Answer:
t = 2.26 x 10⁵ s
Explanation:
The energy supplied to the water will be equal to the heat required for the boiling of water:
E = ΔQ
Pt = mL
where,
P = Power = 10 W
t = time = ?
m = mass of water = 1 kg
L = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Therefore,
[tex](10\ W)t = (1\ kg)(2.26\ x\ 10^6\ J/kg)\\\\t = \frac{2.26\ x\ 10^6\ J}{10\ W}\\\\[/tex]
t = 2.26 x 10⁵ s
This time will be less than the actual time taken due to some heat loss during the transmission of this heat energy to the container in which water is held.
How many joules of energy are required to accelerate one kilogram of mass from rest to a velocity of 0.866c?
Answer:
the amount of energy needed is 1.8 x 10¹⁷ J.
Explanation:
Given;
mass of the object, m₀ = 1 kg
velocity of the object, v = 0.866 c
By physics convection, c is the speed of light = 3 x 10⁸ m/s
The energy needed is calculated as follows;
E = Mc²
As the object approaches the speed of light, the change in the mass of the object is given by Einstein's relativity formula;
[tex]M = \frac{M_0}{\sqrt{1- \frac{v^2}{c^2} } } \\\\ M = \frac{1}{\sqrt{1- \frac{(0.866c)^2}{c^2} } }\\\\ M = \frac{1}{\sqrt{1- \frac{0.74996c^2}{c^2} } }\\\\ M = \frac{1}{\sqrt{0.25} } \\\\ M = 2 \ kg[/tex]
The energy required is calculated as;
E = 2 x (3 x 10⁸)²
E = 1.8 x 10¹⁷ J
Therefore, the amount of energy needed is 1.8 x 10¹⁷ J.
A fast moving vehicle travelling at a speed of 25.4 m/s comes up behind another vehicle which is
travelling at a slower constant speed of 13.6 m/s. If the faster vehicle does not begin braking until it
is 11.4 meters away from the car in front of it, what is the minimum acceleration that the faster car
must exhibit if it is to avoid colliding with the car in front? Assume that both cars are travelling in the
positive direction
Answer:
a = 6.1 m / s²
Explanation:
For this kinematics exercise, to solve the exercise we must set a reference system, we place it in the initial position of the fastest vehicle
Let's find the relative initial velocity of the two vehicles
v₀ = v₀₂ - v₀₁
v₀ = 25.4 - 13.6
v₀ = 11.8 m / s
the fastest vehicle
x = v₀ t + ½ a t²
The faster vehicle has an initial speed relative to the slower vehicle, therefore it is as if the slower vehicle were stopped, so the distance that must be traveled in a fast vehicle to reach this position is
x = 11.4 m
let's use the expression
v² = v₀² - 2 a x
how the vehicle stops v = 0
a = v₀² / 2x
a = [tex]\frac{11.8^2}{2 \ 11.4}[/tex]
a = 6.1 m / s²
this velocity is directed to the left
A ball rolled along a horizontal surface comes to rest in a distance of 72m in 6s. Its initial velocity
and deceleration are
and
Answer:
1. Initial velocity = 24 m/s
2. Deceleration = –4 m/s²
Explanation:
From the question given above, the following data were obtained:
Distance travelled (s) = 72 m
Time (t) = 6 s
Final velocity (v) = 0 m/s
1. Determination of the initial velocity.
Distance travelled (s) = 72 m
Time (t) = 6 s
Final velocity (v) = 0 m/s
Initial velocity (u) =?
s = (u + v)t / 2
72 = (u + 0) × 6 / 2
72 = u × 3
Divide both side by 3
u = 72 / 3
u = 24 m/s
2. Determination of the deceleration.
Time (t) = 6 s
Final velocity (v) = 0 m/s
Initial velocity (u) = 24 m/s
Deceleration (a) =?
v = u + at
0 = 24 + (a × 6)
0 = 24 + 6a
Collect like terms
0 – 24 = 6a
–24 = 6a
Divide both side by 6
a = –24 / 6
a = –4 m/s²
What is the net force on the side of the container The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0.850 ATM. A faucet with an opening 1.0 inch diameter is located at the bottom of the container.
Answer:
26467.21 N
Explanation:
Initial height of water ( h1 ) = 5 m
diameter of container ( d1 )= 100 cm
pressure inside the container ( p1 )= 0.850 atm
Diameter of faucet ( d2 )= 1 inch
Calculate the value of the net force on the side of container
lets assume ; pressure outside the container ( p2 ) = 1 atm
Fnet = ( P1A1 + mg ) - ( P2A2 )
= [ ( 0.85 * 101325 ) ( π(1/2)^2 ) + mg ) - [ ( 101325 )( π )(0.0127)^2 ]
= [ 16902.2766 + pvg ] - [ 51.3161 ]
where ; pvg = pAhg = 1000 * π ( 1/2 )^2 * 5 * 9.8 = 9616.25
= [ 16902.2766 + 9616.25 ] - [ 51.3161 ] = 26467.21 N ( downwards )
please help me .finish this paper
Solution-1:-
[tex]\boxed{\sf \dfrac{10\times 1000}{60\times 60}}[/tex]
Solution:-2
[tex]\boxed{\sf Sodium\:and\:Potassium}[/tex]
Solution:-3
[tex]\boxed{\sf 320m}[/tex]
Solution:-4
[tex]\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}[/tex]
Solution:-5
[tex]\boxed{\sf Mg_3N_2}[/tex]
Solution:-6
[tex]\boxed{\sf Grapes\:and\:Rambutan}[/tex]
Solution:-7
[tex]\boxed{\sf {}^{}_{}N}[/tex]
Solution:-8
[tex]\boxed{\sf Galactuse}[/tex]
Solution:-9
[tex]\boxed{\sf Y-X}[/tex]
Solution:-10
[tex]\boxed{\sf Cell\:wall}[/tex]
A hungry monkey is sitting at the top of a tree 69 m above ground level. A person standing on the ground wants to feed the monkey. He uses a tee-shirt cannon to launch bananas at the monkey. If the person knows that the monkey is going to drop from the tree at the same instant that the person launches the bananas, how should the person aim the banana cannon
Answer:
Well if you want to be sure you should just throw it to the ground so then when he lands he can catch it.
If the cannon throws the banana with the same force the monkey falls
(m.g=Fz <=> m.9,81N/kg=...N).
Then the throw will slow down because of the gravitational pull.
Because the banana cannon is selfmade you can choose what mass the bananas in question have, so let that be the same as the monkeys.
The monkey falls with the speed of 9,81m.s => so it takes the monkey 7,1s to land.
If the cannon can shoot the banana at the same speed the monkey falls then they would cross in the middle.
So to do so you need to throw the bananas with a speed of at least 9,81m.s
Soo ... throw them with a force of that is greater then the gravitational pull and things will work out.
I'm sorry I don't know why I wrote all of this irrelevant information it's 2:21 right now and I'm tired.
kind regards
A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? What is the amplitude?
The period and the amplitude of the weight suspended from spring are 0.33 seconds and 10 centimeters, respectively.
1) The period is given by:
[tex] T = \frac{1}{f} [/tex]
Where:
f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz
[tex] T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s [/tex]
Hence, the period is 0.33 s.
2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:
[tex] A = \frac{20 cm}{2} = 10 cm [/tex]
Therefore, the amplitude is 10 cm.
You can learn more about the period and amplitude here: https://brainly.com/question/15169209?referrer=searchResults
I hope it helps you!
I need the help
Please. I’m terrible at physics
Answer:
Explanation:
so opposite and equal , right? forces are. soooo..
528+52= 580 N is the force that is being exerted up on the scale
12. A car travels in a straight line with an average velocity of 80 km/h for 2.5h and then an average velocity of 40 km/h from 1.5 h (a) what is the total displacement for the 4 h trip? (b) What is the average velocity for the total trip?
Answer:
a. Total displacement = 140 km/h
b. Average velocity = 35 km/h
Explanation:
Given the following data;
Average velocity A = 80 km/h
Time A = 2.5 hours
Average velocity B = 40 km/h
Time B = 1.5 hours
a. To find the total displacement for the 4 h trip;
Total time = Time A + Time B
Total time = 2.5 + 1.5
Total time = 4 hours
Next, we would determine the displacement at each velocity.
Mathematically, displacement is given by the formula;
Displacement = velocity * time
Substituting into the formula, we have;
Displacement A = 80 * 2.5
Displacement A = 200 km/h
Displacement B = 40 * 1.5
Displacement B = 60 km/h
Total displacement = Displacement A - Displacement B
Total displacement = 200 - 60
Total displacement = 140 km/h
b. To find the average velocity for the total trip;
Mathematically, the average velocity of an object is given by the formula;
[tex] Average \; velocity = \frac {total \; displacement}{total \; time} [/tex]
Substituting into the formula, we have;
[tex] Average \; velocity = \frac {140}{4} [/tex]
Average velocity = 35 km/h
The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns
A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.
Answer:
The answer is continuity ( D )
Explanation:
PLZ MARK AS BRAINLIEST
A person pushes horizontally on a heavy box and slides it across the level floor at constant velocity. The person pushes with a 60.0 N force for the first 16.4 m at which time he begins to tire. The force he exerts then starts to decrease linearly from 60.0 N to 0.00 N across the remaining 6.88 m. How much total work did the person do on the box
Over the first 16.4 m, the person performs
W = (60.0 N) (16.4 m) = 984 J
of work.
Over the remaining 6.88 m, they perform a varying amount of work according to
F(x) ≈ 60.0 N + (-8.72 N/m) x
where x is in meters. (-8.72 is the slope of the line segment connecting the points (0, 60.0) and (6.88, 0).) The work done over this interval can be obtained by integrating F(x) over the interval [0, 6.88 m] :
W = ∫₀⁶˙⁸⁸ F(x) dx ≈ 206.4 J
(Alternatively, you can plot F(x) and see that it's a triangle with base 6.88 m and height 60.0 N, so the work done is the same, 1/2 (6.88 m) (60.0 N) = 206.4 J.)
So the total work performed by the person on the box is
984 J + 206.4 J = 1190.4 J ≈ 1190 J
An object carries a charge of -8.5 µC, while another carries a charge of -2.0 µC. How many electrons must be transferred from the first to the second object so that both objects have the same charge?
Answer:
Approximately 2*10^13 electrons must be transferred
Explanation:
Below is the given information:
First object carries charge = -8.5 µC
Number of electrons in 1st = 8.5 x 10^-6/(1.6 x 10^-19) = 5.3125 x 10^13
Second object carries a charge = -2.0 µC
The number of electrons in 2nd = 2*10^-6/(1.6*10^-19) = 1.25 x 10^13
so, approximately 2 x 10^13 electrons must be transferred
Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank the objects according to the magnitude of their momentum.
Required:
Rank from smallest to largest.
Answer:
Momentum of object A = Momentum of object C < momentum of B.
Explanation:
The momentum of an object is equal to the product of mass and velocity.
Object A has a mass m and a speed v. Its momentum is :
p = mv
Object B has a mass m/2 and a speed 4v. Its momentum is :
p = (m/2)×4v = 2mv
Object C has a mass 3m and a speed v/3. Its momentum is :
p = (3m)×(v/3) = mv
So,
Momentum of object A = Momentum of object C < momentum of B.
A 59.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 0.250 mins, what is the spring constant (in N/m) of the bungee cord, assuming it has negligible mass compared to that of the jumper
Answer:
The spring constant of the spring is 10.3 N/m.
Explanation:
Given that,
Mass of a bungee jumper, m = 59 kg
The period of oscillation, T = 0.25 min = 15 sec
We need to find the spring constant of the bungee cord. We know that the period of oscillation is given by :
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Where
k is the spring constant
[tex]T^2=4\pi^2\times \dfrac{m}{k}\\\\k=4\pi^2\times \dfrac{m}{T^2}\\\\k=4\pi^2\times \dfrac{59}{(15)^2}\\\\k=10.3\ N/m[/tex]
So, the spring constant of the spring is 10.3 N/m.
A covalent bond is formet by of electrons..?
Answer:
The covalent bond is formed by pairs of electrons that are shared between two atom
Explanation:
The covalent bond is formed by pairs of electrons that are shared between two atoms, in general the electrons must have opposite spins to have a lower energy state.
In this bond, the electrons are between the two atoms and are shared between them in such a way that there is a configuration of eight electrons in the orbit.
The same constant force is used to accelerate two carts of the same mass, initially at rest, on horizontal frictionless tracks. The force is applied to cart A for twice as long a time as it is applied to cart B. The work the force does on A is WA; that on B is WB. Which statement is correct?
a. WA = WB
b. WA = 2WB.
c. WA=4WB
d. WB= 2WA
Answer:
Option (c).
Explanation:
Let the mass of each cart is m and the force is F.
Time for cart A is 2t and for cart B is t.
Work done is given by the
W= force x displacement
As the distance is given by
S= u t +0.5 at^2
So, when the time is doubled the distance is four times.
So, WA = F x 4 S
WB = F x S
WA= 4 WB
uniform electric field of magnitude 365 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.00 cm. (a) What is the work done by the field on the electron? 1.753e-18 J (b) What is the change in potential energy associated with the electron? J
Answer:
a) W = - 1.752 10⁻¹⁸ J, b) U = + 1.752 10⁻¹⁸ J
Explanation:
a) work is defined by
W = F . x
the bold letters indicate vectors, in this case the force is electric
F = q E
we substitute
F = q E x
the charge of the electron is
q = - e
F = - e E x
let's calculate
W = - 1.6 10⁻¹⁹ 365 3 10⁻²
W = - 1.752 10⁻¹⁸ J
b) the change in potential energy is
U = q ΔV
the potential difference is
ΔV = - E. Δs
we substitute
U = - q E Δs
the charge of the electron is
q = - e
U = e E Δs
we calculate
U = 1.6 10⁻¹⁹ 365 3 10⁻²
U = + 1.752 10⁻¹⁸ J
A compact disk with a 12 cm diameter is rotating at 5.24 rad/s.
a. What is the linear speed _______m/s
b. What is the centripetal acceleration of a point on its outer rim _______
c. Consider a point on the CD that is halfway between its center and its outer rim. Without repeating all of the calculations required for parts (a) and (b), determine the linear speed of this point. _______m/s
d. Determine the centripetal acceleration of this point. _______
Answer:
(a) 31.44 m/s (b) 164.74 m/s²
Explanation:
Given that,
The diameter of a disk, d = 12 cm
Radius, r = 6 cm
Angular speed = 5.24 rad/s
(a) Linear speed,
[tex]v=r\omega\\\\v=6\times 5.24\\\\v=31.44\ m/s[/tex]
(b) Centripetal acceleration,
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{31.44^2}{6}\\\\a=164.74\ m/s^2[/tex]
TRUE or FALSE: The acceleration of projectile is 0 m/s/s at the peak of the trajectory. Identify the evidence which supports your answer.
The vertical acceleration of the projectile is at 0 m/s while the horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration ( i.e. statement in the question is False )
Projectile motion follows a parabolic path with x and y components of its velocity and acceleration. also the acceleration of a projectile is subject only to the acceleration due to gravity unlike other kinds of motions.
In a parabolic motion an object ( projectile ) is thrown into the air and left to move through a parabolic path under the effect of acceleration due to gravity.
Hence we can conclude that the statement is false, because horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration
Learn more : https://brainly.com/question/24658194
What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 3.22 x 104 V
Answer:
[tex]E=3.22*10^6 N/C[/tex]
Explanation:
From the question we are told that:
Separation Distance [tex]d=1.0cm =0.01m[/tex]
Potential difference [tex]V=3.22 * 10^4 V[/tex]
Generally the equation for Electric Field strength is mathematically given by
[tex]E=\frac{v}{d}[/tex]
[tex]E=\frac{3.22*10^4}{0.01}[/tex]
[tex]E=3.22*10^6 N/C[/tex]
what is the prefix notation of 0.0000738?
Answer:
The scientific notation of 738 is 7.38 x 1002.
HEELLPPPPPpppppppppppppppp
Explanation:
Given:
[tex]A_1[/tex] = 4.5 cm[tex]^2[/tex]
[tex]v_1[/tex] = 40 cm/s
[tex]v_2[/tex] = 90 cm/s
[tex]A_2[/tex] = ?
a) The continuity equation is given by
[tex]A_1v_1 = A_2v_2[/tex]
Solving for [tex]A_2[/tex],
[tex]A_2 = \dfrac{v_1}{v_2}A1 = \left(\dfrac{40\:\text{cm/s}}{90\:\text{cm/s}}\right)(4.5\:\text{cm}^2)[/tex]
[tex]= 2\:\text{cm}^2[/tex]
b) If the cross-sectional area is reduced by 50%, its new area [tex]A_2'[/tex] now is only 1 cm^2, which gives us a radius of
[tex]r = \sqrt{\dfrac{A_2'}{\pi}} = 0.564\:\text{cm}[/tex]
Una bala de 10 g se dispara contra un bloque de madera de 102 g inicialmente en reposo sobre una superficie horizontal. Después del impacto el bloque se desliza 8 m antes de detenerse. Si el coeficiente de fricción entre el bloque y la superficie es 0,5, ¿Cuál es la velocidad de la bala inmediatamente antes del impacto?
Answer:
una ess abola cola sola answer
Which person ha the most freedom to make his or her own lifestyle decisions?
A. Frieda is 10 years old and lives with her grandmother
B. Vladimir is 3 years old and attends preschool
C. Quincy is 16 years old and lives in a dormitory
D. Lucinda is 32 years old and has two children
Answer:
C Quincy
Explanation:
Both Frieda and Vladimir are too young to be making their own decisions. Lucinda has limited freedom due to having two children, while Quincy is just now becoming an adult and has his whole life still ahead of him. Therefore, Quincy has the most freedom to make their own lifestyle decisions.
I hope this helps!
Answer:
The correct answer is C. Quincy is 16 years old and lives in a dormitory
Explanation:
Quincy being 16 may be able to get a job and hold a membership at a gym even if not at a gym he still has the most freedom period with Lucinda having children and Frieda and Vladimir being to young to make the choices of exercising on their own.
Please tell me if I'm wrong so I may give you the correct answer!
Happy Holidays!!
What distance do I cover if I travel 10 m E, then 6 mW, then 12 m E?
A. 16 m
B. 28 m
C. 16 m E
D. 28 m E
Answer:
C. 16 m E
Explanation:
Applying,
The law of addition of vector: Vector in the same direction are added while vector in opposite direction are substracted
From the question above,
Step 1: Total distance covered towards east = 10+12 = 22 m E
Step2: Total distance covered towards west = 6 m W
Therefore, the resultant distance traveled = 22-6 = 16 m E
Hence the right option is C. 16 m E
A 2 nC point charge is embedded at the center of a nonconducting sphere of radius = 4.8 cm which has a charge of 1 nC distributed uniformly throughout its volume. What is the magnitude of the electric field at a point that is 2.4 cm from the center of the sphere? Write your answer in terms of kN/C.
Answer: [tex]3.32\times 10^4\ kN/C[/tex]
Explanation:
Given
Charge at the center of the sphere is [tex]q=2\ nC[/tex]
Charge distributed over the entire sphere [tex]Q=1\ nC[/tex]
Radius of sphere [tex]r=4.8\ cm[/tex]
Using Guass law
[tex]\Rightarrow \oint \vec{E}\cdot \vec{dA}=\dfrac{q_{enc}}{\epsilon_o}\\\\\Rightarrow E\cdot 4\pi r^2=\dfrac{1}{\epsilon}(q+\dfrac{Q}{4\pi R^2}\times 4\pi r^2)\\\\\Rightarrow E\cdot 4\pi r^2=\dfrac{1}{\epsilon}(q+Q\dfrac{r^2}{R^2})\\\\\Rightarrow E\cdot (2.4\times 10^{-2})^2=9\times 10^9(2+1\cdot \dfrac{1}{4})\times 10^{-9}\\\\\Rightarrow E=3.32\times 10^4\ kN/C[/tex]
write physical quantities and its unit
length= metre
mass= kg
time= second
temperature = kelvin
current= ampere
luminous intensity= candela
Amount of substance = mole
etc
I hope this will help you
stay safe
A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on
either side of the canal. Each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal. Find the sum of these
two forces on the barge.
answer in ___kN
Answer:
1.621 kN
Explanation:
Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).
The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).
So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N
So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N
The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)
= √(R² + 0²) (since R' = 0)
= √R²
= R
= 1620.82 N
= 1.62082 kN
≅ 1.621 kN
So, the sum of these two forces on the barge is 1.621 kN
Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statement correctly describes the change in momentum of the two balls?
a. |ΔpBl<|ΔPA|
b. |ΔpBl=|ΔPA|
c. |ΔpB|>|ΔPA|
d. ΔpB > ΔPA
Answer:
Option A
Explanation:
From the question we are told that:
Mass [tex]m=0.20kg[/tex]
Velocity [tex]v=4m/s[/tex]
Generally the equation for momentum for Ball A is mathematically given by
Initial Momentum
[tex]M_{a1}=mV[/tex]
[tex]M_{a1}=0.2*4[/tex]
[tex]M_{a1}=0.8[/tex]
Final Momentum
[tex]M_{a2}=-0.8kgm/s[/tex]
Therefore
[tex]\triangle M_a=-1.6kgm/s[/tex]
Generally the equation for momentum for Ball B is mathematically given by
Initial Momentum
[tex]M_{b1}=mV[/tex]
[tex]M_{b1}=0.2*4[/tex]
[tex]M_{b1}=0.8[/tex]
Final Momentum
[tex]M_{b2}=-0 kgm/s[/tex]
Therefore
[tex]|\triangle M_a|>|\triangle Mb|[/tex]
Option A
A force of 15 N toward the WEST is applied to a 4.0 kg box. Another force of 42 N toward the EAST is also applied to the 4.0 kg box. The net force on the 4.0 kg box
is
[tex]\implies F_1 < F_2[/tex]
[tex] \implies F_{net} = F_2 - F1[/tex]
[tex]\implies F_{net} = 42 -15[/tex]
[tex]\implies \underline{ \boxed{ F_{net} = 27 \: N}}[/tex]
The net force on the 4.0 kg box is 27 N towards EAST.
