Answer:
e) A changing magnetic field produces an electric field.
Explanation:
Ok, we start with a magnetic field and let's study how it affects the motion of a single electron. As the magnetic field changes, it will cause an electromotive force, that moves the electron, and because now we have a moving electron, now we will have an electric field. (Such that the direction of the electromotive force opposes the direction in which the magnetic field changes). This also can be deduced if we look at the third Maxwell's equation:
dE/dx = -dB/dt
This says that the spatial change in an electric field depends on how the magnetic field changes as time pass.
Then the correct option is e) A changing magnetic field produces an electric field.
What is the difference between a positively and negatively charged object?
Answer:
Positively charged objects have electrons; they simply possess more protons than electrons. Negatively charged objects have protons; it's just their number of electrons is greater than their number of protons.
The difference between a positively charged object and a negatively charged object is the number of protons and electrons. The imbalance in charge results into formation of charged objects.
What are Charged objects?
Charged objects have an imbalance of charge that is either more negative electrons than the positive protons or more positive protons than the negative electrons in the object. The neutral objects are those species which have a balance of charge with equal number of protons and electrons.
A positively charged object is formed when an atom has more protons than electrons. And, a negatively charged object is formed when an atom has more electrons than protons. As, electrons have a negative charge and protons have a positive charge.
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An amusement park ride swings back and forth once every 60.0 s. What is the ride's frequency?
The Answer choices are in the image above
Answer:
Option B. 0.016 Hz
Explanation:
From the question given above, the following data were obtained.
Period (T) = 60 s
Frequency (f) =?
Frequency and period are related according to the following equation:
Frequency = 1 / period
f = 1/T
With the above formula, we can obtain the frequency of the ride as follow:
Period (T) = 60 s
Frequency (f) =?
f = 1/T
f = 1/60
f = 0.016 Hz
Thus, the rider's frequency is 0.016 Hz
What is the height of a copper cylinder ( ρCu = 8.96 gcm-3) of diameter 10 cm with a mass of 10 kg ?
Answer:
h=0.142m=14.2cm
Explanation:
ρ=8.96 g/cm3=8960 kg/m3
d=0.1m
ρ=m/V - - >ρ=m/[π*((d/2)^2) *h] - - >
8960=10/[π*((0.1)^2) *h] - - >
h=0.142m
Which one of the following
statements about electric
current is correct?
- Current is calculated as the amount of charge that pass
a point on a circuit per time.
- Current is calculated as the amount of resistance a
charge encounters in a given time.
- Current is calculated as the amount of energy a charge
loses per unit of time.
- Current is calculated as the distance that a charge
moves per unit of time
Answer:
Current is calculated as the amount of charge that pass a point on a circuit per time.
Explanation:
Electric current is defined as the amount of charge that passes a point on a circuit per unit time. The mathematical definition of electric current is given by :
[tex]I=\dfrac{q}{t}[/tex]
Where
q is a charge (Coulomb)
t is time (in seconds)
The SI unit of electric current is A. It is equivalent to C/s. So, the correct option is (A).
Why must you bend forward when carrying a
heavy load on your back?
1. The gravitational force has decreased.
2. Angular momentum has decreased.
3. The center of gravity has shifted.
4. Inertia has changed.
Answer:
hi I thinks its number 3
Explanation:
hope you have a nice day
A proton (with charge of 1.6 x 10^-19 C and mass of 1.7*10^-27 kg) traveling at a speed of 57,600,630 m/s in the + x-direction enters a region of space where there is a magnetic field of strength 0.5 T in the - z-direction. What would be the radius of the circular motion that the proton would go into if it is "trapped" in this magnetic field region?
Answer:
r = 1,224 10⁻² m
Explanation:
For this exercise let's use Newton's second law
F = m a
the force is magnetic
F = q v x B
The bold letters indicate vectors, the module of this expesion is
F = q v B
The direction of the force is found by the right hand rule
thumb points in the direction of the velicad + x
fingers extended in the direction of B -z
the palm is in the direction of the force + and
the acceleration of the proton is cenripetal
a = v² / r
we substitute
q v B = m v² / r
r = [tex]\frac{m \ v}{q \ B}[/tex]
let's calculate
r = [tex]\frac{1.7 \ 10^{-27} \ 5.760063 \ 10^7 }{1.6 \ 10^{-19} \ 0.5 }[/tex]
r = 1,224 10⁻² m
Scroll over the answer choices to see the images. Choose all of the true statements concerning the corresponding image
The picture depicts an electric motor which turns electrical energy into mechanical energy.
The picture depicts a generator which turns electrical energy into mechanical energy
A series circuit is a good example of an electromagnet
F more wite is wound around this iron mail, the strength of the electromagnetic is increased
This image devices that the direction of the magnetic field does not depend on the direction of the current
Flow of electron |
Answer:
Explanation:
I jus did it on usatestprep
A simple generator is used to generate a peak output voltage of 23.0 V . The square armature consists of windings that are 5.1 cm on a side and rotates in a field of 0.500 T at a rate of 55.0 rev/s .
How many loops of wire should be wound on the square armature?
Answer: 51
Explanation:
Given
Output is 23 V
The square armature side is [tex]a=5.1\ cm[/tex]
Magnetic field [tex]B=0.5\T[/tex]
Rate of revolution [tex]n=55\ rev/s[/tex]
Angular speed
[tex]\omega =2\pi n\\\omega=2\pi \times 55=110\pi\ rad/s[/tex]
Peak voltage is given by
[tex]E_{peak}=NB\omega A\quad [\text{N=Number of windings; A=area of cross-section}]\\\\N=\dfrac{E_{peak}}{B\omega A}\\\\N=\dfrac{23}{0.5\times 110\pi\times (0.051)^2}\approx 51[/tex]
So, there are approximate 51 loops
A rifle fires a pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by from its unstrained length. The pellet rises to a maximum height of 6.10 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.
This question is incomplete, the complete question is;
A rifle fires a 2.10 × 10⁻² kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 9.10 × 10⁻² m from its unstrained length. The pellet rises to a maximum height of 6.10 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.
Answer:
the spring constant is 303.5 N/m
Explanation:
Given the data in the question;
There is a potential energy associated with the spring;
we know that potential energy = kinetic energy
mgh = [tex]\frac{1}{2}[/tex]kx²
where k is the spring constant and x is the compression
as the pallet is fired, the spring gives kinetic energy which is converted into gravitational potential energy
so
m = 2.10 × 10⁻²
g = 9.81 m/s²
h = 6.10 m
x = 9.10 × 10⁻² m
we substitute
mgh = [tex]\frac{1}{2}[/tex]kx²
2.10 × 10⁻² × 9.81 × 6.10 = [tex]\frac{1}{2}[/tex] × K × ( 9.10 × 10⁻² )²
1.256661 = 0.0041405 × K
K = 1.256661 / 0.0041405
K = 303.5 N/m
Therefore, the spring constant is 303.5 N/m
Question 8: Unspooling Thread (100 points) A 110 g spool of thread with a 4.2 cm radius is held up by a peg through its center and allowed to freely rotate. Assume the thread is ideal (i.e., it does not stretch or slip, and its mass is negligibly small). A 160 g needle is tied to the loose end of the thread. The needle is dropped, and it accelerates to the floor as the thread unwinds. Find the tension in the thread and the magnitude of the acceleration of the needle as it falls.
Answer:
a = 7.29 m / s², T = 0.40 N
Explanation:
To solve this exercise we must apply Newton's second law to each body
The needle
W -T = m a
mg - T = ma
The spool, which we will approach by a cylinder
Σ τ = I α
T R = I α
the moment of inertia of a cylinder with an axis through its center is
I = ½ M R²
angular and linear variables are related
a = α R
α = a / R
we substitute
T R = (½ M R²) a / R
T = ½ M a
we write our system of equations together
mg - T = m a
T = ½ M a
we solve
m g = (m + ½ M) a
a = [tex]\frac{m}{m + \frac{1}{2} M} \ g[/tex]
let's calculate
a = [tex]\frac{0.160}{0.160 + \frac{1}{2} 0.110} \ 9.8[/tex]
a = 7.29 m / s²
now we can look for the tension
T = ½ M a
T = ½ 0.110 7.29
T = 0.40 N
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 3.00 s for the boat to travel from its highest point to its lowest, a total distance of 0.700 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, how fast are the waves traveling ?
d. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, what is the amplitude of each wave?
Answer:
a) v = 0.9167 m / s, b) A = 0.350 m, c) v = 0.9167 m / s, d) A = 0.250 m
Explanation:
a) to find the velocity of the wave let us use the relation
v = λ f
the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength
λ = x
λ = x
the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period
T / 2 = t
T = 2t
period and frequency are related
f = 1 / T
f = 1 / 2t
we substitute
v = x / 2t
v = 5.50 / 2 3
v = 0.9167 m / s
b) the amplitude is the distance from a maximum to zero
2A = y
A = y / 2
A = 0.700 / 2
A = 0.350 m
c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same
v = 0.9167 m / s
d) the amplitude is
A = 0.500 / 2
A = 0.250 m
a student starts his lawnmower by applying a constant tangential force of 150 N to the 0.3 kg disk-shaped flywheel. the radius of the flywheel is 18 cm. what is the flywheels angular acceleration? b. what is the angular speed of the wheel after it has turned through one revolution,( neglect friction and motor compression.) c. what is the tangent speed of a point on the rim of the flywheel?
Answer:
okay
Explanation:
please I don't know
Astronomers define the __________ as all of space and everything in it. It is enormous, almost beyond imagination. Question 2 options: galaxy none of these universe solar system
Answer:
Universe
Explanation:
I took the quiz.
A standard 1kilogram weight is a cylinder 50.5mm in height and 52.0mm in diameter. What is the density of the meterial?(kg/m^3)
Answer:
The correct answer is - 93.24×10^4 kg/m^3.
Explanation:
Given:
height of cylinder: 50.5 mm
diameter = 52.0
then radius will be diameter/2 = 52/2 = 26
Formula:
Density = mass/ volume
Volume = πr^2h
solution:
Now the volume of a cylinder is v = (22/7)×r^2×h
= 22/7×26×26×50.5
= 107261.59 mm^3
Now volume in cubic meter V =10.7261 ×10^(-5) m^3
So density d = m/V = 1/(10.7261 ×10^(-5))
Or d = 93.24×10^4 kg/m^3
1. Although mercury is a metal, it is a liquid at room temperature. Mercury melts at about -39°C. If
the temperature of a block of mercury starts at -54°C and increases by 22°C, does the mercury melt?
Explain your answer.
Answer:
I think so because if it starts at a low temperature for that material, it should melt when you bring it up to that temperature.
Answer: Mercury
Explanation: Mercury is kinda like the opposite of regular semi-liquid. For ice to melt you need Fahrenheit to melt ice into water while you need Celsius to melt mercury.
Consider a pulley of mass mp and radius R that has a moment of inertia 1/2mpR2. The pulley is free to rotate about a frictionless pivot at its center. A massless string is wound around the pulley and the other end of the rope is attached to a block of mass m that is initially held at rest on frictionless inclined plane that is inclined at an angle β with respect to the horizontal. The downward acceleration of gravity is g. The block is released from rest .
How long does it take the block to move a distance d down the inclined plane?
Write your answer using some or all of the following: R, m, g, d, mp,
Answer:
a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex] , t = [tex]\sqrt{ \frac{2d}{a} }[/tex]
Explanation:
To solve this exercise we must use Newton's second law
For the block
let's set a reference system with the x axis parallel to the plane
X axis
Wₓ - T = m a
Y axis
N- W_y = 0
N = W_y
for pulley
∑τ = I α
T R = (½ m_p R²) α
let's use trigonometry for the weight components
sin β = Wₓ / W
cos β = W_y / W
Wx = W sin β
angular and linear variables are related
a = α R
α = a / R
we substitute and group our equations
W sin β - T = m a
T R = ½ m_p R² (a / R)
W sin β - T = m a
T = ½ m_p a
we solve the system of equations
W sin β = (m + ½ m_p) a
a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex]
let's find the time to travel the distance (d) through the block
x = v₀ t + ½ a t²
d = 0 + ½ a t²
t = [tex]\sqrt{ \frac{2d}{a} }[/tex]
Which one of the following statements is not true of free falling object
Answer:
FORCE as for my answer....
A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal disk is placed in contact with a wall, and the disk comes to rest after 10s. Which of the following situations associated with linear impulse is analogous to the angular impulse that is described?
a. A 3kg block is initially at rest. An applied force of 3N is applied to the block, but the block does not move.
b. A 3kg block is initially at rest. A net force of 3N is applied to the block until it has a speed of 10m/s.
c. A 3kg block is initially traveling at 10m/s. An applied force of 3N is applied to the block in the direction of its velocity vector for 10s.
d. A 3kg block is initially traveling at 10m/s. The block encounters a 3N frictional force until the block eventually stops.
Answer:
D
Explanation:
From the information given:
The angular speed for the block [tex]\omega = 50 \ rad/s[/tex]
Disk radius (r) = 0.2 m
The block Initial velocity is:
[tex]v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s[/tex]
Change in the block's angular speed is:
[tex]\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s[/tex]
However, on the disk, moment of inertIa is:
[tex]I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2[/tex]
The time t = 10s
∴
Frictional torques by the wall on the disk is:
[tex]T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m[/tex]
Finally, the frictional force is calculated as:
[tex]F = \dfrac{T}r{}[/tex]
[tex]F= \dfrac{0.6}{0.2} \\ \\ F = 3N[/tex]
I WILL REPORT YOU IF YOU DON'T ANSWER QUESTION OR IF YOU PUT A LINK
Which of the following statements are true
What is characteristic of a good insulator?
A. Electrons are usually not moving at all.
B. Electrons are free to move around.
C. Electrons are semi-free to move around.
D. Electrons are tightly bound to the nuclei.
Answer:
D. Electrons are tightly bound to the nuclei.
Explanation:
In an insulator, the electrons of the outer most shell are bound with a very high electrostatic forces coming from the nucleus of each atom so electrons cannot flow around all atoms making up the material as in a conductor.
The characteristic of a good insulator is Electrons are tightly bound to the nuclei. (option d)
In a good insulator, electrons are tightly bound to the nuclei of their atoms. This means that they are not free to move around within the material, unlike conductors where electrons are relatively loosely bound and can move freely. Due to this strong binding, electrons in insulating materials cannot carry an electric charge or energy easily from one atom to another.
When an electric field is applied to an insulator, the electrons may experience a small displacement within their respective atoms, but they generally do not move from one atom to another or flow through the material like they would in a conductor. As a result, insulators prevent the flow of electric current and are used to isolate or protect conductive elements from accidental contact.
So, the correct answer is D. Electrons are tightly bound to the nuclei.
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Which one of the statements below is true about mechanical waves?
They must travel in empty space.
They can travel in a vacuum.
Both sound and light are examples of mechanical waves.
They require a medium to travel through.
Answer:
D) Mechanical waves require a medium for transmission (wire, air, etc.) as opposed to electromagnetic which travel through empty space - light, radio, etc.)
Two solenoids of equal length are each made of 2000 turns of copper wire per meter. Solenoid I has a 5.00 cm radius; solenoid II a 10.0 cm radius. When equal currents are present in the two solenoids, the ratio of the magnitude of the magnetic field BIalong the axis of solenoid I to the magnitude of the magnetic field BIIalong the axis of solenoid II, BI/BII, is
Answer:
BI/BII = 1
Explanation:
The magnetic field due to a solenoid is given by the following formula:
[tex]B = \mu nI\\[/tex]
where,
B = Magnetic Field due to solenoid
μ = permeability of free space
n = No. of turns per unit length
I = current passing through the solenoid
Now for the first solenoid:
[tex]B_1 = \mu n_1I_1 \\[/tex]
For the second solenoid:
[tex]B_2 = \mu n_2I_2\\[/tex]
Dividing both equations:
[tex]\frac{B_1}{B_2} = \frac{\mu n_1I_1}{\mu n_2I_2}\\[/tex]
here, no. of turns and the current passing through each solenoid is same:
n₁ = n₂ and I₁ = I₂
Therefore,
[tex]\frac{B_1}{B_2} = \frac{\mu nI}{\mu nI}\\[/tex]
BI/BII = 1
Vector A is 3 units length and points along the positive x-axis vector B is 4 units in length and points along negative y-axis .find magnitude and direction of the vector (a) A+B (b)A-B
Answer:
a) < 3 , -4 >
b) < 3 , -4 >
Explanation:
a) If you can imagine this, adding vectors is like putting them "tip to tail", where you put the beginning point of vector B to the end point of vector A (or vice versa). Your new vector (A+B) would be from the "tip" of vector A to the "tail" of vector B.
Mathematically, this is the same as adding the x-components of each vector together as well as the y-components.
Vector A: 3 units along the positive x-axis: < 3 , 0 >
Vector B: 4 units along the negative y-axis: < 0 , -4 >
A+B = < 3 , 0 > + < 0 , -4 > = < (3+0) , (0+(-4)) > = < 3 , -4 >
b) Subtracting is like adding a negative, so you could use the same "tip to tail" visual by adding the negative of vector B instead (which is B in the opposite direction).
Vector A: < 3 , 0 >
Vector B: < 0 , -4 >
Vector -B: < 0 , 4 >
A-B = A+(-B) = < 3 , 0 > + < 0 , 4 > = < (3+0) , (0+4) > = < 3 , 4 >
What does a step-up transformer do?
A. It steps up the energy.
B. It steps up the power.
C. It steps up the voltage.
D. It steps up the current.
PLEASE HELP and actually help plz
The position of masses 4kg, 6kg, 7kg, 10kg ,and 3kg are (0,1), (4,2), (3,5), (5,6), and (-2,4) respectively. Where must you place a mass of 13kg if you want the center of mass to be at (-1,-3)?
Answer:
iEvaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)Evaluate for \(x=2.\)
Explanation:
A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of 0.20 m and released from rest. Find the speed with which the object passes through its original position on the way up.
Answer:
e = 0.0898m
v = 2.07m/s
Explanation:
a) According to Hooke's law
F = ke
e is the extension
k is the spring constant
Since F = mg
mg = ke
e = mg/k
Substitute the given value
e = 1.1(9.8)/120
e = 10.78/120
e = 0.0898m
Hence it is stretched by 0.0898m from its unstrained length
2) Total Energy = PE+KE+Elastic potential
Total Energy = mgh +1/2mv²+1/2ke²
Substitute the given value
5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²
Solve for v
5.0 = 2.156+0.55v²+0.48338
5.0-2.156-0.48338= 0.55v²
2.36 =0.55v²
v² = 2.36/0.55
v² = 4.29
v ,= √4.29
v = 2.07m/s
Hence the required velocity is 9.28m/s
A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result of a directed force combined with a friction force. As a result of the applied torque the angular speed of the wheel increases from 0 to 10.3 rad/s. After 6.10 s the directed force is removed, and the wheel comes to rest 60.6 s later.
(a) What is the wheel's moment of inertia (in kg m2)? kg m
(b) What is the magnitude of the torque caused by friction (in N m)? N m
(c) From the time the directed force is initially applied, how many revolutions does the wheel go through?
______ revolutions
Answer:
[tex]21.6\ \text{kg m}^2[/tex]
[tex]3.672\ \text{Nm}[/tex]
[tex]54.66\ \text{revolutions}[/tex]
Explanation:
[tex]\tau[/tex] = Torque = 36.5 Nm
[tex]\omega_i[/tex] = Initial angular velocity = 0
[tex]\omega_f[/tex] = Final angular velocity = 10.3 rad/s
t = Time = 6.1 s
I = Moment of inertia
From the kinematic equations of linear motion we have
[tex]\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2[/tex]
Torque is given by
[tex]\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2[/tex]
The wheel's moment of inertia is [tex]21.6\ \text{kg m}^2[/tex]
t = 60.6 s
[tex]\omega_i[/tex] = 10.3 rad/s
[tex]\omega_f[/tex] = 0
[tex]\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2[/tex]
Frictional torque is given by
[tex]\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}[/tex]
The magnitude of the torque caused by friction is [tex]3.672\ \text{Nm}[/tex]
Speeding up
[tex]\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}[/tex]
Slowing down
[tex]\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}[/tex]
Total number of revolutions
[tex]\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}[/tex]
[tex]\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}[/tex]
The total number of revolutions the wheel goes through is [tex]54.66\ \text{revolutions}[/tex].
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel during 30 seconds of rotation?
Distance travelled in one round will be equal to the circumference of the disc i.e [tex] 2 \pi r[/tex]
Radius=50cm
Circumference= [tex] 2 \times \frac{22}{7} \times 50cm=> \frac{2200}{7} cm[/tex]
If the disc rotates at speed of 100rpm that means it completes 100 rotation in a minute(60 second)
So, in 30 seconds it will complete 50 rotation.
1 rotation = [tex] \frac{2200}{7} cm [/tex]
[tex] 50 rotation=\frac{2200}{7} \times 50cm \\\\ \frac{110000}{7} cm[/tex]
Break it in decimal.
Your answer will be 15714.2 cm
A physicist at a respected research laboratory reports a startling new
discovery. Her conclusions are based on data from many trials. However,
other scientists are unable to reproduce the results of the experiment
Which statement tells why the original experiment might be faulty?
A. The conclusion are based on multiple trials
B. The results are announced to the public
C. Experimental results cannot be produced
D. The experiment does not include sophisticated equipment
Does latitude has an effect on weight? PLEASE HELP!
Answer:
yes
it does you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.