Answer:
Hewo Otaku Kun Here! (UwU)
Explanation:
1. A rock sitting at the edge of a cliff has potential energy. If the rock falls, the potential energy will be converted to kinetic energy.
2. Tree branches high up in a tree have potential energy because they can fall to the ground.
3. A stick of dynamite has chemical potential energy that would be released when the activation energy from the fuse comes into contact with the chemicals.
4. The food we eat has chemical potential energy because as our body digests it, it provides us with energy for basic metabolism.
5. A stretched spring in a pinball machine has elastic potential energy and can move the steel ball when released.
6. When a crane swings a wrecking ball up to a certain height, it gains more potential energy and has the ability to crash through buildings.
7. A set of double "A" batteries in a remote control car possess chemical potential energy which can supply electricity to run the car.
happy to help!
from: Otaku Kun ^^
4. A vehicle
accelerates from
0 km/h to
100 km/h in 10 s.
What is its average
acceleration?
Answer:
Average acceleration, [tex]a=2.77\ m/s^2[/tex]
Explanation:
Given that,
Initial velocity, u = 0 km/h
Final velocity, v = 100 km/h = 27.77 m/s
Time, t = 10 s
We need to find the average acceleration of the vehicle. It is given by the change in velocity divided by time. Som,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{27.77-0}{10}\\\\a=2.77\ m/s^2[/tex]
So, its average acceleration is [tex]2.77\ m/s^2[/tex].
(7) A 2500 lbm car moving at 25 mi/hr is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/hr. What is the force and total time required?
Answer:
The force is [tex]F = 1164.6\ lbf[/tex]
The time is [tex]\Delta t = 2.44 \ s[/tex]
Explanation:
From the question we are told that
The mass of the car is [tex]m = 2500 \ lbm[/tex]
The initial velocity of the car is [tex]u = 25 \ mi/hr[/tex]
The final velocity of the car is [tex]v = 50 \ mi/hr[/tex]
The acceleration is [tex]a = 15 ft/s^2 = \frac{15 * 3600^2}{ 5280} = 36818.2 \ mi/h^2[/tex]
Generally the acceleration is mathematically represented as
[tex]a = \frac{v-u}{\Delta t}[/tex]
=> [tex]36818.2 = \frac{50 - 25 }{ \Delta t}[/tex]
=> [tex]t = 0.000679 \ hr[/tex]
converting to seconds
[tex]\Delta t = 0.0000679 * 3600[/tex]
=> [tex]\Delta t = 2.44 \ s[/tex]
Generally the force is mathematically represented as
[tex]F = m * a[/tex]
=> [tex]F = 2500 * 15[/tex]
=> [tex]F = 37500 \ \frac{lbm * ft}{s^2}[/tex]
Now converting to foot-pound-second we have
[tex]F = \frac{37500}{32.2}[/tex]
=> [tex]F = 1164.6\ lbf[/tex]
When charges qa, qb, and qc are placed respectively at the corners a, b, and c of a right triangle, the potential at the midpoint of the hypotenuse is 20 V. When the charge qa is removed, the potential at the midpoint becomes 15 V. When, instead, the charge qb is removed (qa and qc both in place), the potential at the midpoint becomes 12 V. What is the potential at the midpoint if only the charge qc is removed from the array of charges?
Answer:
8v
Explanation:
First we apply super position principle
Vt= v1 + v2+ v3
Remove qa
But vt= 20v
So V = v2+v3
V1= 20-15
= 5v
Remove qb
V= v1+v3
V=8v
So the potential when qa and qc are remove is the potential due to qb
Which is 8v
Finally, consider the expression (6.67 x 10^-11)(5.97 x 10^24)/(6.38 x 10^6)^2 Determine the values of a and k when the value of this expression is written in scientific rotation. Enter a and k, separated by commas.
Explanation:
We need to find the value of following expression :
[tex]\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.38 \times 10^6)^2}[/tex]
Firstly, solving the numerator of the above expression :
[tex]=\dfrac{39.8199\times 10^{-11+24}}{40.7044\times 10^{12}}\\\\=\dfrac{39.8199\times 10^{13}}{40.7044\times 10^{12}}\\\\=9.7827[/tex]
Rounding off the result = 9.78
In scientific notation : [tex]9.78\times 10^0[/tex]
The value of a = 9.78 and k = 0.
Which scientist determined that electrons had predicted zones orbiting the nucleus?
Answer:
Rutherford
Explanation:
Because
Schrödinger I just took the unit review
2. If a cyclist in the Tour de France traveled southwest a distance of 12,250 meters in one hour, what would the velocity of the cyclist be?
12,25 km/h
≈ 3,4 m/s
Explanation:v = d/t
= 12250m/h
= 12,25km/h
or
v = d/t
= 12250m/h
1h = 60m×60s = 3600s
= 12250m/3600s
≈ 3,4 m/s
Two 100kg bumper cars are moving towards eachother in oppisite directions. Car A is moving at 8 m/s and Car B at -10 m/s when they collide head on. If the resulting velocity of Car B after the collision is 8 m/s, what is the velocity of Car A after the collision
Answer:
[tex]-10 m/s[/tex]
Explanation:
When two cars collide then the momentum of two cars will remains conserved
Mass of two cars = 100 kg Speed of car A = 8 m/s Speed of car B = - 10 m/s After collision the speed of car B = +8 m/sBy momentum conservation equation
[tex]m1v1i+m2v2i=m1v1f + m2v2f[/tex]
[tex](100)(8)+(100)(-10)=(100v)+(100)(8)\\ v=-10 m/s[/tex]
An 85 kg skydiver is falling through the air at a constant speed of 195 km h-1. At what rate does air resistance remove energy from the skydiver?
Answer:
46041J
Explanation:
Using Energy lost= mgh
Changing to standard its we have
= 195*1000/3600=54.2m/s
So = 85*54.2*10= 46041J
Answer:
45167.15 J/s
Explanation:
mass of the man = 85 kg
The man's speed = 195 km/h = 195 x 1000/3600 = 54.167 m/s
The man's weight = mg
where
m is the mass
g is acceleration due to gravity = 9.81 m/s^2
weight = 85 x 9.81 = 833.85 N
The rate at which energy is removed from the man = speed x weight
==> 54.167 x 833.85 = 45167.15 J/s
How does sleep affect your ability to handle stress?
Answer: Stress can adversely affect sleep quality and duration, while insufficient sleep can increase stress levels. Both stress and a lack of sleep can lead to lasting physical and mental health problems.
Explanation:
Many report that there stress increases when the length and quality of their sleep decreases. When you do not get enough sleep, 21 percent of adults report feeling more stressed.
If one object (a) is moving at 60m/s^2, and the other object (b) is moving at 65m/s^2, at what time will the faster moving object be 10m ahead of the other object?
Answer:
a is moving at 60m and the other object
The speed of a bus increases uniformly from
15 ms- to 60 ms in 20 seconds. Calculate
a. the average speed,
b. the acceleration,
C. the distance travelled during the entire
period The speed of a bus increases uniformly from
15 ms- to 60 ms in 20 seconds. Calculate
a. the average speed,
b. the acceleration,
C. the distance travelled during the entire
period
Explanation:
a. For constant acceleration:
v_avg = ½ (v + v₀)
v_avg = ½ (60 m/s + 15 m/s)
v_avg = 37.5 m/s
b. a = (v − v₀) / t
a = (60 m/s − 15 m/s) / 20 s
a = 2.25 m/s²
c. x = v_avg t
x = (37.5 m/s) (20 s)
x = 750 m
An electron with a speed of 0.95c is emitted by a supernova, where cc is the speed of light. What is the magnitude of the momentum of this electron?
Answer:
2.59×10¯²² Kgm/s
Explanation:
Data obtained from the question include:
Velocity of electron = 0.95c
Momentum =?
Next, we shall determine the velocity of the electron. This can be obtained as follow:
Velocity of electron = 0.95c
Velocity of Light (c) = 3×10⁸ m/s
Velocity of electron = 0.95c
Velocity of electron = 0.95 × 3×10⁸
Velocity of electron = 2.85×10⁸ m/s
Finally, we shall determine the mometum of the electron.
Momentum is simply defined as the product of mass and velocity. Mathematically, it is expressed as:
Momentum = mass x Velocity
Thus, with the above formula, we calculate the momentum of the electron as follow:
Mass of electron = 9.1×10¯³¹ Kg
Velocity of electron = 2.85×10⁸ m/s
Momentum of electron =?
Momentum = mass x Velocity
Momentum = 9.1×10¯³¹ × 2.85×10⁸
Momentum = 2.59×10¯²² Kgm/s
Therefore, the momentum of the electron is 2.59×10¯²² Kgm/s
A -5.40nC point charge is on the x axis at x = 1.25m . A second point charge Q is on the x axis at -0.625m.
A) What must be the charge Q for the resultant electric field at the origin to be 50.0N/C in the +x direction?
B) What must be the charge Q for the resultant electric field at the origin to be 50.0N/C in the -x direction?
Answe
a) Q = 0.820 10⁻⁹ C , b) Q = -3.52 10⁻⁹ C
Explanation:
The electric field is given by the formula
E = k q / r²
where E is a vector quantity, so it must be added as a vector
E_total = E₁ + E₂
let's look for the two electric fields
E₁ = k q₁ / r₁²
E₁ = 9 10⁹ 5.4 10⁻⁹ / 1.25²
E₁ = 31.10 N / C
E2 = k Q / r₂²
E2 = 9 10⁹ Q / 0.625²
E2 = 23.04 10⁹ Q N / C (1)
now we can solve the two cases presented
a) The total field is
E_total = 50.0 N / C towards + x
since the test charge is positive the electric field E1 points to the right in the direction of the + x axis, so the equation is
E_total = E1 + E₂
E₂ = E_toal - E₁
E₂ = 50.0 -31.10
E2 = 18.9 N /C
With the value of the electric field we can calculate the charge (Q) using equation 1
E₂ = 23.04 10⁹ Q
Q = E₂ / 23.04 10⁹
Q = 18.9 / 23.04 10⁹
Q = 0.820 10⁻⁹ C
the charge on Q is positive
b) E_total = -50.0 N / C
E_total = E₁ + E₂
E₂ = E_total - E₁
E2 = -50.0 - 31.10
E2 = -81.10 N /C
we calculate the charge
Q = E2 / 23.04 10⁹
Q = -81.1 / 23.04 10⁹
Q = -3.52 10⁻⁹ C
for this case the charge is negative
The energy radiated per unit surface area (across all wavelengths) for a black body with temperature 2200. Use 5.67 x 10-8 for the Stefan-Boltzmann constant.The ____________ describes the power radiated from a black body in terms of its temperature. Specifically, the total energy radiated per unit surface area of a black body across all wavelengths per unit time is proportional to the fourth power of the black body's thermodynamic temperature
Answer:
Stefan-Boltzmann Law describes the power radiated from a black body in terms of its temperature. Specifically, the total energy radiated per unit surface area of a black body across all wavelengths per unit time is proportional to the fourth power of the black body's thermodynamic temperature
Explanation:
So we now know that
energy per unit area is = zigma x T^4
From this
T = temperature in kelvin
zigma is stefan boltzman constant
So
E/A = 5.67 x 10^-8 x (2200)⁴
= 1.33x 10^6 W/m^2
Explain why a Merry-Go-Round and a Ferris Wheel have a constant acceleration when they are moving?
Answer:
merry go round and Ferris wheel have a constant acceleration due to the change in direction at every point.
Answer:
A merry-go-round is accelerating. Acceleration is a change in speed, direction, or both. Even though the speed of the merry-go-round does not change, its direction constantly changes as it spins.
Explanation:
A paper airplane is thrown horizontally with a velocity of 20 mph. The plane is in the air for 7.63 s before coming to a standstill on the ground. What is the acceleration of the plane?
Answer:
-1.17 m/s²
Explanation:
Given:
v₀ = 20 mph = 8.94 m/s
v = 0 m/s
t = 7.63 s
Find: a
v = at + v₀
0 m/s = a (7.63 s) + 8.94 m/s
a = -1.17 m/s²
The acceleration of the plane will be:
"-1.17 m/s²".
Acceleration and VelocityAccording to the question,
Velocity, v₀ = 20 mph or,
= 8.94 m/s
and,
v = 0 m/s
Time, t = 7.63 s
We know the relation,
→ v = at + v₀
By substituting the values,
0 = a × 7.63 + 8.94
7.63a = - 8.94
a = -[tex]\frac{8.94}{7.63}[/tex]
= - 1.17 m/s²
Thus the response above is correct.
Find out more information about velocity here:
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Gwen releases a rock at rest from the top of a 40-m tower. If g = 9.8 m/s2 and air resistance is negligible, what is the speed of the rock as it hits the ground?
Answer:
[tex]28\; \rm m \cdot s^{-1}[/tex].
Explanation:
Short ExplanationApply the SUVAT equation [tex]\left(v^2 - u^2) = 2\, a \, x[/tex], where:
[tex]v[/tex] is the final velocity of the object,[tex]u[/tex] is the initial velocity of the object, [tex]a[/tex] is the acceleration (should be constant,) and[tex]x[/tex] is the displacement of the object while its velocity changed from [tex]v[/tex] to [tex]u[/tex].Assume that going downwards corresponds to a positive displacement. For this question:
[tex]v[/tex] needs to be found.[tex]u = 0[/tex] because the rock is released from rest.[tex]a = g = 9.8 \; \rm m\cdot s^{-2}[/tex].[tex]x = 40\; \rm m[/tex].Solve this equation for [tex]v[/tex]:
[tex]\displaystyle v = \sqrt{2\, a\, x + u^2} = \sqrt{2\times 9.8 \times 40} = 28\; \rm m \cdot s^{-1}[/tex].
In other words, the rock reached a velocity of [tex]28\; \rm m\cdot s^{-1}[/tex] (downwards) right before it hits the ground.
ExplanationLet [tex]v[/tex] be the velocity (in [tex]\rm m \cdot s^{-1}[/tex]) of this rock right before it hits the ground. Under the assumptions of this question, it would take a time of [tex]t = (v / 9.8)[/tex] seconds for this rock to reach that velocity if it started from rest and accelerated at [tex]9.8\; \rm m \cdot s^{-2}[/tex].
Note that under these assumptions, the acceleration of this rock is constant. Therefore, the average velocity of this rock would be exactly one-half the sum of the initial and final velocity. In other words, if [tex]u[/tex] denotes the initial velocity of this rock, the average velocity of this rock during the fall would be:
[tex]\displaystyle \frac{u + v}{2}[/tex].
On the other hand, [tex]u = 0[/tex] because this stone is released from rest. Therefore, the average velocity of this rock during the fall would be exactly [tex](v / 2)[/tex].
The displacement of an object over a period of time is equal to the length of that period times the average velocity over that period. For this rock, the length of this fall would be [tex]t = (v / 9.8)[/tex], while the average velocity over that period would be [tex](v / 2)[/tex]. Therefore, the displacement (in meters) of the rock during the entire fall would be:
[tex]\displaystyle \left(\frac{v}{2}\right) \cdot \left(\frac{v}{9.8}\right) = \frac{v^2}{19.6}[/tex].
That displacement should be equal to the change in the height of the rock, [tex]40\; \rm m[/tex]:
[tex]\displaystyle \frac{v^2}{19.6} = 40[/tex].
Solve for [tex]v[/tex]:
[tex]v = 28\; \rm m \cdot s^{-1}[/tex].
Once again, the speed of the rock would be [tex]28\;\rm m \cdot s^{-1}[/tex] right before it hits the ground.
A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat three people. The lift is operating at a steady speed of 10 km/h. Neglecting friction and air drag, and assuming that the average mass of each loaded chair is 250 kg, determine the power required to operate this ski lift. Also, estimate the power required to accelerate this ski lift in 17 s to its operating speed when it is first turned on.
Answer:
Explanation:
The question states that the chairs are spaced 20 m apart through a length of 1 km, or say, 1000 m.
It also does say that each chair weighs 250 kg, and as such the load is
M = 50 * 250
M = 12500.
Taking into consideration, the initial and final heights, we have
h1 = 0, h2 = 200 m
The work needed to raise the chairs,
W = mgh, where h = h2 - h1
W = 12500 * 9.81 * (200 - 0)
W = 2.54*10^7 J
The work is done at a rate of 10 km/h, and at a distance of 1 km, time taken would be
t = 1/10 = 0.1 h or say, 360 s
The power needed thus, is
P = W/t
P = 2.54*10^7 / 360
P = 68125 W, or 68 kW
Initial velocity, u = 0 m/s
Final velocity, v = 10 km/h = 2.78 m/s
Startup time, t is 17 s
Acceleration during the startup then is
a = (v - u)/t
a = 2.78/17
a = 0.163 m/s²
The power needed for the acceleration is
P = ½m [(v² - u²)/t]
P = ½ * 12500 * [2.78²/17]
P = 6250 * 0.455
P = 2844 W
A runner jumps off the ground at a speed of 16m/s. At what angle did he jump from the ground if he landed 8m away?
Answer:lol I don’t know good qeustion
Explanation:
Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 777 kW. Assume that the wave spreads out uniformly into a hemisphere above the ground. At a home 5.00 km away from the antenna,
(a) what average pressure does this wave exert on a totally reflecting surface,
(b) what are the amplitudes of the electric and magnetic fields of the wave, and
(c) what is the average density of the energy this wave carries?
(d) For the energy density in part (c), what percentage is due to the electric field and what percentage is due to the magnetic field?
Answer:
A) P = 3.3 × 10^(-11) Pa
B) Amplitude of electric field = 1.931 N/C
Amplitude of magnetic field = 6.44 × 10^(-9) T
C) μ_av = 1.65 × 10^(-11) J/m³
D) 50% each for the electric and magnetic field
Explanation:
A) First of all let's calculate intensity.
I = P_av/A
We are given;
P_av = 777 KW = 777,000 W
Distance = 5 km = 5000 m
Thus;
I = 777000/(2π × 5000²)
I = 0.00495 W/m²
Now, the average pressure would be given by the formula;
P = 2I/C
Where C is speed of light = 3 × 10^(8) m/s
P = (2 × 0.00495)/(3 × 10^(8))
P = 3.3 × 10^(-11) Pa
B) Formula for the amplitude of the electric field is gotten from;
E_max = √[2I/(εo•c)].
Where εo is the Permittivity of free space with a constant value of 8.85 × 10^(−12) c²/N.mm²
I and c remain as before.
Thus;
E_max = √[(2 × 0.00495)/(8.85 × 10^(−12) × 3 × 10^(8))]
E_max = √3.72881355932
E_max = 1.931 N/C
Formula for amplitude of magnetic field is gotten from;
B_max = E_max/c
B_max = 1.931/(3 × 10^(8))
B_max = 6.44 × 10^(-9) T
C) Formula for average density is;
μ_av = εo(E_rms)²
Now, E_rms = E_max/√2
Thus;
E_rms = 1.931/√2
μ_av = 8.85 × 10^(−12) × (1.931/√2)²
μ_av = 1.65 × 10^(-11) J/m³
D) The energy density for the electric and magnetic field is the same. So both of them will have 50% of the energy density.
A commuter backs her car out of her garage with an acceleration of 1.40 m/s2.A) How long does it take her to reach a speed of 2.00 m/s?B) If she then breaks to a stop in 0.800 s, what is her deceleration?
Answer:
(A) 1.43secs
(B) -2.50m/s^2
Explanation:
A commuter backs her car out of her garage with an acceleration of 1.40m/s^2
(A) When the speed is 2.00m/s then, the time can be calculated as follows
t= Vf-Vo/a
The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0
= 2.00-0/1.40
= 2.00/1.40
= 1.43secs
(B) The deceleration when the time is 0.800secs can be calculated as follows
a= Vf-Vo/t
= 0-2.00/0.800
= -2.00/0.800
= -2.50m/s^2
4. Lead has a density of 11.5g/cmº. A rectangular block of lead measures 7cm x5cmx2cm.
a) Find the volume of the block of lead.
b) Find the mass of the block of lead
Answer:
(a) 70cm³
(b) 805 grams
Explanation:
(a) V = L×B×H
= 7cm×5cm×2cm
= 35cm×2cm
= 70cm³
(b) Mass = Volume × Density
= 70cm³ × 11.5g/cm³
= 805 grams
it is easier to swim in Ocean than in a water
Answer:
Explanation:
Ocean contains salty water which causes the object to float rather than sinking, So yes it's easy to swim in ocean than water without salt.
which water molecules have the greatest kinetic energy
A ball of mass m moving with speed V collides with another ball of mass 2m (e= 1/2) in a horizontal smooth fixed circular tube of radius R (R is sufficiently large R>>>d). The time after which next collision will take place is:________
Answer:
[tex]$ \frac{4\pi R}{V}$[/tex]
Explanation:
Given :
Mass of ball 1 = m
Mass of ball 2 = 2m
Since, R>>>d, the collision is head on.
Therefore, we get
[tex]$ \frac{v_1 -v_2}{V}=\frac{1}{2}$[/tex]
[tex]$ \therefore \frac{\text{velocity of seperation}}{\text{velocity of approach}}= v_1-v_2 = \frac{V}{2}$[/tex]
Relative velocity is given by V/2. So, we get the time when the masses will again collide as
[tex]$ t = \frac{2\pi R}{\frac{V}{2}}=\frac{4\pi R}{V} $[/tex]
if a cart goes around a turn at 20 km/h ,what remains constant
1.position
2.velocity
3.direction
4.speed
Answer: 4.speed
Explanation:
In this case, we know that the cart remains at a constant 20km/h.
Now, one could say that "the velocity remains constant, because it always is 20km/h"
But remember that velocity is a vector, so this has a direction, and if the cart is going around a turn, then the direction of motion is changing, which tell us that there is acceleration.
But the module of the velocity, the speed, remains constant at 20km/h.
Then the correct option is 4, speed.
A satellite dish has the shape of a parabola when viewed from the side. The dish is inches wide and inches deep. How far is the receiver from the bottom of the dish if the receiver is placed at the focus
Complete question is;
A satellite dish has the shape of a parabola when viewed from the side. The dish is 60 inches wide and 45 inches deep. How far is the receiver from the bottom of the dish if the receiver is placed at the focus?
Answer:
the receiver should be put 40 inches from the bottom of the dish on the concave side of the dish
Explanation:
The base of the dish would simply be the vertex of parabola.
Since we want to find how far the receiver is from the bottom, the place where we'll place the receiver is simply the focus of the parabola.
Now, for example, if this is a parabola that opens upward and has it's vertex at the origin, then half of the diameter at a height of 45 inches gives the two points (60, 22.5) and (-60, 22.5)
Standard form equation of parabola with vertex at origin and pointing upwards is given by;
x² = 4ay
Plugging in the values of x and y gives;
60² = 4a(22.5)
3600/90 = a
a = 40 inches
Thus, the receiver should be put 40 inches from the vertex on the concave side of the dish
A Lotus will travel 275 meters in 4.71 seconds. What is this car's average speed?
Which of these terms is defined as the ability to cause motion or create change? A. efficiency B. energy C. force D. sound
The term that defines the ability to cause motion or create change is called force. Details about force can be found below.
What is force?Force is the physical quantity that denotes ability to push, pull, twist or accelerate a body and which has a direction.
Force is measured in Newtons and can be calculated by multiplying the mass of the body by its acceleration.
Therefore, it can be said that the term that defines the ability to cause motion or create change is called force.
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A car stops in 120 m. If it has an acceleration of –5m/s 2 , how long did it take to stop
Answer:
t=240s
Explanation:
Distance=120m
Acceleration=-5m/s^2
v=0
Let u=x m/s
Using equation v^2-u^2=2as:-
0-x=2(-5)(120)
-x=-1200
x=1200m/s
Using now equation v=u+at:-
0=1200+-5t
5t=1200
t=240s
If a car stops at 120 meters. If it has an acceleration of –5 meters/second², then it would take 6.928 seconds to stop.
What is acceleration?The rate of change of the velocity with respect to time is known as the acceleration of the object.
As given in the problem a car stops at 120 meters. If it has an acceleration of –5 meters/second², then we have to find out how long it would take seconds to stop.
By using the second equation of motion,
s = ut + 1/2at²
The distance traveled by car before stopping = 120 meters
acceleration = –5 meters/second²
-120 = 0 + 0.5×( –5)t²
t² = 120/2.5
t² =48
t = 6.928 seconds
Thus, the time taken by the car before stopping would be 6.928 seconds.
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