Answer:
It is direct proportionality. The greater the mass, the greater is the gravitational potential energy. The equation for GPE is : GPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height above the ground. As you can see GPE is directly proportional to mass, and height. KT.
Explanation:
Gravitational potential energy is a function of both the mass of your system and the mass of the thing generating the gravity field around your system.
The relationship is linear, which means that if you multiply or divide one of the masses by some number but leave everything else the same, you multiply or divide the potential energy by the same number. A 3kg mass has three times the gravitation potential energy of a 1kg mass, if placed in the same location.
What’s the physics of a football
Answer:
There are many forces involved in the game of football. These are: Force of Gravity, Normal Force, Force of Friction, and Applied Force. Force of Gravity applies to football when the football is thrown or kicked, when a player jumps in the air to avoid a tackle or catch a ball, and is constantly being applied.
Explanation:
How physics is used in football?
When you throw a football across the yard to your friend, you are using physics. You make adjustments for all the factors, such as distance, wind and the weight of the ball. The farther away your friend is, the harder you have to throw the ball, or the steeper the angle of your throw.
what is value of 2.37 in sl units
about conserving and protecting our coral reefs. 3. Punch a hole and put ribbon or string on it. Rubrics will serve as your guide in creating your bookmark. (SAMPLE BELOW) Protect Keep it our clean! Corals! de Questions: 1. What was your awareness message all about in your bookmark? 5. Do you believe there is a need to protect our coral reefs after doing this activity? Why? The Tubbataha Reef in Palawan is our country's pride. It is one of UNESCO World Heritage Sites. What would you suggest our government do to protect this?
The above question requires a personal answer about your perception of corals, environmental conservation, and the activity you did in the classroom. In that case, I can't answer your question, but I'll show you how to answer it.
Please consider the following information to answer your questions:
The awareness message about corals refers to their preservation, due to their environmental and economic importance.After doing your activity you should be able to understand the importance of preserving corals. That's because they are important for fish spawning and habitat, blocking wave erosion, and protecting coastal communities.To protect coral reefs, the government must map reef areas, reduce marine pollution, monitor fish exploitation, protect coral areas and encourage environmental education and awareness of nature conservation.In summary, coral reefs must be protected so that the natural marine balance is maintained.
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Question:
The US Navy is developing a railgun that uses magnetic field repulsion. The railgun can propel a 981 N projectile at seven times the speed of sound. How much energy is stored in this magnetic field? Assume a complete conversion of energy and a speed of sound of 340 m/s.
Answer choices:
A. 119 MJ
B. 283 MJ
C. 578 MJ
D. 2780 MJ
The relation of the kinetic energy allows to find the correct result for the energy stored in the electromagnet is:
B) 283 MJ
Kinetic energy is the energy due to the movement of bodies.
K = ½ m v²
Where K is the kinetic energy, m is the mass and v the spped.
They indicate that the weight of the bodye is W = 981 N and its final velocity is v = 7 [tex]v_s[/tex].
W = m g
Since the projectile starts from rest, its initial velocity is zero, therefore the change in energy is
ΔK = [tex]K_f - K_o = K_f[/tex]
we substitute
ΔK = ½ 981 / 9.8 (7 340) ²
ΔK = 2.835 10⁸ J
They indicate that all the energy of the electromagnet is transformed into the energy of the projectile,
Em = K
When reviewing the results, the correct one is:
B) 283 MJ
In conclusion, using the relationship of kinetic energy we can find the correct result for the energy stored in the electromagnet is:
B) 283 MJ
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the impulse-momentum relationship is a direct result of
Newton's second law.
__________________
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The impulse-momentum relationship is a direct result of Newton's second law of motion.
What is the impulse-momentum?The impulse-momentum theorem states that the impulse applied to an object is equal to the change in its momentum. It proves that the change in momentum of an object depends not only on the amount of force applied but also on the duration of force applied.
Newton's second law states that the force acting on an object is equal to the rate of change of its momentum.
This means that a force applied to an object will cause a change in its momentum. The impulse-momentum relationship describes the relationship between the force applied to an object and the resulting change in its momentum.
The impulse-momentum relationship states that the impulse acting on an object is equal to the change in its momentum.
Impulse is defined as the force applied to an object over a period of time, while momentum is the product of an object's mass and velocity. Therefore, the impulse-momentum relationship can be expressed as:
Impulse = Change in momentum
This relationship is important in understanding the behavior of objects in motion, particularly in collisions or other situations where forces act over a period of time.
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The magnetic field 0. 02 m from a wire is 0. 1 T. What is the magnitude of the magnetic field 0. 01 m from the same wire? 0. 01 T 0. 05 T 0. 1 T 0. 2 T.
The magnitude of the magnetic field 0.01 m from the same wire is 0.2 T.
Given to us:
Magnetic field, [tex]B_1 = 0.1\ T[/tex]
Radius of wire, [tex]R_1 = 0.02\ m[/tex]
To find out the magnitude of the magnetic field 0. 01 m from the same wire, we need to find out current first. we will use the formula,
[tex]B = \dfrac{\mu_oI }{2\pi R},\\\rn\\where,\\B= magnetic\ field\\\mu_o = 4\pi\times 10^{-7} m\cdot kg\cdot s^{-2} A^{-2}\ is\ the\ magnetic\ constant\\I= current\\R= radius\ of\ the\ wire[/tex]
Putting the values,
[tex]B_1 = \dfrac{\mu_oI }{2\pi R_1},\\\rn\\\\0.1= \dfrac{4\times \pi \times 10^{-7}\times I}{2\times \pi\times0.02}\\\\I=10,000\ A[/tex]
Now, for [tex]B_2[/tex]
[tex]B_2 = \dfrac{\mu_oI }{2\pi R_2},\\\rn\\\\B_2= \dfrac{4\times \pi \times 10^{-7}\times 10,000}{2\times \pi\times0.01}\\\\B_2= 0.2\ T[/tex]
Hence, the magnitude of the magnetic field 0. 01 m from the same wire is 0.2 T.
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what is the cost of monthly (30 days) electric bill of ana if her city's cost of electricity is 0.05$ per kwh and she uses three refrigerators running in 600-watt power rating and open 24 hours
Answer:
1kW = 1000W
600W = 0.6kW
Cost of electric bill = 0.6kWh × 24 × 30 × $0.05
= $21.60
anyone to help with this question
Answer:
i can
Explanation:
i know what's the answer
Why aren’t the Appalachian Mountains still as tall as the Himalayas?
Answer:
mountains are limited in their theoretical height by several processes. First is isostasy: the bigger a mountain gets, the more it weighs down its tectonic plate, so it sinks lower. ... Bottom line: mountains can get taller than Mount Everest in earth gravity, like the Appalachians probably did—but not much taller.
Answer:
It probably won't get any taller, though. From a geological standpoint, the Appalachians haven't seen much growth in quite a while. Since the dawn of the dinosaurs about 225 million years ago, this range has been getting whittled down by weathering forces.
Explanation:
A 2.0 kg block rests on a level surface. The coefficient of static friction is, and the coefficient of kinetic friction is A horizontal force, X, is applied to the block. As X is increased, the block begins moving. Describe how the force of friction varies as X increases from the moment the block is at rest to when it begins moving. Indicate how you could determine the force of friction at each value of X―before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
ps. I had to change F to X because of brainly.
By Newton's second law, the net force acting on the block in the vertical direction is
∑ F [ver] = n - mg = 0
where n = magnitude of normal force and mg = weight of the block. It follows that n = mg.
When the block is at rest, the applied force X will not be enough to move the box until it can overcome the maximum mag. of static friction. If µ[s] is the coefficient of static friction, then the maximum mag. of the frictional force is
f = µ[s] n = µ[s] mg
The net horizontal force would be
∑ F [hor] = X - µ[s] mg = 0
so a minimum force of X = µ[s] mg is required to get the block moving. Any mag. smaller than this and the block stays at rest/in equilibrium.
Once the mag. of X exceeds µ[s] mg, the block will begin to move. At that point, if the coefficient of kinetic friction is µ[k], then the net force on the block is
∑ F [hor] = X - µ[k] mg = 0
so a minimum force of X = µ[k] mg would be needed to keep the block moving at constant speed, or otherwise X = µ[k] mg + ma if the block is accelerating with mag. a.
The principles here are captured in the attached plot.
do forces that act a distance come in pairs?
Answer:
Forces come in pairs.
Explanation:
Distances are between two points. The distance from A to B is equal but opposite the distance from B to A.
18 through 22 multiple choice please help
Answer:
18:equilibrium
19:10.8
20:separating the surfaces with a layer of air
21:it remains motionless
22:cm/s
Under the Big Top elephant. Ella [2500 kg]. is attracted to Phant, the 3,000 kg elephant. They are separated by 8 m. What is the gravitational attraction between them? G=6.67×10^-11 (-11 is an exponent)
Hi there!
We can use the same equation for Gravitational Force:
[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]
Fg = force due to gravity (N)
G = gravitational constant
m1,m2 = masses of objects (kg)
r = distance between objects (m)
Plug in the values provided:
[tex]F_g = (6.67*10^{-11})\frac{(2500)(3000)}{8^2} = \large\boxed{7.814 * 10^{-6}N }[/tex]
[tex]\huge\bf\underline{\underline{\pink{A}\orange{N}\blue{S}\green{W}\red{E}\purple{R:-}}}[/tex]
Here we've been given,
Universal gravitational constant (G) = [tex] \sf{6.67 \times {10}^{ - 11} }[/tex]Mass of object 1 (m1) = 2500 kg Mass of object 2 (m2) = 3000 kg Distance between two objects (r) = 8 mWe have to find the gravitational attraction force (Fg) = ?
The standard formula to solve is given by,
[tex]:\implies\tt{F_g = g \frac{m_1m_2}{ {r}^{2} } } [/tex]
[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{(2500 )(3000)}{ {8}^{2} } }[/tex]
[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{7500000}{64} }[/tex]
[tex]:\implies\tt{F_g = 7.814 \times {10}^{ - 6} }[/tex]
Gravitational force of attraction is 7.814 × 10^-6 N.what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three signficant figures.
Answer:
Approximately [tex]5.11 \times 10^{-19}\; {\rm J}[/tex].
Explanation:
Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.
Look up the Rydberg constant for hydrogen: [tex]R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}[/tex].
Look up the speed of light in vacuum: [tex]c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}[/tex].
Look up Planck's constant: [tex]h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}[/tex].
Apply the Rydberg formula to find the wavelength [tex]\lambda[/tex] (in vacuum) of the photon in question:
[tex]\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}[/tex].
The frequency of that photon would be:
[tex]\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}[/tex].
Combine this expression with the Rydberg formula to find the frequency of this photon:
[tex]\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}[/tex].
Apply the Einstein-Planck equation to find the energy of this photon:
[tex]\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}[/tex].
(Rounded to three significant figures.)
if the speed of a car is increased by 50%, bywhat factor will its minimum braking distance be increased, assuming all else is the same?
When the speed of a car increases by a factor of 50%, the minimum braking distance will also be increased by a factor.
Braking distance will increase by a factor 2.25
Solution
From Newton's equation of motion, we can say that;
v² = u² + 2as
Where initial velocity is zero, we have;
v² = 2as
s = v²/2a
s is the distance and v is the final speed.
50% increase in speed means it has increased by a factor of 1.50
Hence we have
1.50²v² is directly proportional to the 1.50²d
Distance = 1.50²v² = 2.25d
Hence, breaking distance will increase by a factor 3.0625
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How do you un rewrap something?
A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with
the racket for 0.04 s, what is the average force on the ball by the racket?
A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with
the racket for 0.04 s, what is the average force on the ball by the racket?
0.57 N
32 N
98 N
0.00005 N
From Newton's second law of motion, the average force on the ball by the racket is 98 Newtons. The correct answer is option C
Given that a 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. And the time for contact with the racket is 0.04 s, that is,
mass m = 0.07 kg
velocity v = 56 m/s
time t = 0.04 s
force f = ?
To calculate the average force on the ball by the racket, let us apply Newton's second law of motion.
Impulse = change in momentum
ft = mv
Substitute all the parameters into the equation above
0.04f = 0.07 x 56
make f the subject of the formula
f = 3.92 / 0.04
f = 98 N
Therefore, the average force on the ball by the racket is 98 Newtons. The correct answer is option C
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The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N
From the question, we are to determine the average force on the ball by the racket.
From the formula,
[tex]F = \frac{mv}{t}[/tex]
Where F is the force
m is the mass
v is the velocity
and t is the time
From the given information
m = 0.07 kg
v = 56 m/s
t = 0.04 s
Putting the parameters into the formula,
we get
[tex]F = \frac{0.07 \times 56}{0.04}[/tex]
[tex]F = \frac{3.92}{0.04}[/tex]
F = 98 N
Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N
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A 0.15 kg baseball moving at 20 m/s is stopped by a player in 0.010 s. What is the average force of the ball?
Answer: 300N
Explanation:
Impulse= Mass * Velocity
F.T = M * V
F= MV/T
F= (0.15*20)/ 0.01
F= 300N
Which branch of science deals with the study of the structures shown here?
FISH
AMPHIBIAN
REPTILE
BIRD?
Answer:
Vertebrate zoology
Explanation:
Have a great day!
how did you identify the layer that belonged next to the cambrian layer?
Answer:
Morphology and phylogenetics revealed by fossils. Perhaps the strongest evidence to support the Cambrian evolutionary explosion of animal forms is the first clear appearance, in the Early Cambrian, of skeletal fossils representing members of many marine bilaterian animal phyla
The decomposition of dinitrogen pentoxide is described by the reaction: 2N2O5(g)→4NO2(g)+O2(g). If O2 is being formed at a rate of 2.40 mol/min, what is the rate at which N2O5 is being used up?
Answer:
4.80 mol N2O5/min
Explanation:
The balanced equation tells us that 2 moles of N2O5 are required for every 1 mole of O2. Therefore:
(2.40 mol O2/min)*(2 mol N2O5/mol O2) = 4.80 mol N2O5/min
the smallest division value of electronic balance
Answer:
0.1g to 0.0000001g hope it helps uu
An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.
a) Express the initial velocity of the package as a vector. What
observer on the ground.
b) How long will it take for the package to reach the ground?
c) How fast will it be going as it lands? Express your answer as
a vector.
d) Describe the path of the package as seen by a stationary
assumptions have you made about the frame of reference?
e) Describe the path of the package as seen by someone in the
aeroplane.
The motion of the package can be described as the motion of a projectile,
given that it has an horizontal velocity and it is acted on by gravity.
a) [tex]\vec{v}[/tex] = 70·ib) The package will reach ground in approximately 12.77 seconds.c) The speed of the package as it lands is approximately 145.51 m/s.d) The path of the package based on a stationary frame of reference is parabolice) The path of the package as seen from the plane is directly vertical downwards
Reasons:
Velocity of the aircraft = 70 m/s
Direction of flight of the aircraft = Eastward
Height from which the aircraft drops the package, h = 800 m
a) The initial velocity of the package, [tex]\vec{v}[/tex] = 70·i
b) The time it will take the package to reach the ground, t, is given by the formula;
[tex]\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}[/tex]
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
Therefore;
[tex]\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}[/tex]
Which gives;
[tex]\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}[/tex]
The time it will take the package to reach the ground, t ≈ 12.77 seconds
c) The vertical velocity just before the package reaches the ground, [tex]v_y[/tex], is given as follows;
[tex]v_y^2[/tex] = 2·g·h
Therefore;
[tex]v_y[/tex] = √(2·g·h)
Which gives;
[tex]v_y[/tex] = √(2 × 9.81 × 800) ≈ 125.28
[tex]v_y[/tex] ≈ 125.28 m/s
Which gives; [tex]\vec{v}[/tex] = 70·i - 125.28·j
Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51
The speed of the package as it lands, |v| ≈ 143.51 m/s
d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.
Therefore;
The path of the package is the path of a projectile, which is a parabolic shape.
e) As seen by someone on the aeroplane, the horizontal velocity will be
zero, therefore, the package will appear as accelerating directly vertical
downwards.
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how do you find average velocity (average) from acceleration) and time (t)?
Average velocity is defined as the ratio in change in position to change in time,
v[ave] = ∆x/∆t
which on its own doesn't have anything to do with acceleration.
If acceleration is constant, the average velocity is the literal average of the initial and final velocities,
v[ave] = (v[final] + v[initial]) / 2
If this constant acceleration has magnitude a, the final velocity can be expressed in terms of the initial velocity by
v[final] = v[initial] + a*t
and plugging this into the previous equation gives
v[ave] = (v[initial] + a*t + v[initial])/2
v[ave] = v[initial] + 1/2*a*t
If the body in consideration is initially at rest, then
v[ave] = 1/2*a*t
which might be the relation you're looking for. But bear in mind the conditions I've underlined.
If acceleration is not constant and changes over time, so that the acceleration is some function of time a(t), then you can determine the velocity function v(t) by using the fundamental theorem of calculus. You need to know a particular velocity for some time to completely characterize v(t), though. For example, if you're given the initial velocity v[initial] = v(0), then
[tex]\displaystyle v(t) = v(0) + \int_0^t a(u) \, du[/tex]
or if you know any other velocity for some time t₀ > 0,
[tex]\displaystyle v(t) = v(t_0) + \int_{t_0}^t a(u) \, du[/tex]
Someone plz give me some boy advice I really need it.
what does the slope of the curve on a velocity vs. time graph represent?
Answer:
the slope of velocity-time graph represent an object acceleration
if vector b is added to vector a under what conditions does the resultant vector has magnitude a+b?
Explanation:
hope it's useful for you knows
Please please help
Matter is made up of particles which
are too small to see without the aid of
a microscope. What do scientists call
these particles?
A.bonds
B.mixtures
C.energy
D.atoms
Answer:
D
Explanation:
Tim is pushing a heavy box across the floor. He is using 300N of force and can accelerate at 2m/s/s. What is the mass of the box?
-298 kg
-600 kg
-0.006 kg
-150 kg
Using Newton's second law
[tex]\\ \sf\Rrightarrow F=ma[/tex]
[tex]\\ \sf\Rrightarrow 300=2m[/tex]
[tex]\\ \sf\Rrightarrow m=150kg[/tex]
Hey there!
The formula of “mass” in physics is:
m = F/a
Whereas “f” is ‘force’, “a” is ‘acceleration’, & “m” is your ‘mass’ of course.
mass = 300 Net force/2 acceleration
300 Net force/2 acceleration = m
mass = 150
Therefore, your answer is: 150 kg
Good luck on your assignment and enjoy your day!
~Amphitrite1040:)
Yoo what it do I don’t know the answer tho
Answer:
You should try your best to answer the question.