Answer:
a) Δp = -2.0 kgm / s, b) Δp = -4 kg m / s
Explanation:
In this exercise the change in moment of a ball is asked in two different cases
a) clay ball, in this case the ball sticks to the door and we have an inelastic collision where the final velocity of the ball is zero
Δp = p_f - p₀
Δp = 0 - m v₀
Δp = - 0.100 20
Δp = -2.0 kgm / s
b) in this case we have a bouncing ball, this is an elastic collision, as the gate is fixed it can be considered an object of infinite mass, therefore the final speed of the ball has the same modulus of the initial velocity, but address would count
v_f = - v₀
Δp = p_f -p₀
Δp = m v_f - m v₀
Δp = m (v_f -v₀)
Δp = 0.100 (-20 - 20)
Δp = -4 kg m / s
PLZ help asap :-/
............................
Explanation:
[16][tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]
Here,
[tex]\rm { R_1} [/tex] = 2Ω[tex]\rm { R_2} [/tex] = 2Ω[tex]\rm { R_3} [/tex] = 2Ω[tex]\rm { R_4} [/tex] = 2ΩWe have to find the equivalent resistance of the circuit.
Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,
[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]
Reciprocating both sides,
[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]
Henceforth, Option A is correct.
_________________________________[17][tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]
Here, we have to find the amount of flow of current in the circuit. By using ohm's law,
[tex] \longrightarrow [/tex] V = IR
[tex] \longrightarrow [/tex] 3 = I × 3.33
[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I
[tex] \longrightarrow [/tex] 0.90 Ampere = I
Henceforth, Option B is correct.
____________________________[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]
How are elastic and inelastic collisions different?
A: Elastic collisions occur when the colliding objects move separately after the collision; after inelastic collisions, the objects are connected and move together.
B: Elastic collisions occur when the objects are going the same direction when they collide; inelastic collisions occur when the objects are going in opposite directions when they collide.
C: Momentum is conserved in elastic collisions; momentum is not conserved in inelastic collisions.
D: Elastic collisions occur between objects of the same mass; inelastic collisions occur between different masses.
Answer:
a
Explanation:
Answer:
the answer is c
'
Explanation:
A rigid tank contains 10 lbm of air at 30 psia and 60 F. Find the volume of the tank in ft3. The tank is now heated until the pressure doubles. Find the heat transfer in Btu.
Answer:
Hence the amount of heat transfer is 918.75 Btu.
Explanation:
Now,
A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels, from left to right along a long, horizontal stretched string with a speed of 36.0 m s. I Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement,
(a) What is the frequency of the wave?
(b) What is the angular frequency of the wave?
(c) What is the wave number of the wave?
(d) What is the function y(x,t) that describes the wave?
(e) What is y(t) for a particle at the left end of the string?
(f) What is y(t) for a particle 1.35 m to the right of the origin?
(g) What is the maximum magnitude of transverse velocity of any particle of the string?
(h) Find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
(i) Find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
Explanation:
Given that,
Amplitude, A = 2.5 nm
Wavelength,[tex]\lambda=1.8\ m[/tex]
The speed of the wave, v = 36 m/s
At time t = 0 the left end of the string has its maximum upward displacement.
(a) Let f is the frequency. So,
[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]
(b) Angular frequency of the wave,
[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]
(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]
[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]
The correct equation for the x component of a vector named A with an angle measured from the x axis would be which of the following?
Answer:
Acosθ
Explanation:
The x-component of a vector is defined as :
Magnitude * cosine of the angle
Maginitude * cosθ
The magnitude is represented as A
Hence, horizontal, x - component of the vector is :
Acosθ
Furthermore,
The y-component is taken as the sin of the of the angle multiplied by the magnitude
Vertical, y component : Asinθ
Your cell phone typically consumes about 300 mW of power when you text a friend. If the phone is operated using a lithium-ion battery with a voltage of 3.5 V, what is the current (in A) flowing through the cell-phone circuitry under these circumstances
Answer:
I = 0.0857 A
Explanation:
Given that,
Power consumed by the cellphone, P = 300 mW
The voltage of the battery, V = 3.5 V
Let I is the current flowing through the cell-phone. We know that,
P = VI
Where
I is the current
So,
[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]
So, the current flowing the cell-phone is 0.0857 A.
A 1,200kg roller coaster car starts rolling up a slope at a speed of 15m/s. What is the highest point it could reach
Answer: 11.36 m
Explanation:
Given
Mass of roller coaster is m=1200 kg
Initial speed of roller coaster is v=15 m/s
Energy at bottom and at the top is same i.e.
[tex]\Rightarrow \dfrac{1}{2}mv^2=mgh\\\\\Rightarrow \dfrac{1}{2}\times 1200\times 15^2=1200\times 9.8\times h\\\\\Rightarrow h=\dfrac{15^2}{2\times 9.8}\\\\\Rightarrow h=11.36\ m[/tex]
Thus, the highest point reach by the roller coaster is 11.36 m
Answer:
11.36m
Explanation:
Is the actual height the puck reached greater or less than your prediction? Offer a possible reason why this might be.
Answer:
Answer to the following question is as follows;
Explanation:
The puck's real altitude is lower than ones projection. That's because the mechanism may not be completely frictionless. Electricity is nevertheless wasted owing to particle interactions such as friction, which might explain why the present the results is lower than predicted.
A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the generator has an area of 0.016 m2. If the magnetic field used in the generator has a magnitude of 0.052 T, how many turns of wire are needed
Answer:
The number of turns of wire needed is 573.8 turns
Explanation:
Given;
maximum emf of the generator, = 190 V
angular speed of the generator, ω = 3800 rev/min =
area of the coil, A = 0.016 m²
magnetic field, B = 0.052 T
The number of turns of the generator is calculated as;
emf = NABω
where;
N is the number of turns
[tex]\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s[/tex]
[tex]N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns[/tex]
Therefore, the number of turns of wire needed is 573.8 turns
as the ball rises the vertical component of it's velocity_____. explain
Answer:
Decreases
Explanation:
because its moving against gravitational attraction and at maximum height its velocity will be and it will decrease until it reaches maximum height and the start to increase again
What are the differences among elements, compounds, and mixtures?
Answer:
Elements have a characteristic number of electrons and protons.Both Hydrogen(H) and oxygen(O) are two different elements.
••••••••••••••••
Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of .Hydrogen(H) and Oxygen(O) both qr4 naturally gases,but they react to form water(H2O),which is liquid compound.
•••••••••••••••
A mixture is made of atleast two parts》 solid,liquid or gas.The difference is that it's not a chemical substance that's bonded by other elements.
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Hope it helps...
Have a great day!!!
Answer: Elements have a characteristic number of electrons and protons. Both Hydrogen(H) and oxygen(O) are two different elements. Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of.Hydrogen(H) and Oxygen(O) both qr4 naturally gases, but they react to form water(H2O), which is liquid compound. A mixture is made of at least two parts solid, liquid, or gas. The difference is that it's not a chemical substance that's bonded by other elements.
a microwave operates at a frequency of 2400 MHZ. the height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. assume that microwave energy is generated uniformly on the uipper surface. What is the power output of the oven
Complete question is;
A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly
downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.
Answer:
Power ≈ 600,000 W
Explanation:
We are given;
Frequency; f = 2400 Hz
height of the oven cavity; h = 25 cm = 0.25 m
base area; A = 30 cm by 30 cm = 0.3m × 0.3m = 0.09 m²
total microwave energy content of the cavity; E = 0.50 mJ = 0.5 × 10^(-3) J
We want to find the power output and we know that formula for power is;
P = workdone/time taken
Formula for time here is;
t = h/c
Where c is speed of light = 3 × 10^(8) m/s
Thus;
t = 0.25/(3 × 10^(8))
t = 8.333 × 10^(-10) s
Thus;
Power = (0.5 × 10^(-3))/(8.333 × 10^(-10))
Power ≈ 600,000 W
convert 56km/h to m/s.
Explanation:
15.556 metres per second
ACCORDING TO NEWTON'S THIRD LAW EVERY ACTION HAS EQUAL AND OPPOSITE REACTION BUT THEN WHY DON'T WE FLY WHEN WE FART??
Answer:
Your fart only has so much force, not nearly enough to launch you into oblivion. Your fart and you still exert a force onto each other, so I guess, hypothetically, you could fly if you really, really try hard enough. Just make sure you don't try too hard and prolapse as a result :)
Assuming the atmospheric pressure is 1 atm at sea level, determine the atmospheric pressure at Badwater (in Death Valley, California) where the elevation is 86.0 m below sea level.
Answer:
Atmospheric pressure at Badwater is 1.01022 atm
Explanation:
Data given:
1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa
Elevation (h) = 86m
gravity (g) = 9.8 m/s2
Density of air P = 1.225 kg/m3
Therefore pressure at bad water Pb = Pi + Pgh
Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)
Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa
hence:
Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm
A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction
Answer:
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
Explanation:
By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:
[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)
[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]f_{s}[/tex] - Static friction force, in newtons.
[tex]f_{k}[/tex] - Kinetic friction force, in newtons.
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravitational constant, in meters per square second.
If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:
[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{s} = 0.273[/tex]
[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 0.181[/tex]
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
A car accelerates at 2 meters/s/s. Assuming the car starts from rest how far will it travel in 10 seconds
Answer:
Distance = velocity x time, so 10 m/s X 10 s = 100 m
Explanation:
If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.
V(f) = V(i) + a*t
Final velocity = initial velocity + acceleration x time
Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.
Distance = velocity x time, so 10 m/s X 10 s = 100 m
Answer:
100 m
Explanation:
Given,
Initial velocity ( u ) = 0 m/s
Acceleration ( a ) = 2 m/s^2
Time ( t ) = 10 sec s
To find : Displacement ( s ) = ?
By 2nd equation of motion,
s = ut + at^2 / 2
= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2
= 0 + ( 2 ) ( 100 ) / 2
= 200 / 2
s = 100 m
Assume that I = E/(R + r), prove that 1/1 = R/E + r/E
[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \frac{1}{I} = \frac{R}{E} + \frac{r}{E} }}}}}}[/tex]
[tex]\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]
[tex]I = \frac{ E}{ R + r} \\[/tex]
[tex] ➺\:\frac{I}{1} = \frac{E}{R + r} \\[/tex]
Since [tex]\frac{a}{b} = \frac{c}{d} [/tex] can be written as [tex]ad = bc[/tex], we have
[tex]➺ \: I \: (R + r) = E \times 1[/tex]
[tex]➺ \: \frac{1}{I} = \frac{R + r}{E} \\ [/tex]
[tex]➺ \: \frac{1}{I} = \frac{R}{E} + \frac{r}{E} \\ [/tex]
[tex]\boxed{ Hence\:proved. }[/tex]
[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}[/tex]
A basketball of mass 0.608 kg is dropped from rest from a height of 1.37 m. It rebounds to a height of 0.626 m.
(a) How much mechanical energy was lost during the collision with the floor?
(b) A basketball player dribbles the ball from a height of 1.37 m by exerting a constant downward force on it for a distance of 0.132 m. In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.37 m, what is the magnitude of the force?
Answer:
a)[tex]|\Delta E|=4.58\: J[/tex]
b)[tex]F=61.90\: N[/tex]
Explanation:
a)
We can use conservation of energy between these heights.
[tex]\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})[/tex]
[tex]\Delta E=0.608*9.81(0.6026-1.37)[/tex]
Therefore, the lost energy is:
[tex]|\Delta E|=4.58\: J[/tex]
b)
The force acting along the distance create a work, these work is equal to the potential energy.
[tex]W=\Delta E[/tex]
[tex]F*d=mgh[/tex]
Let's solve it for F.
[tex]F=\frac{mgh}{d}[/tex]
[tex]F=\frac{0.608*9.81*1.37}{0.132}[/tex]
Therefore, the force is:
[tex]F=61.90\: N[/tex]
I hope is helps you!
The angular velocity of an object is given by the following equation: ω(t)=(5rads3)t2\omega\left(t\right)=\left(5\frac{rad}{s^3}\right)t^2ω(t)=(5s3rad)t2 What is the angular displacement of the object (in rad) between t = 2 s and t = 4 s?
Answer:
The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.
Explanation:
The angular velocity of the object ([tex]\omega[/tex]), in radians per second, is given by the following expression:
[tex]\omega(t) = 5\cdot t^{2}[/tex] (1)
Where [tex]t[/tex] is the time, measured in seconds.
The change in the angular displacement ([tex]\Delta \theta[/tex]), in radians, is found by means of the following definite integral:
[tex]\Delta \theta = \int\limits^{4}_{2} {5\cdot t^{2}} \, dt[/tex] (2)
Then we proceed to integrate on the function in time:
[tex]\Delta \theta = \frac{5}{3}\cdot (4^{2}-2^{2})[/tex]
[tex]\Delta \theta = 20\,rad[/tex]
The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.
1. Draw four illustrations of a globe and paper that are positioned to yield equatorial, transverse, oblique, and polar aspect projections. Label the equator in each. Use your textbook or lecture material if you need a reference.2. On any map, why is there distortion at areas that do not fall on lines of tangency or secancy?
Answer:
1) attached below
2) assumption that the earth is spherical
Explanation:
1) Four illustrations of a globe
attached below
2) Reason for distortions at areas that do not fall on lines of tangency or secancy
The reason for distortion on areas outside the lines of tangency or secancy is because of the assumption that the earth is spherical which is not true hence map projections on the areas that fall on the lines of tangency do not experience distortion and are true
The bulk modulus of water is B = 2.2 x 109 N/m2. What change in pressure ΔP (in atmospheres) is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C?
Answer:
A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.
Explanation:
The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:
[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)
Where:
[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.
[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.
[tex]\Delta P[/tex] - Pressure change, in pascals.
If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:
[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]
[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]
[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]
A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.
If an electrical component with a resistance of 53 Q is connected to a 128-V source, how much current flows through the component?
Answer:
the current that flows through the component is 2.42 A
Explanation:
Given;
resistance of the electrical component, r = 53 Ω
the voltage of the source, V = 128 V
The current that flows through the component is calculated using Ohm's Law as demonstrated below;
[tex]V = IR\\\\I = \frac{V}{R} = \frac{128 \ V}{53 \ ohms} = 2.42 \ A[/tex]
Therefore, the current that flows through the component is 2.42 A
Define relative density.
Relative density is the ratio of the density of a substance to the density of a given material.
You are driving to the grocery store at 20 m/s. You are 150 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.
Required:
a. How far are you from the intersection when you begin to apply the brakes?
b. What acceleration will bring you to rest right at the intersection?
c. How long does it take you to stop?
Hi there!
a.
Use the formula d = st to solve:
d = 20 × 0.5 = 10m
150 - 10 = 140m away when brakes are applied
b.
Use the following kinematic equation to solve:
vf² = vi² + 2ad
Plug in known values:
0 = 20² + 2(150)(a)
Solve:
0 = 400 + 300a
-300a = 400
a = -4/3 (≈ -1.33) m/s² required
c.
Use the following kinematic equation to solve:
vf = vi + at
0 = 20 - 4/3t
Solve:
4/3t = 20
Multiply both sides by 3/4 for ease of solving:
t = 15 sec
Three spheres (water, iron and ice) of the exact same volume are submerged in a tub of water. After the spheres are lined up, they are released. The spheres are made of plastic with the same density as water, ice, and iron.
Required:
a. Compare the weights of the three spheres.
b. Compare the buoyant forces on the three spheres.
c. What direction does the net force push on each of the spheres?
d. What happens to each sphere after it is released?
Answer:
(a) Iron > plastic > ice
(b) Same on all
(c) Iron downwards, plastic net force zero, ice upwards.
(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.
Explanation:
Three spheres have same volume , plastic, ice and iron.
(a) The weight is given by
Weight = mass x gravity = volume x density x gravity
As the density of iron is maximum and the density of ice is least so the order of the weight is
Weight of iron > weight of plastic > weight of ice
(b) Buoyant force is given by
Buoyant force = Volume immersed x density of fluid x g
As they have same volume, density of fluid is same so the buoyant force is same on all the spheres.
(c) Net force is
F = weight - buoyant force
So, the net force on the iron sphere is downwards
On plastic sphere is zero as the density of plastic sphere is same as water. On ice sphere it is upwards.
(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.
Four equal-value resistors are in series with a 5 V battery, and 2.23 mA are measured. What isthe value of each resistor
Answer:
560.54 Ω
Explanation:
Applying,
V = IR'............... Equation 1
Where V = Voltage of the battery, I = currrent, R' = Total resistance of the resistors
make R' the subject of the equation
R' = V/I............ Equation 2
From the question,
Given: V = 5 V, I = 2.23 mA = 2.23×10⁻³ A
Substitute these values into equation 2
R' = 5/(2.23×10⁻³ )
R' = 2242.15 Ω
Since the fours resistor are connected in series and they are equal,
Therefore the values of each resistor is
R = R'/4
R = 2242.15/4
R = 560.54 Ω
A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q
Answer:
0.3 N
Explanation:
Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.
The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N
Coulomb's law equationF = Kq₁q₂ / r²
Where
F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart Data obtained from the question Initial distance apart (r₁) = rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =? How to determine the final forceFrom Coulomb's law,
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ = constant
F₁r₁² = F₂r₂²
With the above formula, we can obtain the final force as follow:
F₁r₁² = F₂r₂²
1.2 × r² = F₂ × (2r)²
1.2r² = F₂ × 4r²
Divide both side by 4r²
F₂ = 1.2r² / 4r²
F₂ = 0.3 N
Learn more about Coulomb's law:
https://brainly.com/question/506926
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?
Answer:
The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".
Explanation:
According to the question,
The work will be:
⇒ [tex]Work=-\frac{kQq}{R}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]
[tex]=\frac{0.3978}{\varepsilon }[/tex]
[tex]=4.49\times 10^{10} \ joules[/tex]
Thus the above is the correct answer.
We have that the workdone is mathematically given as
W=4.49*10e10 J
From the question we are told
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?WorkdoneGenerally the equation for the workdone is mathematically given as
W=-kQq/R
Therefore
0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2
Hence
W=4.49*10e10 JFor more information on Charge visit
https://brainly.com/question/9383604
find out the odd one and give reason (length, volume, time, mass
Answer:
Time
Explanation:
The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.
