Answer:
you could put the iron filings on the peace of paper and hover a magnet over top of the paper and the iron filings would stand up, or even stick to the magnet
Explanation:
QUICK SOMEONE PLEASE HELP!!!! I’LL MARK BRAINLIEST!!!
Define the following soil conservation technique. Make sure to include pro’s and con’s of this method.
(Terraces)
Answer: Terraces on moderate to steep irregular slopes pro- ... sure of infertile or toxic soils. ... Following are terms used to define distances mea- ... the soil in the entire field will be disturbed to con-.
Explanation:
Two small charged objects repel each other with a force F when separated by a distance d. If the charge on each object is reduced to 1/ 14 of its original value and the distance between them is reduced to d/ 24 the force becomes Group of answer choices F F * 24 / 196 F * 576 / 14 F * 576 / 196 F * 196 / 576
Answer:
Ff = F₀ *(576/196)
Explanation:
Assuming that both charges are equal each other, we can express the repulsion force between the charges (assuming that we can treat them as point charges) using Coulomb's Law, as follows:[tex]F_{o} = \frac{k*q^{2} }{d^{2}} (1)[/tex]
Now, if q reduces to q/14, and d is reduced to d/24, the new value of the force will be:[tex]F_{f} = \frac{k*(q/14)^{2} }{(d/24)^{2}} = \frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} (2)[/tex]
⇒[tex]F_{f} =\frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} = F_{o} * \frac{576}{196} (3)[/tex]
A girl jogs around a horizontal circle with a constant speed. She travels one fourth of a revolution, a distance of 25 m along the circumference of the circle, in 5.0 s. The magnitude of her acceleration is
Answer:
The centripetal acceleration of the girl is 2.468 m/s²
Explanation:
Given;
number of turns, = ¹/₄ Revolution
distance traveled by the girl, d = 25 m
time of motion, t = 5.0 s
The linear speed of the of the girl is calculated as;
[tex]v = \omega \ r\\\\v =(\frac{1}{4}rev \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1}{5 \ s} ) (25 \ m)\\\\v = (0.3142 \ \frac{rad}{s} )(25 \ m)\\\\v = 7.855 \ m/s[/tex]
The centripetal acceleration of the girl is calculated as;
[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(7.855)^2}{25} \\\\a_c = 2.468 \ m/s^2[/tex]
Therefore, the centripetal acceleration of the girl is 2.468 m/s²
Describe Kinetic Energy and Potential Energy (in your own words please!!)
Answer:
Energy stored in an object due to its position is Potential Energy. · Energy that a moving object has due to its motion is Kinetic Energy.
Explanation:
A student pulls on a cart and applies a 20N force at an angle of 25 degrees above the horizontal to
move a crate a distance of 10m across the floor. How much work does the student do?
Answer:
W = 181.26 J
Explanation:
Given that,
The force acting on the cart, F = 20 N
It is at an angle of 25 degrees above the horizontal to move a crate a distance of 10m across the floor.
We need to find work done by the student. The work done by the student is given by :
[tex]W=Fd\cos\theta\\\\W=20\times 10\times \cos25\\W=181.26\ J[/tex]
So, the required work done is 181.26 J.
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 3.05 m/sm/s . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 40.8 mm above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 7.00 ss after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.
Required:
a. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?
b. Where is Henrietta when she catches the bagels?
Answer:
v = 10.46 m/s
x = 30.134 m from house
Explanation:
given data
speed = 3.05 m/s
time = 7 s
height = 40.8 mm
solution
we get here first time required to fall that is t
t = [tex]\sqrt{\frac{2\times 40.8}{9.8}}[/tex] ..................1
t = 2.88 s
now we take here initial speed that is v
so v × t = 3.05 × ( t+ 7)
v = [tex]\frac{3.05\times (2.88 + 7)}{2.88}[/tex]
v = 10.46 m/s
and
when she catch the bagel henrietta was at
x = 3.05 × ( 2.88 + 7)
x = 30.134 m from house
3. A 10-centimeter diameter solid sphere made of a conducting material has 10 micro-Coulombs of charge placed upon it. What is the potential difference between a point on one side of the sphere to a point on the exact opposite side of the sphere
Answer:
zero
Explanation:
For a solid conducting sphere, charges are present on the surface of the sphere due to a phenomenon known as electrostatic sheilding. This affects the charge present in the body and makes it zero. However, the electrostatic potential appears to be equal to the whole present point that shows on the surface. The surface of a spherical conducting solid sphere is known as an equipotential surface. Thus, the potential difference between the two opposite points on the surface of the sphere will also be zero.
Please Help!!!!
When energy is transferred in a system, the total amount of energy before the transfer is _____________ after the transformation is complete, just in different forms.
Group of answer choices
different
lost
transformed into light
the same
Define position
i am not sure?
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antinode is 2.20 mm and the speed of propagation of transverse waves on the string is 260 m/s. The string extends along the x-axis, with one of the fixed ends at x= 0, so that there is a node at x =0. The smallest value of x where there is an antinode is x= 0.150m.
Required:
a. What is the maximum transverse speed of a point on the string at an antinode?
b. What is the maximum transverse speed of a point on the string at x = 0.075 m?
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so [tex]V_{max1}[/tex] = A × ω
[tex]V_{max1}[/tex] = 2.2×10⁻³ × 2722.69 m/s
[tex]V_{max1}[/tex] = 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
[tex]V_{max2}[/tex] = A' × ω
[tex]V_{max2}[/tex] = 1.555×10⁻³ × 2722.69
[tex]V_{max2}[/tex] = 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Collisions question plzz help mee
I need help will mark brainliest
Answer: ITS 1 TRUST ME MAN BYE K
Explanation: OK BYE TRUST YEAH
two objects are moving in the xy plane. no external forces are acting on the objects. object a has a mass of 3.2 kg and has a velocity of v= (2.3m/s)i + (4.2m/s)j and object b has a mass of 2.9kg and has a velocity of v=(-1.8m/s)i = (2.7 m/s)j. sometime later object a is seen to have a velocity va=(1.7m/s)i+(3.5m/s)j what is the velocity of object b at that instant
Answer:
58.469 kg.m/s are moving in the xy plane
a car travels 10 miles east in 30 minutes. what is its velocity in miles per hour. what is its velocity in miles per hour?
Answer:
popu
Explanation:
2u2uwju2i2je82jei
A monk is sitting atop a mountain in complete rest in meditation. What is the kinetic Energy of the monk? (assume mass of 65 kg and the mountain's height was 1000 m)
Answer:
no kinetic energy
hope this helps! :-D
Explanation:
the monk is not moving
I need help please will mark brainliest
Answer: IT B TRUST ME MANN OK
Explanation: OK BYE TRUST
For this assignment, you should mathematically solve and record a video testing your solution for the following prompt: Two rolls of toilet paper, of equal mass and radius, are dropped from different heights so that they hit the ground at the same time. One roll of toilet paper is dropped normally while the other is dropped while a person holds onto a sheet of toilet paper such that the roll unravels as it descends. Determine the ratio of heights h1/h2, where h1 represents the height of the toilet paper dropped normally and h2 represents the height of the toilet paper that unravels, so that both rolls hit the ground at the same time.
Answer:
h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
Explanation:
Using two rolls of tissue paper : One roll dropped normally while the other drops as some holds onto a sheet of the toilet paper ( I.e. the tissue paper drops rotating about its axis )
Determine the ratio of heights h1/h2
mass of tissues = same
radius of tissues = same
h1 = height of tissue 1
h2 = height of tissue 2
For the first tissue ( Tissue that dropped manually )
potential energy = kinetic energy
mgh = 1/2 mv^2
therefore the final velocity ( v^2 ) = 2gH ----- ( 1 )
second tissue ( Tissue that dropped while rotating )
gh = [tex]\frac{v^2}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] ) ------ ( 2 )
To determine the ratio of heights we will equate equations 1 and 2
hence :
gh = [tex]\frac{2gH}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] )
∴ h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
Certain neutron stars (extremely dense stars) are believed to be rotating at about 500 rev/s. If such a star has a radius of 17 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation
Answer:
7.22 × 10²⁹ kg
Explanation:
For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.
So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km
The centripetal force F' = mRω² where R = radius of neutron star and ω = angular speed of neutron star
So, since F = F'
GMm/R² = mRω²
GM = R³ω²
M = R³ω²/G
Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²
Substituting the values of the variables into M, we have
M = R³ω²/G
M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²
M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²
M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²
M = 7217.66 × 10²⁶ kg
M = 7.21766 × 10²⁹ kg
M ≅ 7.22 × 10²⁹ kg
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field towards the other end. If they move fast enough when they strike the positive electrode at the other end, they will give up their energy as X-Rays
(a) Through what potential difference should electrons be accelerated so that their speed is 1% of the speed of light?
(b) What potential difference would be needed to give the protons same kinetic energy as electrons?
(c) What speed would this potential difference give to the protons, both in m/s and as a % of the speed of light.
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV = [tex]\frac{1}{2}[/tex] mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV = [tex]\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }[/tex]
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
[tex]K_e = K_p[/tex]
K_p = ½ m v_e²
K_p = [tex]\frac{1}{2}[/tex] 9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² = [tex]\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }[/tex]
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c = [tex]\frac{7 \ 10^4 }{ 3 \ 10^8}[/tex]
v/c= 2.33 10⁻⁴
what is the angle between 3i-2j-3k and the negative x axis
Answer:
Um its the vbuck card on the 3 thrid level
Explanation:
Bc its a vbuck card you know sihdg;aig
PHYSICS QUESTION PLS HELP
The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is
mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J
The total energy is the same, 970,200 J.
Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.
At point B, the coaster has dropped to a height of 10 m, so it has PE
mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J
which means it must have KE
970,200 J = KE + 294,000 J → KE = 676,200 J
which gives the coast a speed v at point B of
1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J → v ≈ 21.2 m/s
At point C, the coaster has a speed of 16.0 m/s, so it has KE
1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J
and hence PE
970,200 J = 384,000 J + PE → PE = 586,200 J
This lets us determine the height h at C:
mgh = (3000 kg) (9.80 m/s²) h = 586,200 J → h ≈ 19.939 m
which means the loop has diameter h - 10 m ≈ 9.94 m.
At point D, the coaster is 15 m above the ground so its PE at D is
mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J
and so its KE is
970,200 J = KE + 441,000 J → KE = 529,200 J
and hence has speed v at D
1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J → v ≈ 18.9 m/s
A spring with a spring constant of 22 N/m is stretched from equilibrium to 2.9 m. How much work is done in the process?
O A. 186 )
OB. 47 J
O C. 933
OD. 121 )
what is the answer ?
Answer:
using W=1/2kW2
k=22N/m w=2.9
w=1/2×22×2.9×2.9
w=92.51Joules
Approximately 93J answer is C
You push a box across the floor at a constant speed of 1 m/s, applying a horizontal constant force of magnitude 20 N. Your friend pushes the same box across the same floor at a constant speed of 2 m/s, applying a horizontal force. What is the magnitude of the force that your friend applies to the box
Answer:
the force your friend applied on the box is 40 N.
Explanation:
Given;
speed of the box, v₁ = 1 m/s
force applied to the box, F₁ = 20 N
the speed of the box when your friend pushes it, v₂ = 2 m/s
then your friends applied force, F₂ = ?
Assuming the time, t, through which both forces were applied and mass of the box, m, to be constant;
[tex]F_1 = \frac{mv_1}{t} \\\\\frac{m}{t} = \frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
[tex]F_2 = \frac{F_1v_2}{v_1} \\\\F_2 = \frac{20\times 2}{1} \\\\F_2 = 40 \ N[/tex]
Therefore, the force your friend applied on the box is 40 N.
Connecting math to physics
Answer:
wat
Explanation:
Heeeeeeeeelp please
OK please your picture not perfect please try again
Which hormone do ovaries release?
A. estrogen
B. glucagon
C. insulin
D. testosterone
Answer:
A. estrogen
Explanation:
This is released in the female reproductive organ.
A circuit has 12 Amps and 220 Volts. What is the Resistance of the circuit?
Answer:
:To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)
To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)
To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)
To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)
Do you believe you can create a Controlled experiment without an Observational Study? Why or Why not. Include scientific evidence to support your response. PLEASE HELP I BEG YOU.
Answer:
No, it is very unlikely to perform a controlled experiment, because you need to observe the amount or anything from something. Consider someone on the busy street of a New York neighborhood asking random people that pass by how many pets they have, then taking this data and using it to decide if there should be more pet food stores in that area.
1.0 kg clay ball traveling straight down at -10 m/s hits the floor and and sticks on it
Answer:
What am I suppose to solve
Explanation: