Answer:
10J
Explanation:
In this question we have the following information
The charge of the particle is q = 5 C
The equipotenetial level is V1 = 5.5 v
and also the
equipotenetial level is V2 = 3.5 v
So we calculate the
work done W=q x (v1-v2)
workdone = 5 x (5.5-3.5)
= 5x2
=10 J
Workdone = 10 J
So we conclude that the workdone on a particle with these information is 10j
Is Solar Energy renewable? Why or why not? Use in your own words.
Answer:
Solar energy is renewable.
Explanation:
If something is renewable, it is generated faster than it can be reasonably used or won't run out for longer than it would be used. Solar falls into the latter category. Using solar panels won't deplete the sun and the sun will likely be around for much longer than we will.
If a 4N weight is hung on a spring, and it extends by 0.2m, what is the spring constant (k)?
Answer: 200 N/m
Explanation:
The Gravitational spring energy(Us) is equal to 1/2kx^2. So we have x as .2 m and Us as 4 N. So 4 N = 1/2 * k * .2^2. So now we solve for K and get 200 N/m.
Answer:
20 N/m
Explanation:
4/0.2 = 20 N/m
/ = divide
REEEEEEEEEEEEEEEEEE
Answer:
eeeeeeeeeeeeeeeeeeeeeee
How does increasing the width of a wire affect a circuit?
A. It restricts the flow of electrons.
B. It reduces the voltage
C. It allows electrons to flow more easily
D. It increases the resistance
Whoever gets this right I’ll give brainliest. Be sure that the answer is right. I’d love a explanation too if you could include one.
Answer:
The resistance of a wire decreases with increasing thickness.
Explanation:
Hope this helped!
Answer: C it allows electrons to flow more easily
Explanation:i got it right i hope this helps you
A small block of mass m1 = 0.4 kg is placed on a long slab of mass m2 = 2.8 kg. Initially, the slab is stationary and the block moves at a speed of vo = 3 m/s. The coefficient of kinetic friction between the block and the slab is 0.15 and there is no friction between the slab and the surface on which it moves.
Determine the speed v1.
Determine the distance traveled by the slab before it reaches the speed v1.
Answer:
v₁ = 0.375 m / s , x = 0.335 m
Explanation:
Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.
We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.
initial moment. Right at the start of the movement
p₀ = m v₀ + 0
final moment. Just when it comes to equilibrium
[tex]p_{f}[/tex] = (m + M) v₁
how the forces are internal
p₀ =p_{f}
m v₀ = (m + M) v₁
v₁ = m /m+M v₀
let's calculate
v₁ = 0.4 /(0.4 + 2.8) 3
v₁ = 0.375 m / s
Let's apply Newton's second law to the Block, to find the friction force
Y axis
N - W = 0
N = W
N = m g
where m is the mass of the block
the friction force has the formula
fr = μ N
fr = μ m g
We apply Newton's second law to slab
X axis
fr = M a
where M is the mass of the slab
μ m g = M a
a = μ g m / M
let's calculate
a = 0.15 9.8 0.4 / 2.8
a = 0.21 m / s²
With kinematics we can find the position
v²= v₀²+2 a x
as the slab is initially at rest, its initial velocity is zero
v² = 2 a x
x = v2 / 2a
let's calculate
x = 0.375²/2 0.21
x = 0.335 m
How much is the velocity of a body when it travels 600m in 5 minutes?
There are 60 minutes in an hour. Whatever distance a person crosses in 5 mins, he/she can cross 12 times that distance in 1 hour. So, in one hour the person could have crossed 12 X 600 = 7200 meters.
But, 1000 meters equal one kilometer. So, 7200m = 7.2 km. Thus, this person travels 7.2 km in one hour, and hence has a speed of 7.2 km/hr
Answer:
2m/s
Explanation:
V=S/t
v=distance ÷ time
v=600m. × 5mins
v=600/5 ×60
600/300
2m/s
A car’s brakes decelerate it at a rate of -2.40 m/s2. If the car is originally travelling at 13 m/s and comes to a stop, then how far, in meters, will the car travel during that time?
Answer:
Approximately [tex]35.2\; \rm m[/tex].
Explanation:
Given:
Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].
Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)
Implied:
Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)
Required:
Displacement, [tex]x[/tex].
Not required:
Time taken, [tex]t[/tex].
Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:
[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].
In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.
Do you think there is a temperature at which sound cannot be heard?
Answer:
The definition would be defined in the clarification section downwards and according to the given question.
Explanation:
This would be considered a lapse of temperature. The rate of acceleration sometimes gradually decreases, because as the temperature is lowered through height. The latter suggests that perhaps the layer of water nearest to either the surface moves the quickest, as well as the layer of water and the furthest just above the surface, travels the fastest pace, with shock energy passing underground.The waves move and curve vertically as a consequence. This will create an area of the "shadow zone" where the sound couldn’t reach into. Although it could be possible to have seen the origin, a guy standing throughout the loss impairment may not remember the sound. The charged particles are refracted vertically and it's never going to reach the listener.A car accelerates at a rate of 5.0 m/s2 when the engine supplies a net force of 5500 N. What is the mass of the car?
1100 kg
550 kg
2200 kg
27500 kg
Answer:
1100 kgExplanation:
The mass of the car can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{5500}{5} \\ [/tex]
We have the final answer as
1100 kgHope this helps you
True or false. when objects collide , some momentum is lost
Answer:
It is neither false nor true. When they collide some of one of the objects goes to the other object.
Explanation:
Answer: True
Explanation:
The cardinal, central, and secondary traits are all part of __________ categorized traits. A. Gordon Allport’s B. Robert McCrae’s C. Paul Costa’s D. Hans Eysenck’s
Answer:
Gordon Allport’s
Explanation:
edge2o2o
The cardinal, central, and secondary traits are all part of Gordon Allport’s categorized traits. The Correct option is A
Who was Gordon Allport ?
Gordon Willard Allport was born on 11 November 1897 and died 9 October 1967. He was an American psychologist. Allport was first psychologists who studied on personality. he has developed theory of personality. which was one of the greatest finding in the study of personality psychology. He was Appointed as a social science instructor at Harvard University in 1924,
Gordon Allport was a great trait theorist who categorized personality traits into three categories cardinal, central, and secondary.
Hence option A is Correct.
To know more about traits, click :
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While traveling North along a highway a driver slows from 25 m/s to 15 m/s in 12 seconds. What is the
automobile's acceleration?
Answer:
- 0.8333 m/s^2
Explanation:
the equation for calculating acceleration that I have used is
(V2-V1)/t
where V2 is the final recorded velocity and V1 is the initial velocity and t is the duration of the acceleration.
which you plug the numbers in (15 - 25)/12 = 10/12 and because you are slowing down, it should be negative
I am in highschool honors physics so I am not college leveled, I might be wrong but I hope this is useful,
Xavier is roller skating at 14 km/h and tosses a set of keys forward on the ground at 8 km/h. The speed of the keys relative to the ground is
Answer:
22 km/h
Explanation:
Given that,
Speed of Xavier, v = 14 km/h
He tosses a set of keys forward on the ground at 8 km/h, v' = 8 km/h
We need to find the speed of the keys relative to the ground. Let it is V.
As both Xavier and the keys are moving in same diretion. The relative speed wrt ground is given by :
V = v+v'
V= 14 + 8
V = 22 km/h
So, the speed of the keys relative to the ground is 22 km/h.
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude of the gravitational force exerted on the satellite by the earth? (The mass of the earth is 6.0 x 1024 kg and G = 6.67 x 10-11Nm2/kg2.
Answer:
A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.
Explanation:
At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:
[tex]\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}[/tex] (1)
Where:
[tex]T[/tex] - Period of rotation of the satellite, measured in seconds.
[tex]r[/tex] - Distance of the satellite with respect to the center of the planet, measured in meters.
[tex]G[/tex] - Gravitational constant, measured in newton-square meters per square kilogram.
[tex]M[/tex] - Mass of the Earth, measured in kilograms.
Now we clear the distance of the satellite with respect to the center of the planet:
[tex]r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}[/tex]
[tex]r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }[/tex] (2)
If we know that [tex]G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]M = 6.0\times 10^{24}\,kg[/tex] and [tex]T = 25800\,s[/tex], then the distance of the satellite is:
[tex]r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }[/tex]
[tex]r \approx 18.897\times 10^{6}\,m[/tex]
The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:
[tex]F = \frac{G\cdot m\cdot M}{r^{2}}[/tex] (3)
Where:
[tex]m[/tex] - Mass of the satellite, measured in kilograms.
[tex]F[/tex] - Force exerted on the satellite by the Earth, measured in newtons.
If we know that [tex]G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]M = 6.0\times 10^{24}\,kg[/tex], [tex]m = 6105\,kg[/tex] and [tex]r \approx 18.897\times 10^{6}\,m[/tex], then the gravitational force is:
[tex]F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}[/tex]
[tex]F = 6841.905\,N[/tex]
A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.
If a dog has a mass of 2.5 kg, what is its weight and what is the normal force that it feels.
I
Answer:
Weight = normal force = 24.5 N
Explanation:
Given that,
Mass of a dog, m = 2.5 kg
We need to find its weight and the normal force that it feels.
The weight of an object is given by :
W = mg
Where g is the acceleration due to gravity
[tex]W=2.5\times 9.8\\\\=24.5\ N[/tex]
The normal force is balanced by the weight of an object. So,
Weight = normal force = 24.5 N
If the net force acting on an object is not zero, then the object is definitely
at rest
moving with a constant velocity
being accelerated
losing mass
Answer:
C
Explanation:
Since the net force is changing, and not the speed, it is C
Hope this helped!
3. If the bus stop is 0.68 km down the street from the museum and it takes
you 9.5 min to walk north from the bus stop to the museum entrance,
what is your average velocity?
Answer:
0.07km/minExplanation:
Step one:
given data
distance between museum and bus top= 0.68km
time taken to walk from the bus stop to the museum = 9.5min
Required
The average velocity is defined as the rate at which you are moving
The expression for the average velocity is
Average velocity = distance/time taken
Average velocity= 0.68km/9.5
Average velocity= 0.07km/min
Hence your average velocity is 0.07km/min
During the stretching routine who used the medicine ball for support.
The guy
The guy
The Girl
The Girl
Both people used a ball for support
Both people used a ball for support
No one used it
No one used it
a 1000kg car uses a breaking force of 10,000N to stop in two second. What is the change in momentum of the car?
Answer:
ΔP = 20000 N s
Explanation:
To solve this problem we use the relation between momentum and moment
I = Δp
let's calculate the momentum
I = ∫F dt
if we use the average force
I = F t
I = 10000 2
I = 20000 N s
therefore with the first equation
ΔP = I = 20000 N s
1. How far away must you be from a 675 kHz radio station with power 50.0 kW for there to be only one photon per second per square meter? Assume no reflections or absorption, as if you were in deep outer space. 2. Discuss the implications for detecting intelligent life in other solar systems by detecting their radio broadcasts.
a) 0.321 ly
b) 0.321 light years is not far in astronomical terms. Alien life would need to transmit at tremendous power in order for their radio transmissions to be detectable. Their radio signal also needs to be stronger than background noise in order to be distinguishable. Therefore it is unlikely that radio transmissions from alien life will ever be detected.
A cube of wood having an edge dimension of 18.0 cm and a density of 651 kg/m3 floats on water.(a) What is the distance from the horizontaltop surface of the cube to the water level?(b) How much lead weight has to be placed on top of the cube sothat its top is just level with the water?
Answer:
A. 6.282
B. 2.03kg
Explanation:
A.
We solve using archimedes principle
L³pwood = L²dwater
We make d subject of the formula
d = Lpwood/pester
= 18x651/1000
= 18x0.651
= 11.718cm
Distance from horizontal top to water level
= 18-11.718
= 6.282cm
B.
When we place lead block
WL + L³pwoodg = L³pwaterg
WL = L³g(Pwater-Pwood)
= 0.18³x9.8(1000-651)
= 19.94N
19.94/9.8
= 2.03kg
The mass m is therefore 2.03kg
(a) The distance will be "6.282 cm".
(b) Mass will be "2.03 kg".
(a)
From Archimedes' principle, we get
→ [tex]L^3 \rho_{Wood} = L^2 d \rho_{Water}[/tex]
[tex]d = L \frac{\rho_{Wood}}{\rho_{Water}}[/tex]
[tex]= 18\times \frac{651}{1000}[/tex]
[tex]= 11.72 \ cm[/tex]
So,
The distance from horizontal top to the water level will be:
= [tex]18-11.72[/tex]
= [tex]6.282 \ cm[/tex]
(b)
After placing the lead block of weight [tex]W_L[/tex], we have
→ [tex]W_L +L^3 \rho_{Wood} g = L_3 \rho_{Water} g[/tex]
[tex]W_L = L^3 g(\rho_{Water}-\rho_{Wood})[/tex]
[tex]= 0.18^3\times 9.8\times (1000-651)[/tex]
[tex]= 19.94 \ N[/tex]
Mass,
m = 2.03 kg
Thus the above answer is right.
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A ball is thrown vertically upward with an initial velocity of 23 m/s. What are its position and velocity after 2 s?
Answer:
The position of the ball after 2 s is 26.4 mThe velocity of the ball after 2 s is 3.4 m/sExplanation:
Given;
initial velocity of the ball, u = 23 m/s
time of motion, t = 2 s
The position of the ball after 2 s is given by;
h = ut - ¹/₂gt²
h = (23 x 2) - ¹/₂ x 9.8 x 2²
h = 46 - 19.6
h = 26.4 m
The velocity of the ball after 2 s is given by;
v² = u² + 2(-g)h
v² = u² - 2gh
v² = 23² - (2 x 9.8 x 26.4)
v² = 529 - 517.44
v² = 11.56
v = √11.56
v = 3.4 m/s
A small box slides down a ramp on a friction with surface. If the total energy of the system is conserved, which computational model expresses the kinetic energy of the box?
B. KEsubbox = Esubtotal - mgh
If you are modeling a system in which an object’s total energy is conserved, you could express potential energy as the difference between total and kinetic, or, alternatively, you could model kinetic energy as the difference between total and potential.
student measures the weight of a bag of bananas with a spring balance.
Describe what is inside a spring balance and explain how it works.
A spring balance measures the weight of an object by opposing the force of gravity acting with force of an extending spring. May be used to determine mass as well as weight by recalibrating the scale. Some spring balances are available in gram or kilogram markings and are used to measure the mass of an object. Spring balances consist of a cylindrical tube with a spring inside. One end (at the top) is fixed to an adjuster which can be used to calibrate the device. The other end is attached to a hook on which you can hang masses etc.
Lisa throws a stone horizontally from the roof edge of a 50 meter high dormitory. It hits the ground at a point 60 m from the building. Find the time of flight.
Answer:
Explanation:
Time of flight is the time of takes to hit the ground
Given
Height H = 50m
Acceleration due to gravity g = 9.8m/s³
Using the equation of motion;
S = ut+1/2gt²
u = 0m/s
Substitute and get time t
50 = 0(t)+1/2(9.8)t²
50 = 4.9t²
t² = 50/4.9
t² = 10.204
t = √10.204
t = 3.19secs
Hence the time of flight is 3.19secs
a current of 200mA through a conductor converts 40 joules of electrical energy into heat in 30 seconds determine the potential drop across the conductor
Answer:
V = 6.65 [volt]
Explanation:
We must first find the power generated, power is defined as the amount of energy consumed or generated in a given time.
[tex]P=\frac{E}{t}[/tex]
where:
P = power [w]
E = energy = 40 [J]
t = time = 30 [s]
[tex]P =40/30\\P = 1.33[w][/tex]
Now we can calculate the voltage or potential drop by means of the power, the power is calculated by means of the product of the voltage by the current.
[tex]P =V*I[/tex]
where:
V = voltage [volts]
I = current = 200mA = 0.2 [A]
[tex]V = P/I\\V = 1.33/0.2\\V = 6.65 [Volt][/tex]
A relaxed spring of length 0.15 m stands vertically on the floor; its stiffness is 1070 N/m. You release a block of mass 0.5 kg from rest, with the bottom of the block 0.6 m above the floor and straight above the spring. How long is the spring when the block comes momentarily to rest on the compressed spring?
Answer:
x' = 0.085 m = 8.5 cm
Explanation:
The law of conservation of energy says that:
Potential Energy Stored in Spring = Loss in Gravitational Potential Energy of Block
(1/2)kΔx² = mgh
where,
k = stiffness of spring = 1070 N/m
Δx = change in length of spring = ?
m = mass of block = 0.5 kg
g = 9.8 m/s²
h = height of block above spring = 0.6 m - 0.15 m = 0.45 m
Therefore,
(1/2)(1070 N/m)Δx² = (0.5 kg)(9.8 m/s²)(0.45 m)
Δx = √[2(2.205 Nm)/(1070 N/m)]
Δx = 0.064 m
but,
Δx = x - x' = 0.15 m - x' = 0.064 m
x' = 0.15 m - 0.064 m
x' = 0.085 m = 8.5 cm
Why are weathering, erosion and deposition a NECESSARY process in the rock cycle?
ANSWER THIS NOW PLEASE! AND YOU GET 225 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
YESS well it is partly nessary but it depends on the situation
Explanation:
Answer:
Weathering, erosion, and deposistion are necessary processes in the rock cycle because:
Explanation:
First, start with igneous rocks. magma erupts (Extrusive igneous rocks) or solidifies in the sub-surface of the earth (Intrusive igneous rock). when they are exposed Weathering and erosion occur which is a slow breakdown of rock through the wind, water, or other processes. The weathered pieces (sediments) move to other places by wind or water and get deposited someplace else. When there are enough sediments and there is overburden pressure on these sediments, they become a sedimentary rock. Due to overburden pressure, they become metamorphic rocks. now the thing to understand here is that when metamorphic rocks are exposed, they too undergo weathering and erosion and their pieces also become sedimentary rocks.
Two objects attract each other with a gravitational force of magnitude 1.01 10-8 N when separated by 19.9 cm. If the total mass of the two objects is 5.11 kg, what is the mass of each?
Answer:
m₂ = 1.17 kg
Explanation:
Given that,
Force between two objects, [tex]F=1.01\times 10^{-8}\ N[/tex]
Mass of object 1, [tex]m_1=5.11\ kg[/tex]
Distance between masses, r = 19.9 cm = 0.199 m
The gravitational force between two masses is given by :
[tex]F=\dfrac{Gm_1m_2}{r^2}[/tex]
m₂ is the mass of object 2
[tex]m_2=\dfrac{Fr^2}{Gm_1}\\\\m_2=\dfrac{1.01\times 10^{-8}\times (0.199)^2}{6.67\times 10^{-11}\times 5.11}\\\\=1.17\ kg[/tex]
So, the mass of second object is 1.17 kg.
An object in FREE-FALL on the MOON would experience which of the following
FORCES?
O a. Weight
O b. Normal
O c. Air Resistance
d. a and c
O e. None of these
Answer:
e. none of these
Explanation:
An object in FREE-FALL on the MOON would experience only acceleration