A
B) Formula for power dissipated is;
P = I²R
P = 1.379² × 10000
P = 19016.41 W
C) Now, we are told that the maximum current should be I_max = 1.00mA or less. So, I_max = 0.001 A
Thus, from I = E/(R + r) and making r the subject, we have;
r = (E/I) - R
r = (16000/0.001) - 10000
r = 15990000 Ω
PLEASE HELP EASY MULTIPLE CHOICE!!!!!!!!!!!
Answer:
options C is correct
Explanation:
asking questions is super in this education life
Answer:
option c should be the answer
Two charged objects are separated by distance, d. The first charge has a larger magnitude (size) than the second charge. Which one exerts the most force?
Answer:
The two charged objects will exert equal and opposite forces on each other.
Explanation:
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of charges on the objects and inversely proportional to the square of the distance between the two objects.
This force of attraction or repulsion between the two charged objects is always equal and opposite.
Therefore, the two charged objects will exert equal and opposite forces on each other.
Help me out on this?
If a rock is skipped into a lake at 24 m/s2, with that what force was the rock thrown if it was 1.75kg?
Answer: f= M×A
1.75kg×24= 42N
Explanation:
Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!
the radius of the earth social
An object moving 20 m/s
experiences an acceleration of 4 m/s' for 8
seconds. How far did it move in that time?
Variables:
Equation and Solve:
Answer:
We are given:
initial velocity (u) = 20m/s
acceleration (a) = 4 m/s²
time (t) = 8 seconds
displacement (s) = s m
Solving for Displacement:
From the seconds equation of motion:
s = ut + 1/2 * at²
replacing the variables
s = 20(8) + 1/2 * (4)*(8)*(8)
s = 160 + 128
s = 288 m
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
Answer:
303 Ω
Explanation:
Given
Represent the resistors with R1, R2 and RT
R1 = 633
RT = 205
Required
Determine R2
Since it's a parallel connection, it can be solved using.
1/Rt = 1/R1 + 1/R2
Substitute values for R1 and RT
1/205 = 1/633 + 1/R2
Collect Like Terms
1/R2 = 1/205 - 1/633
Take LCM
1/R2 = (633 - 205)/(205 * 633)
1/R2 = 428/129765
Take reciprocal of both sides
R2 = 129765/428
R2 = 303 --- approximated
Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis. What happens as you move away from the center axis toward the coil? What happens above the coil? Outside the coil? Below the coil?
Answer:
Please find the answer in the explanation
Explanation:
Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis.
What happens as you move away from the center axis toward the coil? The direction of the magnetic compass needle will move in an opposite direction since the direction of the induced voltage is reversed.
What happens above the coil?
the needle on the magnetic compass will be deflected. Since compasses work by pointing along magnetic field lines
Outside the coil? The magnetic compass needle will experience no deflection. Since there is no induced voltage or current.
Below the coil?
The needle will move in an opposite direction.