Complete question:
A person holds a 0.300 kg pomegranate at the top of a tower that is 96m high. Another person holds a 0.800 kg melon next to an open window 32m up the tower.
Calculate the gravitational potential energy of the pomegranate and melon.
Answer:
The gravitational potential energy of pomegranate is 282.24 J
The gravitational potential energy of melon is 250.88 J
Explanation:
Given;
mass of the pomegranate, m₁ = 0.3 kg
height of the pomegranate, h₁ = 96 m
mass of the melon, m₂ = 0.8 kg
height of the melon, h₂ = 32 m
The gravitational potential energy of pomegranate is calculated as;
P.E₁ = m₁gh₁
P.E₁ = 0.3 x 9.8 x 96
P.E₁ = 282.24 J
The gravitational potential energy of melon is calculated as;
P.E₂ = m₂gh₂
P.E₂ = 0.8 x 9.8 x 32
P.E₂ = 250.88 J
the one that answered this is correct so give him credit not me
Answer:
Complete question:
A person holds a 0.300 kg pomegranate at the top of a tower that is 96m high. Another person holds a 0.800 kg melon next to an open window 32m up the tower.
Calculate the gravitational potential energy of the pomegranate and melon.
The gravitational potential energy of pomegranate is 282.24 J
The gravitational potential energy of melon is 250.88 J
Given;
mass of the pomegranate, m₁ = 0.3 kg
height of the pomegranate, h₁ = 96 m
mass of the melon, m₂ = 0.8 kg
height of the melon, h₂ = 32 m
The gravitational potential energy of pomegranate is calculated as;
P.E₁ = m₁gh₁
P.E₁ = 0.3 x 9.8 x 96
P.E₁ = 282.24 J
The gravitational potential energy of melon is calculated as;
P.E₂ = m₂gh₂
P.E₂ = 0.8 x 9.8 x 32
P.E₂ = 250.88 J
the one that answered this is correct so give him credit not me
Young's double slit experiment is one of the quintessential experiments in physics. The availability of low cost lasers in recent years allows us to perform the double slit experiment rather easily in class. Your professor shines a green laser (560 nm) on a double slit with a separation of 0.108 mm. The diffraction pattern shines on the classroom wall 3.0 m away. Calculate the fringe separation between the third order and central fringe.
Answer:
y = 4.666 10⁻² m
Explanation:
The constructive interference experiment for the double slit
d sin sin θ = m λ
Let's use trigonometry to find a sine relationship.
Tan θ = y / L
tan θ = sin θ/ cos θ
in these experiments the angles are very small
tan θ = sin θ
sin θ = y / L
[tex]d \frac{y}{L}[/tex] = m λ
y = [tex]\frac{ m \lambda \ L}{d}[/tex]
we replace the values
y = 3 560 10⁻⁹ 3.0 / 0.108 10⁻³
y = 4.666 10⁻² m
C
D
7
The sun is the original source of
energy for many of our energy
resources
Which energy resource does not
originate from the sun? *
(1 Point)
.
A. Geothermal
B. Hydroelectric
C. Waves
D. Win
Answer:
geothermal
Explanation:
geothermal energy is the heat energy obtained from within the Earth. Hence not derived from Sun's energy.
A 1.6 kg ball is attached to the end of a 0.40 m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.80 kg block that is at rest on a horizontal frictionless surface. The speed of the block just after the collision is 3 m/s. What is the speed of the ball just after the collision
Answer:
the speed of the ball just after the collision is 1.5 m/s.
Explanation:
Given;
mass of the ball, m₁ = 1.6 kg
initial velocity of the ball, u₁ = 0
mass of the block, m₂ = 0.8 kg
initial velocity of the block, u₂ = 0
final velocity of the block, v₂ = 3 m/s
let the final velocity of the ball after collision = v₁
Apply the principle of conservation of linear momentum for elastic collision;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
1.6 x 0 + 0.8 x 0 = 1.6 x v₁ + 0.8 x 3
0 = 1.6v₁ + 2.4
-1.6v₁ = 2.4
v₁ = -2.4 / 1.6
v₁ = - 1.5 m/s
v₁ = 1.5 m/s (in opposite direction of the block)
Therefore, the speed of the ball just after the collision is 1.5 m/s.
As Carlos and Manual both push a 4 kg box to the right across a rough floor, the box speeds up at a rate of 2.0 m/s2.(Hint, this is the acceleration.) If you assume that Manual pushes with a force strength of 8.0 N and the friction force of the floor on the box is 5.0 N, then what is the strength of the force that Carlos pushes on the box
Answer: 5 N
Explanation:
Given
Mass of box [tex]m=4\ kg[/tex]
acceleration of box [tex]a=2\ m/s^2[/tex]
Force applied by Manual [tex]F_m=8\ N[/tex]
Friction force [tex]f=5\ N[/tex]
The net force on the block is [tex]F_{net}=m\times a=4\times 2=8 \N[/tex]
Suppose [tex]F_c[/tex] is the force applied by Carlos
[tex]\Rightarrow F_m+F_c-f=F_{net}\\\Rightarrow F_c=F_{net}-F_m+f\\\Rightarrow F_c=8-8+5=5\ N[/tex]
The strength of the force that Carlos pushes on the box will be [tex]F_c=5\ N[/tex]
What is force?The force is defined as the external effort applied on any object to move it or to restrict it.
Here following information is given in the question:
Mass of the box m=4 kg
The acceleration of the box [tex]a=2\ \frac{m}{s^2}[/tex]
The force applied by the Manual [tex]F_m=8\ N[/tex]
The frictional force is [tex]F_f=5\ N[/tex]
The net force on the body will be calculated by
[tex]F_{net}=m\times a=4\times 2=8\ N[/tex]
The force that Carlos pushes on the box will be calculated by the equilibrium of the forces:
[tex]F_m+F_c-F_f=F_{net}[/tex]
[tex]8+F_c-5=8[/tex]
[tex]F_c=5\ N[/tex]
Thus the strength of the force that Carlos pushes on the box will be [tex]F_c=5\ N[/tex]
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10. John does 176 J of work lifting himself a distance of 0.40 m. How
much force must John apply to accomplish this?
O 440 m
O 440 J
O 440 N
O 70.4 J
how could you increase the gravitational potential energy of an object without changing its mass and gravity
Explain why the sound waves always reach the observer after the light waves
Answer:
I think it's because the light waves travel faster than the sound waves.
The speed of light is far greater than the speed of sound hence, sound waves always reach the observer after the light waves.
What is the speed of light?Light occurs in the electromagnetic spectrum. Recall that light can be transmitted through vaccuum unlike sound.
The speed of light is far greater than the speed of sound hence, sound waves always reach the observer after the light waves.
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Define electromagnetism
Answer:
the phenomenon of the interaction of electric currents or fields and magnetic fields.
Which is a primary energy source used by power plants to generate electricity?
coal
O wood
o gasoline
batteries
Answer:
Your answer is wood that is what they used
The primary energy source used by power plants to generate electricity is coal.
What are power plants?Power plants are plants which use fuels to generate electricity for use by homes and industries.
Power plants have different energy sources for their fuel.
The primary energy source used by power plants to generate electricity is coal.
Therefore, coal is a primary source of fuel for power plants.
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A baseball is traveling (+30m/s) and is hit by a bat. It leaves the bat traveling (−40m/s). What is the change in its velocity?
Unlike some others, this is any choice, so not just 4 choices!
Answer:
Explanation:
The change must be 30 - - 40 which means it came in a 30 meters / second and went out in the opposite direction at 40 meters / second
The change is 70 m/sec.
You could show it to be - 70 meters per second as well. That's done by making the outgoing direction minus.
Delta v = vf - vi.
Now it depends on which way you define vf and vi.
The bus lay 40 km at a speed of 72 km / h, and then another 60 km at a speed of 30 m / s. Determine the average speed of the bus along the way.
Answer:
25 m/s
Explanation:
From the question,
Average speed = Total distance /total time.
S' = D/T........................... Equation 1
D = 40+60 = 100 Km = 100000 m.
T = t₁+t₂
t₁ = (40×3600/72) s = 2000 s
t₂ = 60000/30 = 2000 s
T = 2000+2000 = 4000 s.
SUbstitute the values of T and D into equation 1
S' = 100000/4000
S' = 25 m/s
10. A 50 kg bicyclist on a 10 kg bicycle speeds up from 5.0 m/s to 10 m/s.
(a) What was the total kinetic energy before accelerating? Full working out
Answer:
T.K.E = 750 Joules.
Explanation:
Given the following data;
Initial velocity, u = 5m/s
Final velocity, v = 10m/s
Mass of bicyclist = 50kg
Mass of bicycle = 10kg
Total mass, Tm = 50 + 10 = 60kg
Kinetic energy can be defined as an energy possessed by an object or body due to its motion.
Mathematically, kinetic energy is given by the formula;
[tex] K.E = \frac{1}{2}mv^{2}[/tex]
Where;
K.E represents kinetic energy measured in Joules.
m represents mass measured in kilograms.
v represents velocity measured in metres per seconds square.
To find the total kinetic energy before accelerating simply means the kinetic energy due to the initial velocity and total mass;
[tex] T.K.E = \frac{1}{2}T_{m}U^{2}[/tex]
Substituting into the equation, we have;
[tex] T.K.E = \frac{1}{2}*60*5^{2}[/tex]
[tex] T.K.E = 30*25 [/tex]
T.K.E = 750 Joules.
Which of the following is an example of the transformation of gravitational potential energy into motion energy (kinetic energy)?
Answer:
c
Explanation:
because a drop of water is falling and that is gravitational potential energy into motion energy
Why do we use copper wiring?
Explanation:
The lower the level of resistivity the more electrical conductivity a metal has. Copper has low resistivity, and therefore is an excellent conductor. Copper is also less oxidative than other metals
5j-Tj=6j+3 Tj
Determine the value of T from the given vector
3. A backpack weighs 8.2 N and has a mass of 5 kg on the moon. What is t
strength of gravity on the moon?
8.2 N
Need a little help here, ASAP please. The subject is simple science, not physics.
4. If the cooling of magma takes place slowly beneath Earth's surface, the rock is that is formed
is called
a. extrusive igneous rock
b. intrusive igneous rock
c. sedimentary rock
d. metamorphic rock
Assume a uniformly charged ring of radius R and charge Q produces an electric field Ering at a point P on its axis, at distance x away from the center of the ring. Now the charge Q is spread uniformly over the circular area the ring encloses, forming a flat disk of charge with the same radius. How does the field Edisk produced by the disk at P compare to the field produced by the ring at the same point?Assume a uniformly charged ring of radius R and charge Q produces an electric field Ering at a point P on its axis, at distance x away from the center of the ring. Now the charge Q is spread uniformly over the circular area the ring encloses, forming a flat disk of charge with the same radius. How does the field Edisk produced by the disk at P compare to the field produced by the ring at the same point?
Answer:
* E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]
*E_ disk= 2kQ [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]
Explanation:
Let's start by finding the electric field of the charged ring
in the attachment we can see a diagram of the system. Due to circular symmetry, the electric field perpendicular to the axis is canceled and only the electric field remains parallel to the axis.
Eₓ = E cos θ (1)
E = k ∫ [tex]\frac{dq}{r^2}[/tex]
cos θ = x / r
using the Pythagorean theorem
r = [tex]\sqrt{x^2 + y^2}[/tex]
we substitute
Eₓ = k ∫ [tex]\frac{dq}{x^2+y^2} \ \frac{x}{\sqrt{ x^2+y^2} }[/tex]
Eₓ = [tex]k \frac{x}{(c^2+y^2)^{3/2} }[/tex] ∫ dq
Eₓ = k \frac{x}{(c^2+y^2)^{3/2} } Q
the ring's electric field
E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]
Now let's find the electric field of the disk
The charge is distributed over the entire disk, so let's use the concept of charge density
σ = [tex]\frac{dq}{dA}[/tex]
Let's approximate the disk as a group of rings, the width of each ring is dr, the area is
dA = 2πr dr
we substitute
σ = [tex]\frac{1}{2\pi r} \ \frac{dq}{dr}[/tex]
dq = 2π σ r dr
we substitute in equation 1, where the electrioc field is of each ring
Eₓ = [tex]k \int\limits^R_0 \ { \frac{x}{(x^2+r^2)^{3/2} } \ 2\pi \sigma \ r } \, dr[/tex]
if we use a change of variable
dv = 2rdr
v = r²
Eₓ = [tex]k x \pi \sigma \int\limits^a_b { \frac{1}{(x^2+v)^{3/2} } } \, dv[/tex]
we integrate
Eₓ = k x π σ [tex][ \frac{ (x^2 + r^2)^{-1/2} }{-1/2} ][/tex]
we value in the limits from r = 0 to r = R
Eₓ = k π σ x (-2) [ [tex]\frac{1}{ \sqrt{x^2+R^2} } - \frac{1}{x}[/tex]]
Eₓ = 2π k σ ([tex]1 - \frac{x}{(x^2 + R^2 ) ^{1/2} }[/tex] )
σ = Q/πR²
substitute
Eₓ = 2 k Q/R² (1 - \frac{x}{(x^2 + R^2 ) ^{1/2} } )
E_ disk= 2kQ [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]
The two electric fields are
* E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]
*E_ disk= 2kQ [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]
we can see that the functional relationship of the two fields is different
True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then the increase in the velocity as it falls to the ground would increase its potential energy.
If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then the increase in the velocity as it falls to the ground would increase its potential energy. _ This is false statement.
What is projectile motion?When a particle is hurled obliquely near the surface of the earth, it travels along a curved path while accelerating continuously in the direction of the planet's center (we assume the particle stays close to the surface of the globe). Such a particle's motion is known as projectile motion, and its route is referred to as a projectile.
In projectile motion total energy is conserved. Hence, when a rock is thrown into the air, the increase in the height would increase the rock’s potential energy, and then the increase in the velocity as it falls to the ground would increase its kinetic energy.
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A marching band consists of rows of musicians walking in straight, even lines. When a marching band performs in an event, such as a parade, and must round a curve in the road, the musician on the outside of the curve must walk around the curve in the same amount of time as the musician on the inside of the curve. This motion can be approximated by a disk rotating at a constant rate about an axis perpendicular to its plane. In this case, the axis of rotation is at the inside of the curve. Consider two musicians, Alf and Beth. Beth is four times the distance from the inside of the curve as Alf. Knowing that If Beth travels a distance s during time Δt, how far does Alf travel during the same amount of time= (1/4)s
If Alf moves with speed v, what is Beth's speed?
a. 4v
b. v
c. v/4
Answer:
A) Total distance Alf travel during the same amount of time = [tex]\frac{1}{4}s[/tex]
B) Speed of Beth = 4v
Explanation:
Given - Consider two musicians, Alf and Beth. Beth is four times the
distance from the inside of the curve as Alf.
To find - A) Knowing that If Beth travels a distance s during time Δt, how
far does Alf travel during the same amount of time.
B) If Alf moves with speed v, what is Beth's speed?
Proof -
A)
As we know that
Speed = Distance / Time
⇒Distance = Speed ×Time
Now,
Given Speed of Alf = v
⇒Distance of Alf = vt
Also,
Distance of Beth = Speed of beth×time
Given that
Beth is four times the distance from the inside of the curve as Alf. Knowing that If Beth travels a distance s during time Δt, how far does Alf travel during the same amount of time=
⇒Distance of Alf = Speed of Alf×time
= [tex]\frac{1}{4}[/tex] Distance of Beth
⇒Distance of Alf = [tex]\frac{1}{4}s[/tex]
∴ we get
Total distance Alf travel during the same amount of time = [tex]\frac{1}{4}s[/tex]
B)
We know the conversion of angular velocity
ω(Alf) = ω(Beth)
⇒V(Alf)/ r = V(Beth)/4r
⇒V(Alf) = V(Beth) / 4
⇒V(Beth) = 4 V(Alf)
As given, Alf moves with speed v
⇒V(Beth) = 4v
So, the correct option is - a.4v
Which water on earth is the largest,second to the largest,3rd to the largest,fourth to the largest and 5th to the largest?
Answer:
Explanation:
There are a total of 5 differen classification for the types of bodies of water that exist on Earth. These 5 would be the following: Oceans, seas, lakes, rivers, and canals. Oceans are the biggest bodies of water on Earth. If we ranked them by their size the ranking would be the following...
Pacific Ocean - being 155,556,651 sq km. in size
Atlantic Ocean - being 76,761,938 sq km. in size
Indian Ocean - being 68,555,923 sq km. in size
Antarctic / Southern Ocean - being 20,327,001 sq km. in size
Arctic Ocean - being 14,055,930 sq km. in size
Why do scientists perform multiple trials of the same experiment?
Answer:
it's a good idea to do multiple trials, that is, do the same experiment lots of times. When we do multiple trials of the same experiment, we can make sure that our results are consistent and not altered by random events. Multiple trials can be done at one time.
Explanation:
the last one which is to increase the likelihood of accurate experiment results
Explanation:
A 65 kg cart travels at a constant speed of 4.6 m/s. What is its kinetic energy?
Answer:
Explanation:
mass (m) = 65 kg
velocity (v) = 4.6 m/s
Kinetic energy (KE)
= 1/2 * m * v²
= 1/2 * 65 * 4.6²
= 687.7 J
hope it helps :)
What color is a carrot?
Answer:
reddish-orrange
Explanation:
please mark me as brainliest
An instrument that produces an electric current in the presence of
a radioactive substance is called a
a. particle accelerator.
b. Geiger counter.
c. cloud chamber.
d. bubble chamber.
Answer:
Option B, Geiger counter
Explanation:
NEED HELP What color is the container for R-134a refrigerant? A. Light blue B. Yellow C. Dark green D.White and yellow
Answer:
It is A. Light blue
_____________
Hope this helps!
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Can you help me please here is the picture DUE NOW PLEASE.
they are formed when hot
it's dependent on the rate of cooling of the melt, slow cooling allows large crystals to form, fast cooling yields small crystals. They cool too quickly to form crystals.
Hope this helps! <3 :3
PLEASE HELP ME WITH THIS ONE QUESTION
How many joules are needed to vaporize 125 g of liquid water at 100°C? The latent heat of vaporization is 2258 J/g.
Answer:
Q = 282,000 J
Explanation:
Given that,
The mass of liquid water, m = 125 g
Temperature, T = 100°C
The latent heat of vaporization, Hv = 2258 J/g.
We need to find the amount of heat needed to vaporize 125 g of liquid water. We can find it as follows :
[tex]Q=mH_v\\\\Q=125\ g\times 2285\ J/g\\\\Q=282250\ J[/tex]
or
Q = 282,000 J
So, the required heat is 282,000 J .
ball A is dropped from a hot air balloon rising at a costant velocity of 14,7 m.s'1 at a height of 19,7 m above the ground.the ball took 1.5s to reach its maximum height and hits the ground after some time in air.ignore the effects of air resistanceUse the ground as zero reference.3.1.1calculate the maximum height reached by the ball above the ground
Answer:
this slow site thinks the answer is a link
Explanation:
this was a week ago so i dont know if u still need help