Answer:
t = 781.25 s
Explanation:
This is an exercise in velocity composition, if we set a reference system where the x-axis is perpendicular to the river and the y-axis is parallel to the river.
The swimmer has a velocity on the x axis
vx = 1.60 m / s
a velocity on the y axis, created by the current of the river
vy = 0.549 m / s
time is a scalar, therefore the time it takes to cross the river is the same time it creates the displacement in e; Axis y
X axis
vₓ = x / t
t = x / vₓ
t = 1250 / 1.6
t = 781.25 s
in this time a distance has descended
y = v_y t
y = 0.549 781.25
y = 428.9 m
The plates of a parallel plate capacitor each have an area of 0.40 m2 and are separated by a distance of 0.02 m. They are charged until the potential difference between the plates is 3000 V. The charged capacitor is then isolated. Determine the magnitude of the electric field between the capacitor plates.
Answer:
E = -1.5 10⁵ N / C
Explanation:
In a capacitor the electric field is uniform between the blades, therefore we use the expression
V = - E s
E = - V / s
let's calculate
E = - 3000 / 0.02
E = -1.5 10⁵ N / C
the sign indicates that the field and the potential are opposite, when one increases the gold decreases
I have a massive rock weighing 3,000 Newtons but I can only accelerate it to 500 m/s2 what is its mass?
Answer:
6 kg
Explanation:
F=ma
F is Force(newtons)
m is mass(kg)
a is acceleration(m/s^2)
Plug in the numbers
3000 = m(500)
divide both sides by 500 to cancel out the 500.
3000/500=6
6 = m
6kg is the mass
20 points!!!! A 2,00ON steel rod that is 5 meters long is placed in a corner between the floor and a wall, and balanced at an angle using a cord attached to the wall The rod is balanced such that its top end is 2.38 meters away from the wall, The cord is 40 cm long, and it is attached to the wall at a height of 75 cm above the floor. The diagram to the right shows the situation If the lower end of the rod does not slip from the corner, what is the tension in the cord?
Answer:
WE NEED TO ADD ALL 40+2.38 +75+5
Explanation:
PLSE GIVE SOME POINTS DUDE
If rider A is moving 8 meters per second, and rider B is moving 3 meters per second, how far away from rider A was rider B when first observed at the start of the graph?
Answer:
the answer is 32
Explanation:
he was riding pretty far
There is a very long straw of charge that is uniformly charged in electro static equilibrium. It has a charge per unit length of 4.0E-9 C/m (4.0 nC/m) and a radius of 0.5 m. What is the strength of the electric field a distance of 10.0 m from its center outside the straw
Answer:
2880 N/c
Explanation:
Given that:
Charge per unit length ; λ = 4 * 10^-9
radius, r = 10
Radius, R = 0.5m
Using the relation :
2λr / 4πE0R²
Columb's constant, k = 1/4πE0 =. 9* 10^9Nm²/C²
Hence, we have :
2λrk/ R²
(2 * 4 * 10^-9 * 10 * 9 * 10^9) / 0.5^2
(720 ÷ 0.25)
= 2880 N/c
What is the average speed of a car that travels 60 meters in 2
seconds?
Answer:
30 m/s
Explanation:
Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.
What is the potential energy of an object 20 m in the air with a
mass of 600 kg?
Answer:
Ep = 117600 J
Explanation:
Data:
Mass (m) = 600 kgHeight (h) = 20 mGravity (g) = 9.8 m/s²Potential Energy (Ep) = ?Use formula:
Ep = m * g * hReplace:
Ep = 600 kg * 9.8 m/s² * 20 mMultiply operations, and units:
Ep = 117600 JWhat is the potential energy?
The potential energy is 117600 Joules.
Rank the four fundamental forces from strongest to weakest. Use 1 to indicate the strongest force and 4 to indicate the weakest force. The gravitational force: The electromagnetic force: The strong nuclear force: The weak nuclear force:
Answer:
4
2
1
3
Explanation:
Be safe, lovelies <3
help? its a short question
Answer:
i think its ancestor
Explanation:
sry if im wrong
Answer:
scientists compare organisms DNA to support the theory that all species share a common Ancestor.
Did I hear correctly that the speed of light is different in deep space observation?
Answer:
Astronomers can learn about the elements in stars and galaxies by decoding the information in their spectral lines. There is a complicating factor in learning how to decode the message of starlight, however. If a star is moving toward or away from us, its lines will be in a slightly different place in the spectrum from where they would be in a star at rest. And most objects in the universe do have some motion relative to the Sun.
Explain why your PE and KE are usually not both high at the same time (If PE is high then usually KE is low)
Help please ! Ill give brainliest !! ☁️✨
The electric potential inside a living cell is lower than the potential outside. Suppose the electric potential difference between the inner and the outer cell wall is 0.095 V, a typical value. To maintain the internal electrical balance, the cell pumps out sodium ions. How much work must be done to remove a single sodium ion (charge e)
Answer:
1.52 × 10⁻²⁰ J
Explanation:
The electrical potential difference is defined as the amount of work done in carrying a unit charge from one point to another point in an electric field. Electric potential difference is measured in volts. It is given by the formula:
ΔV = ΔU / q
ΔV is electric potential difference between the two points, ΔU is the work done and q is the unit charge.
Given that ΔV = 0.095 V, q = 1.6 × 10⁻¹⁹ C. Hence:
ΔU = ΔV.q
ΔU = 0.095 V * 1.6 × 10⁻¹⁹ C
ΔU = 1.52 × 10⁻²⁰ J
The electric potential difference is the amount of effort done in an electrical field to shift a unit charge from one spot to another.The electric potential difference will be 1.52×10⁻²⁰J.
What is the electric potential difference ?
The electrical potential difference is the amount of effort done in an electrical field to shift a unit charge from one spot to another.
Traditional current flows from positive to negative terminals, signifying positive charge transfer in that direction.
The given data in the problem is
[tex]\triangle V[/tex] is the electric potential difference between the inner and the outer cell wall = 0.095 V
[tex]\triangle U[/tex] is the amount of work done
q is the charge on the electron =1.6×10⁻¹⁹C
Electric potential difference is given by the formula
[tex]\triangle V=\frac{\triangle U}{q} \\\\ \triangle U=\triangle Vq\\\\\triangle U=0.095 V\times1.6\times10^{-19}\\\\ \triangle U=1.52\times10^{-20}J[/tex]
Hence the electric potential difference will be 1.52×10⁻²⁰J.
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What is the period, in seconds, that corresponds to each of
the following frequencies: (a) 10 Hz, (b) 0.2 Hz, (c) 60 Hz?
Answer:
0.1s,5s,0.017s
Explanation:
T=1÷frequency
Answer:
a =
✔ 6
The period is
✔ 2 seconds.
b =
✔ pi
Explanation:
Graph the function using the graphing calculator. Find the least positive value of t at which the pendulum is in the center.
t =
✔0.5 sec
To the nearest thousandth, find the position of the pendulum when t = 4.25 sec.
d =
✔ 4.243 in.
8) A train enters a curved horizontal section of the track at a speed of 100 km/h and slows down with constant deceleration to 50 km/h in 12 seconds. If the total horizontal acceleration of the train is 2 m/s2 when the train is 6 seconds into the curve, calculate the radius of curvature of the track for this instant.
Answer:
the radius of curvature of the track for this instant is 266 m
Explanation:
Given that;
The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s
velocity of the train at t=12 s is;
[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s
now, we calculate the deceleration of the train
[tex]V_{t=12}[/tex] = u + at
13.89 = 27.77 + [tex]a_{t}[/tex]12
[tex]a_{t}[/tex] = (13.89 - 27.77) / 12
[tex]a_{t}[/tex] = -13.88 / 12
[tex]a_{t}[/tex] = - 1.1566 m/s²
Now, the velocity of the train at 6 seconds is;
[tex]V_{t=6}[/tex] = u + at
[tex]V_{t=6}[/tex] = 27.77 + ( - 1.1566 m/s²)6
[tex]V_{t=6}[/tex] = 27.77 - 6.9396
[tex]V_{t=6}[/tex] = 20.83 m/s
The acceleration at t=6 s is;
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
we substitute
2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]
4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²
4 = 1.3377 + ([tex]a_{n}[/tex])²
([tex]a_{n}[/tex])² = 4 - 1.3377
([tex]a_{n}[/tex])² = 2.6623
[tex]a_{n}[/tex] = √2.6623
[tex]a_{n}[/tex] = 1.6316 m/s²
Now the radius of curve is;
a = V² / p
[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² / [tex]a_{n}[/tex]
[tex]p_{t=6}[/tex] = ( 20.83 m/s )² / 1.6316 m/s²
[tex]p_{t=6}[/tex] = 433.8889 / 1.6316
[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m
Therefore; the radius of curvature of the track for this instant is 266 m
how can you rewrite the force formula (f=ma) to solve the acceleration?
The force formula can be rewritten to solve the acceleration as:
acceleration = force/mass.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
From Newton's 2nd law of motion, we can write that:
Force = mass × acceleration
⇒ acceleration = force/mass.
Hence, the force formula can be rewritten to solve the acceleration as:
acceleration = force/mass.
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Analyze the data to identify the mathematical relationship between the
amplitude and energy of a mechanical wave. If mechanical wave A has an
amplitude of 4 cm and mechanical wave B has an amplitude of 5 cm, what
will be the relationship between the energy carried by the two waves?
Amplitude
Energy
A. Wave A has about 1.25 times more energy than wave B.
ОО
B. Wave A has about 1.6 times more energy twan wave B.
C. Wave B has about 1.6 times more energy than wave A.
O D. Wave A has about 1.15 times more energy than wave B.
Answer:
Its C
Explanation:
Because I got it wrong for you
Wave B has about 1.6 times more energy than wave A.
What is energy?
Energy is the ability or capability to do tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.
The amplitude and energy of a mechanical wave. If mechanical wave A has an amplitude of 4 cm and mechanical wave B has an amplitude of 5 cm wave B has about 1.6 times more energy than wave A.
Wave B has about 1.6 times more energy than wave A.
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Which is the weakest of the four fundamental forces?
strong nuclear
weak nuclear
electromagnetic
gravitational
Answer:
Gravitational
Explanation:
gravitational
Answer:
Gravitational
Explanation:
In order from strongest to weakest.
Strong nuclear
Electromagnetic
Weak nuclear
Gravitational
You are inside the Great Hall, 15 m from the north wall with the doors to the RMC, and centered between two open doors that are 3 m apart. Someone is blairing a 200 Hz tone outside the Great Hall so that it enters the doors as a plane wave. You hear a maximum intensity in your current position. As you walk along the direction of the wall with the doors (but maintain a distance 15 m from the wall), how far will you walk (in m) to hear a minimum in the sound intensity
Answer:
Δr = 0.425 m
Explanation:
This is a sound interference exercise, the expression for destructive interference is
Δr = (2n + 1) λ / 2
in this case the movement is in the same direction as the sound, therefore the movement is one-dimensional
let's use the relationship between the speed of sound and its frequency and wavelength
v = λ f
λ = v / f
the first minium occurs for n = 0
Δr = λ / 2
Δr = v / 2f
Δr = [tex]\frac{340}{2 \ 400}[/tex]
Δr = 0.425 m
this is the distance from the current position that we assume in the center of the room
The study of heat is ____?
Explanation:
thermodynamics is the study of heat.
Answer The study of heat and its relationship to useful work is called thermodynamics and involves macroscopic quantities such as pressure, temperature, and volume without regard for the molecular basis of these quantitie
Explanation:
If a 500-pound object is moved 200 feet how much work is being done?
a. 200 FT LB
b. 500 FT LB
c. 1000 FT LB
d. 100,000 FT LB
Answer:
D
Explanation:
Work = Distance x Mass
work done = 100,000 FT LB
What is work done ?
Work is done whenever a force moves something over a distance or The work done by a force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement.
Work done = force * displacement
given :
force = 500 pound
displacement = 200 feet
work done = 500 * 200 = 100,000 FT LB
correct option is d. 100,000 FT LB
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a 90 kilogram dog runs across the dog park at a speed of 6.5 meters per second. what is the magnitude and direction of the average force required to stop the dog in .85 seconds?
Answer:
am not sure about the answer
Explanation:
you need to find out the amount of force it's going in for example 10n or 100n then you need to times it the distance then devide by the time
An electron moves from point i to point f, in the direction of a uniform electric field. During this motion:Group of answer choicesthe work done by the field is positive and the potential energy of the electron-field system increasesthe work done by the field is negative and the potential energy of the electron-field system increasesthe work done by the field is positive and the potential energy of the electron-field system decreasesthe work done by the field is negative and the potential energy of the electron-field system decreasesthe work done by the field is positive and the potential energy of the electron-field system does not change
Answer:
the work done by the field is positive and the potential energy of the electron field system decreases
Explanation:
This exercise asks to find the work and the potential energy of an electron in an electric field.
Work is defined by
W = F .d = F d cos θ
the electric force is
F_e = q E
W = q E d cos θ
since the charge of the electron is negative the force is in the opposite direction to the electric field
W = - e E d
we select the direction to the right is positive, point i is to the left of point f,
therefore the work moving from point i to point F has two possibilities
* The electric field lines go from i to f point , so that point i is on the side of the positive charges, so the electron approaches them, This movement is opposite to that indicated
* the field line reaches point i, this implies that the charges are negative, so the electrioc field is then negativeand the electron charge is negative too. The electron moves away from this point, this is in accordance with the indicated movement
In the latter case the electric field lines go from f to i point, therefore the Work is positive
Now let's examine the potential energy
ΔU = - q E .d
so we see that this definition is related to work,
ΔU = -W
Therefore, as the work is positive, the power energy must decrease
When reviewing the different answers, the correct ones are:
the work done by the field is positive and the potential energy of the electron field system decreases
The work done by the electron while moving from point [tex]i[/tex] to point [tex]f[/tex] in the direction of uniform electric field is negative and the potential energy of the electron increases.
An electron moves from point i to point f, in the direction of a uniform electric field, then the potential energy of the electron can be calculated s given below.
[tex]\Delta V=-qEd[/tex]
Where [tex]\Delta V[/tex] is the potential energy, [tex]E[/tex] is the electric field, [tex]q[/tex] is the charge and [tex]d[/tex] is the displacement of the electron.
The work done by the electron in the uniform electric field can be calculated as,
[tex]W = F\times d \times cos\theta[/tex]
Where [tex]W[/tex]is the work done by electron, [tex]F[/tex] is the electric force, [tex]d[/tex] is the displacement of the electron and for uniform electric field, the value of [tex]\theta[/tex] is zero.
Hence [tex]W=F\times d\times 1\\W=F \times d[/tex]
Electric force [tex]F = q E[/tex]
By substituting the value of electric force on the above formula,
[tex]W = qEd[/tex]
Hence, the relation between the work done the electron in an uniform electric field and potential energy of the electron can be given below.
[tex]W = -\Delta V[/tex]
The work done by the electron is negative and the potential energy of the electron increases.
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The kinetic theory states that the higher the temperature, the faster the
Answer: the higher the kinetic energy
Explanation:
An 80 N rightward force is applied to a 10 kg object to accelerate it to the right.
The object encounters a friction force of 50 N.
net force = 30 N
mass = 8.16 kg
acceleration = 3.68 m/s²
Further explanationGiven
80 N force applied
mass of object = 10 kg
Friction force = 50 N
Required
Net force
mass
acceleration
Solution
net forceNet force = force applied(to the right) - friction force(to the left)
Net force = 80 - 50 = 30 N
massGravitational force(downward) : F = mg
m = F : g
m = 80 : 9.8
m = 8.16 kg
accelerationa = F net / m
a = 30 / 8.16
a = 3.68 m/s²
If an ocean wave has a wavelength of 2 m and a frequency of 1 wave/s, then its speed is m/s Enter the answer Check it CRATCHPAD Improve this questic 트
Answer:
2m/s
Explanation:
v=f×wavelength
v=2×1
=2m/s
A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is falling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target?
Answer:
Answer:
15.67 seconds
Explanation:
Using first equation of Motion
Final Velocity= Initial Velocity + (Acceleration * Time)
v= u + at
v=3
u=50
a= - 4 (negative acceleration or deceleration)
3= 50 +( -4 * t)
-47/-4 = t
Time = 15.67 seconds
We have that the speed must be at the speed below if they want to just hit their target
From the Question we are told that
Distance [tex]d=50m[/tex]
Height [tex]h=70m[/tex]
Constant Velocity [tex]v= 7.0 m/s[/tex]
Generally the equation for the time is mathematically given as
[tex]T=\frac{h}{v}\\\\T=\frac{70}{7}\\\\T=10s[/tex]
Therefore
The velocity required to make horizontal movement is
[tex]V=\frac{d}{T}\\\\V=\frac{50}{10}\\\\V=5m/s[/tex]
Given that
Velocity on the Vertical axis is
[tex]v_y=7m/s[/tex]
Velocity on the horizontal axis is
[tex]v_x=5m/s[/tex]
Therefore resultant speed
[tex]v_r=\sqrt{v_x^2+V_y^2}\\\\v_r=\sqrt{(5)^2+(7)^2}[/tex]
[tex]v_r=8.6023m/s[/tex]
In conclusion
[tex]v_r=8.6023m/s[/tex] must be their total speed if they want to just hit their target
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Explain two reasons why astronomers are continually building larger and larger telescopes. Explain two reasons why astronomers are continually building larger and larger telescopes. Larger telescope mirrors have a larger surface area and can therefore collect more light, which makes faint objects bright enough to detect. Also, larger telescope mirrors produce more scattering of light due to diffraction, which contributes to better angular resolution. Larger telescope mirrors have a larger surface area and can therefore collect more light, which contributes to better angular resolution. Also, larger telescope mirrors produce more scattering of light due to diffraction, which makes faint objects bright enough to detect. Larger telescope mirrors have a larger surface area and can therefore collect more light, which contributes to better angular resolution. Also, larger telescope mirrors produce less scattering of light due to diffraction, which makes faint objects bright enough to detect. Larger telescope mirrors have a larger surface area and can therefore collect more light, which makes faint objects bright enough to detect. Also, larger telescope mirrors produce less scattering of light due to diffraction, which contributes to better angular resolution.
Answer:
* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected
* arger mirrors decreases the angle of dispersion giving a better resolution of the bodies
Explanation:
Refracting telescopes get bigger every day for two main reasons.
* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected
* the diffraction process for circular apertures is given by
θ = 1.22 λ / D
where d is the diameter of the mirror, therefore having larger mirrors decreases the angle of dispersion giving a better resolution of the bodies
A warm hockey puck has a coefficient of restitution of 0.50, while a frozen hockey puck has a coefficient of restitution of only 0.35. In the NHL, the pucks to be used in games are kept frozen. During a game, the referee retrieves a puck from the cooler to restart play but is told by the equipment manager that several warm pucks were just put into the cooler. To check to make sure he has a game-ready puck, the referee drops the puck on its side from a height of 2 m. How high should the puck bounce if it is a frozen puck
Answer:
the required height is 0.2449 m only
Explanation:
Given the data in the question;
Initial height = 2m
so speed of the puck before hitting the ground will be;
u² = 2gh
Initial speed u_ball = √2gh
u_ball = √( 2 × 9.8 × 2 )
u_ball = √39.2
u_ball = 6.26 m/s
given that; FOR THE FROZEN PUCK, coefficient of restitution = 0.35 only
R = - (v_ball - v_ground / u_ball - u_ ground)
so
0.35 = - (v_ball - 0 / 6.26 - 0)
0.35 = -v_ball / - 6.26
-v_ball = 0.35 × (- 6.26)
-v_ball = -2.191 m/s
v_ball = 2.191 m/s
to get the height;
v² = 2gh
h = v² / 2g
we substitute
h = (2.191)² / 2×9.8
h = 4.800481 / 19.6
h = 0.2449 m
Therefore, the required height is 0.2449 m only
At what latitude are there almost no differences between the seasons? Explain
why this occurs?
Answer:
The four-season year is typical only in the mid-latitudes. The mid-latitudes are places that are neither near the poles nor near the Equator. The farther north you go, the bigger the differences in the seasons.
Explanation:
hope this helps have a good day :) ❤
At a latitude equal to zero degrees there is little seasonal variation. This phenomenon is due each day the Sun's rays strike the Earth's surface at approximately the same angle near the Equator.
The Equator is the line of 0° (zero degrees) latitude around the middle of the Earth.
Moroever, the intensity of solar radiation and therefore also the temperature at the Earth's surface largely depends on the angle of incidence of the Sun's rays.
At 0° latitude, there is a very little seasonal variation because all days the Sun's rays strike the Earth's surface at approximately the same angle. At the Equator, the Sun's rays strike the Earth's surface at an angle of 90°, causing warmer temperatures compared to higher latitudes.
In additon, at 0° latitude, all days also have the same number of hours of light and dark (i.e., approximately 12 hours of sunlight).
In conclusion, at a latitude equal to zero degrees (i.e., at the Equator) there is little seasonal variation. This phenomenon is due each day the Sun's rays strike the Earth's surface at approximately the same angle near the Equator.
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