Answer:
B) is perpendicular to the magnetic field
A person of weighing 0.62 kN rides in an elevator that has an upward acceleration of 1 m/s2. What is the magnitude of the force of the elevator floor on the person, in newtons
Answer:
Force = 63.27 Newton
Explanation:
Given the following data;
Weight = 0.62 kN = 0.62 * 1000 = 620 N
Acceleration = 1 m/s²
To find the magnitude of the force of the elevator floor on the person;
First of all, we would determine the mass of the elevator;
Weight = mass * acceleration due to gravity
We know that acceleration due to gravity is equal to 9.8 m/s²
Substituting into the above formula, we have:
620 = mass * 9.8
Mass = 620/9.8
Mass = 63.27 kg
Next, we would determine the force by using the formula;
Force = mass * acceleration
Force = 63.27 * 1
Force = 63.27 Newton
Which of the following processes has a positive value for ΔE and which are endothermic? i. A system absorbs 50 kJ of heat from the surroundings and does 30 kJ of work on the surroundings. ii. A system releases 25 kJ of heat into the surroundings and 45 kJ of work is done on the system. iii. q = 12.5 kJ, w = – 3500 J
Answer:
Explanation:
According to conservation of energy
q = ΔE + w
q is heat given , ΔE is increase in internal energy and w is work done by the gas.
i )
q = 50 kJ ; w = 30 kJ
Substituting the values
50 kJ = ΔE + 30 kJ
ΔE = 20 kJ ; Hence ΔE is positive.
ii )
q = - 25 kJ ; w = 45 kJ
- 25 kJ = ΔE + 45 kJ
ΔE = - 70 kJ .
ΔE is negative. It is endothermic.
iii )
q = 12.5 kJ ; w = - 3.5 kJ
12.5 kJ = ΔE - 3.5 kJ
ΔE = 16 kJ
Hence ΔE is positive.
The area around a magnet containing all of magnetic Lines of force is called
Answer:
Magnetic Field
Explanation:
The area around a magnet containing all of magnetic Lines of force is called the magnetic field.
A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floor and box? Determine it to three significant figures even through that's quite unrealistic. How much work is done in overcoming friction between the object and floor if the box slides 8m along horizontally on the floor?
Answer:
Coefficient of friction is [tex]0.068[/tex].
Work done is [tex]320~J[/tex].
Explanation:
Given:
Mass of the box ([tex]m[/tex]): [tex]60[/tex] kg
Force needed ([tex]F[/tex]): [tex]40[/tex] N
The formula to calculate the coefficient of friction between the floor and the box is given by
[tex]F=\mu mg...................(1)[/tex]
Here, [tex]\mu[/tex] is the coefficient of friction and [tex]g[/tex] is the acceleration due to gravity.
Substitute [tex]40[/tex] N for [tex]F[/tex], [tex]60[/tex] kg for [tex]m[/tex] and [tex]9.80[/tex] m/s² for [tex]g[/tex] into equation (1) and solve to calculate the value of the coefficient of friction.
[tex]40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068[/tex]
The formula to calculate the work done in overcoming the friction is given by
[tex]W=Fd..........................(2)[/tex]
Here, [tex]W[/tex] is the work done and [tex]d[/tex] is the distance travelled.
Substitute [tex]40[/tex] N for [tex]F[/tex] and [tex]8 m[/tex] for [tex]d[/tex] into equation (2) to calculate the work done.
[tex]W=40~N\times8~m\\~~~~= 320~J[/tex]
A point source of light illuminates an aperture 1.70 m away. A 11.0 cm -wide bright patch of light appears on a screen 0.800 m behind the aperture. Part A How wide is the aperture
Answer:
7.48 cm
Explanation:
The diagrammatic representation of information is shown in the image below:
From the image, Δ X₁X₂ and Δ Y₁Y₂ are equivalent triangles.
where;
X₁X₂ = width of the aperture
Using equivalent triangles, It implies that:
[tex]\implies \dfrac{X_1X_2}{Y_1Y_2} = \dfrac{1.70}{1.70+0.800} \\ \\ \implies X_1X_2=\dfrac{1.70}{1.70+0.800}\times 11.0[/tex]
[tex]X_1X_2= \dfrac{1.70}{2.50}\times 11.0 cm[/tex]
X₁X₂ = 7.48 cm
Calculate the linear acceleration of a car, the 0.220-m radius tires of which have an angular acceleration of 13.0 rad/s2. Assume no slippage and give your answer in m/s2. m/s2 (b) How many revolutions do the tires make in 2.50 s if they start from rest
Answer:
a). 2.86 [tex]m/s^2[/tex]
b). 6.465 revolutions
Explanation:
Given :
a). Radius of the tires, r = 0.220 m
Angular acceleration of the tires, α = 13.0 [tex]$rad/s^2$[/tex]
The line acceleration is defined as the rate of change of velocity with changing in direction.
The linear acceleration is equal to the product of the angular acceleration and the radius.
Therefore, linear acceleration is given by :
a = α x r
= 13 x 0.22
= 2.86 [tex]m/s^2[/tex]
b). Given time , t = 2.5 s
The angle moved is given by :
[tex]$\theta =\frac{1}{2}\alpha t^2$[/tex]
[tex]$=\frac{1}{2} \times 13 \times (2.5)^2$[/tex]
= 40.625 rad
Number of revolutions is
[tex]$n=\frac{\theta}{2 \pi}$[/tex]
[tex]$n=\frac{4.625}{2 \pi}$[/tex]
n = 6.465 revolutions
A basketball player shoots toward a basket 4.9 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor at an angle of 60 o above the horizontal, what must the initial speed be if it were to go through the basket
Answer:
v₀ = 6.64 m / s
Explanation:
This is a projectile throwing exercise
x = v₀ₓ t
y = y₀ + v_{oy} t - ½ g t²
In this case they indicate that y₀ = 1.8 m and the point of the basket is x=4.9m y = 3.0 m
the time to reach the basket is
t = x / v₀ₓ
we substitute
y- y₀ = [tex]\frac{ v_o \ x \ sin \theta }{ v_o \ cos \theta} - \frac{1}{2} g \ \frac{x^2 }{v_o^2 \ cos^2 \theta }[/tex]
y - y₀ = x tan θ - [tex]\frac{ g \ x^2 }{ 2 \ cos^2 \theta } \ \frac{1}{v_o^2 }[/tex]
we substitute the values
3 -1.8 = 3.0 tan 60 - [tex]\frac{ 9.8 \ 3^2 }{2 \ cos^2 60 } \ \frac{1}{v_o^2}[/tex]
1.2 = 5.196 - 176.4 1 / v₀²
176.4 1 / v₀² = 3.996
v₀ = [tex]\sqrt{ \frac{ 176.4}{3.996} }[/tex]
v₀ = 6.64 m / s
Calculate the force exerted on a thresher shark's eye by the hydrostatic pressure in ocean water at a depth of 380 m. (Assume the water's mass density at this depth is 1000 kg/m3k.)
Answer:
Explanation:
Hydrostatic pressure due to a water column of height h can be given by the following expression.
P = hρg
where ρ is density of water and g is acceleration due to gravity .
Substituting the values.
P = 380 x 1000 x 9.8
= 3.72 x 10⁶ Pa.
Answer:
[tex]F=\dfrac{3.72\times 10^6\ Pa}{A}\ N[/tex]
Explanation:
Given that,
The density of water, d = 1000 kg/m³
Depth, h = 380 m
We need to find the force exerted on a thresher shark's eye by the hydrostatic pressure in ocean water. The force exerted by the hydrostatic pressure is given by :
[tex]P=\rho gh[/tex]
Put all the values,
[tex]P=1000\times 9.8\times 380\\\\P=3.72\times 10^6\ Pa[/tex]
Force exerted,
F = P/A
So,
[tex]F=\dfrac{3.72\times 10^6\ Pa}{A}\ N[/tex]
Where
A is the area of crosss section
Hence, this is the required solution.
In the series circuit, if the potential difference across the battery is 20 V and the potential difference across R1 is 12 V, what is the potential difference across R2
Answer:
The correct answer is "8 V".
Explanation:
Given:
Potential difference across battery,
= 20 V
Potential difference across R1,
= 12 V
Now,
On applying the Kirchorff loop, we get
⇒ [tex]E-I_1R_1-I_2R_2=0[/tex]
⇒ [tex]E-V_1-V_2=0[/tex]
⇒ [tex]V_2=E-V_1[/tex]
On putting values, we get
⇒ [tex]=20-12[/tex]
⇒ [tex]= 8 \ V[/tex]
The potential difference across the resistance R2 will be "8 Volts.
What is Kirchoff;s law?According to the kirchoff's law in a loop of a circuit when there are number of the resistances so the sum of all the potential differences will be zero.
It is given that:
Potential difference across battery,= 20 V
Potential difference across R1,= 12 V
Now,
On applying the Kirchorff loop, we get
⇒ [tex]\rm E-I_1R_1-I_2R_2=0[/tex]
⇒ [tex]\rm E-V_1-V_2=0[/tex]
⇒ [tex]\rm V_2=E-V_1[/tex]
On putting values, we get
⇒ [tex]V_2=20-12=8\ volt[/tex]
Hence the potential difference across the resistance R2 will be "8 Volts.
To know more about Kirchoff's law follow
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PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
Answer:
Energy = 1.5032*10^(-10) Joules
Explanation:
By Einstein's relativistic energy equation, we know that the energy of a given particle is given by:
Energy = rest energy + kinetic energy.
= m*c^2 + (γ - 1)*mc^2
Where γ depends on the velocity of the particle.
But if the proton is at rest, then the kinetic energy is zero, and γ = 1
Then the energy is just given by:
Energy = m*c^2
Where we know that:
mass of a proton = 1.67*10^(-27) kg
speed of light = c = 2.9979*10^(8) m/s
Replacing these in the energy equation, we get:
Energy = ( 1.6726*10^(-27) kg)*( 2.9979*10^(8) m/s)^2
Energy = 1.5032*10^(-10) kg*m^2/s^2
Energy = 1.5032*10^(-10) J
Two kg of water is contained in a piston-cylinder assembly, initially at 10 bar and 200°C. The water is slowly heated at constant pressure to a final state. If the heat transfer for the process is 1740 kJ, determine the temperature at the final state, in °C, and the work, in kJ. Kinetic and potential energy effects are negligible.
Answer:
Explanation:
Given that:
mass of water = 2kg
Pressure P₁ = 10 bar
temperature T₁ = 200⁰ C
Obtaining the following value from superheated tables at P₁ = 10 bar and T₁ = 200⁰ C
v₁ = 0.2060 m³/kg
h₁ = 2827.9 kJ/kg
The heat transfer Q₁₋₂ = 1740 kJ
Using the energy balance equation:
Q₁₋₂ = m(h₂ - h₁)
1740 = 2(h₂ - 2827.9)
1740 = 2h₂ - 5655.8
1740 + 5655.8 = 2h₂
7395.8 = 2h₂
h₂ = 7395.8/2
h₂ = 3697.9 kJ/kg
Since P₁ = P₂ = 10 bar
Using the superheated table at P₂ = 10 bar and h₂ = 3697.9 kJ/kg;
v₂ = 0.4011 m³/kg
The temperature at final state = 600 °C
The work done [tex]W_{1-2} = \int P dv[/tex]
[tex]W_{1-2} = P\times m (V_2-V_1)[/tex]
= (10×10² kPa)×2kg (0.4011 - 0.2060) m³/kg
= (10×10² kPa)×2kg × 0.1951 m³/kg
= 390.2 kJ
A solid sphere and a hollow sphere each of mass M and radius R are released at the same time from the top of an inclined plane. Which one of the following statements is necessarily true?
A.Both spheres will reach the bottom at the same time.
B. The hollow sphere will reach the bottom with the greater kinetic energy.
C.The solid sphere will reach the bottom first.
D. The hollow sphere will reach the bottom first.
E. The solid sphere will reach the bottom with the greater kinetic energy.
Answer:
C. The solid sphere will reach the bottom first.
E. The solid sphere will reach the bottom with the greater kinetic energy.
Explanation:
To answer this,the sphere that has the smaller mass moment of inertia will be the one that will have the largest acceleration down the plane. And we know that higher acceleration means higher speed.
Now, moment of inertia of both spheres are;
I_solid sphere = (2/5)mr² = 0.4mr²
I_hollow sphere = (2/3)mr² = 0.67 mr²
Now, it is clear that the solid sphere has the smaller mass moment of inertia. This means that the solid sphere will have more acceleration and as well higher speed and will thus reach the bottom first.
Also, the higher the speed, the higher the kinetic energy since K.E = ½mv².
Thus, options C & E are correct.
How does the thickness of a planets atmosphere affect a planets average temperature
Answer:
The atmosphere of planets will block or enable heat coming from the sun
Question 24 of 33 Which of the following is an example of uniform circular motion? A. A car speeding up as it goes around a curve O B. A car slowing down as it goes around a curve 2 C. A car maintaining constant speed as it goes around a curve D. A car traveling along a straight road
Answer:
Uniform Circular Motion is the Movement or Rotation of an Object along a circular Path at constant speed.
OPTION C IS YOUR ANSWER!.
Charge of uniform density (80 nC/m3) is distributed throughout a hollow cylindrical region formed by two coaxial cylindrical surfaces of radii 1.0 mm and 3.0 mm. Determine the magnitude of the electric field at a point which is 2.0 mm from the symmetry axis.
Answer: Magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.
Explanation:
Given: Density = 80 [tex]nC/m^{3}[/tex] (1 n = [tex]10^{-9}[/tex] m) = [tex]80 \times 10^{-9} C/m^{2}[/tex]
[tex]r_{1}[/tex] = 1.0 mm (1 mm = 0.001 m) = 0.001 m
[tex]r_{2}[/tex] = 3.0 mm = 0.003 m
r = 2.0 mm = 0.002 m (from the symmetry axis)
The charge per unit length of the cylinder is calculated as follows.
[tex]\lambda = \rho \pi (r^{2}_{2} - r^{2}_{1})[/tex]
Substitute the values into above formula as follows.
[tex]\lambda = \rho \pi (r^{2}_{2} - r^{2}_{1})\\= 80 \times 10^{-9} \times 3.14 \times [(0.003)^{2} - (0.001)^{2}]\\= 2.01 \times 10^{-12} C/m[/tex]
Therefore, electric field at r = 0.002 m from the symmetry axis is calculated as follows.
[tex]E = \frac{\lambda}{2 \pi r \epsilon_{o}}[/tex]
Substitute the values into above formula as follows.
[tex]E = \frac{\lambda}{2 \pi r \epsilon_{o}}\\= \frac{2.01 \times 10^{-12} C/m}{2 \times 3.14 \times 0.002 m \times 8.85 \times 10^{-12}}\\= 18.08 N/C[/tex]
Thus, we can conclude that magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.
A car moves round a circular track of radius 0.3m of two revolution per/sec find its angular velocity.
Answer:
the angular velocity of the car is 12.568 rad/s.
Explanation:
Given;
radius of the circular track, r = 0.3 m
number of revolutions per second made by the car, ω = 2 rev/s
The angular velocity of the car in radian per second is calculated as;
From the given data, we convert the angular velocity in revolution per second to radian per second.
[tex]\omega = 2 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 4\pi \ rad/s = 12.568 \ rad/s[/tex]
Therefore, the angular velocity of the car is 12.568 rad/s.
It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft is dropping vertically at a speed of 293 m/s. The spacecraft is to make a soft landing -- that is, at the instant it reaches the surface of Mars, its velocity is zero. Assume the spacecraft undergoes constant acceleration from the elevation of 34.7 km until it reaches the surface of Mars. What is the magnitude of the acceleration
Answer: [tex]1.23\ m/s^2[/tex]
Explanation:
Given
At an elevation of [tex]y=34.7\ km[/tex], spacecraft is dropping vertically at a speed of [tex]u=293\ m/s[/tex]
Final velocity of the spacecraft is [tex]v=0[/tex]
using equation of motion i.e. [tex]v^2-u^2=2as[/tex]
Insert the values
[tex]\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2[/tex]
Therefore, magnitude of acceleration is [tex]1.23\ m/s^2[/tex].
A 5.0 A electric current passes through an aluminum wire of 4.0~\times~10^{-6}~m^2 cross-sectional area. Aluminum has one free electron per atom. The density of aluminum is 2.7~grams/cm^3, and the aluminum molar mass is 27 g. What is the electron number density (the number of electrons per unit volume) in the wire
Answer: The electron number density (the number of electrons per unit volume) in the wire is [tex]6.0 \times 10^{28} m^{-3}[/tex].
Explanation:
Given: Current = 5.0 A
Area = [tex]4.0 \times 10^{-6} m^{2}[/tex]
Density = 2.7 [tex]g/cm^{3}[/tex], Molar mass = 27 g
The electron density is calculated as follows.
[tex]n = \frac{density}{mass per atom}\\= \frac{\rho}{\frac{M}{N_{A}}}\\[/tex]
where,
[tex]\rho[/tex] = density
M = molar mass
[tex]N_{A}[/tex] = Avogadro's number
Substitute the values into above formula as follows.
[tex]n = \frac{\rho \times N_{A}}{M}\\= \frac{2.7 g/cm^{3} \times 6.02 \times 10^{23}/mol}{27 g/mol}\\= \frac{16.254 \times 10^{23}}{27} cm^{3}\\= 0.602 \times 10^{23} \times \frac{10^{6} cm^{3}}{1 m^{3}}\\= 6.0 \times 10^{28} m^{-3}[/tex]
Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is [tex]6.0 \times 10^{28} m^{-3}[/tex].
WILL GIVE 50 POINTS TO CORRECT ANSWER‼️
A capacitor has 4.33 x 10-7 C of
charge on it when 3.45 V is applied.
How much energy is stored in the
capacitor?
Answer:
7.47×10¯⁷ J
Explanation:
From the question given above, the following data were obtained:
Charge (Q) = 4.33×10¯⁷ C
Potential difference (V) = 3.45 V
Energy (E) =?
The energy stored in a capacitor is given by the following equation:
E = ½QV
Where
E => is the energy.
Q => is the charge.
V => is the potential difference.
With the above formula, we can obtain the energy stored in the capacitor as follow:
Charge (Q) = 4.33×10¯⁷ C
Potential difference (V) = 3.45 V
Energy (E) =?
E = ½QV
E = ½ × 4.33×10¯⁷ × 3.45
E = 7.47×10¯⁷ J
Thud, the energy stored in the capacitor is 7.47×10¯⁷ J
Answer:
The answer would be 7.47x10^-7J
Explanation:
Put in 7.47 then -7
difference between work done against gravity and friction
The movement of the liquid in a thermometer shows changes in temperature. An increase in temperature indicates the molecules in the liquid *
Answer:
The movement of the liquid in a thermometer shows changes in temperature. An increase in temperature indicates the molecules in the liquid *
Explanation:
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each a distance 0.0429 m from the origin. Where should a third particle with charge 9.03 µC be placed so that the magnitude of the electric field at the origin is zero?
Answer:
The third particle should be at 0.0743 m from the origin on the negative x-axis.
Explanation:
Let's assume that the third charge is on the negative x-axis. So we have:
[tex]E_{1}+E_{3}-E_{2}=0[/tex]
We know that the electric field is:
[tex]E=k\frac{q}{r^{2}}[/tex]
Where:
k is the Coulomb constant q is the charger is the distance from the charge to the pointSo, we have:
[tex]k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0[/tex]
Let's solve it for r(3).
[tex]\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0[/tex]
[tex]r_{3}=0.0743\: [/tex]
Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.
I hope it helps you!
How fast in m/s is the moon orbiting around the earth if it has orbital period of 27 days(2,330,800 seconds) and orbits at a radius of 380,000,000 m?
Answer:
C = 2 pi R circumference of orbit
V = C / P speed of moon by dividing circumference by period of revolution
V = 2 pi R / P_ = 2 pi 3.8 * 10E8 m / 2.3308 * 10E6 sec = 1024 m/s
if a seismic wave has a period of 0.0202s, find the frequency of the wave.
Answer:
49.5 Hz.
Explanation:
From the question given above, the following data were obtained:
Period (T) = 0.0202 s
Frequency (f) =?
The frequency and period of a wave are related according to the following equation:
Frequency (f) = 1 / period (T)
f = 1/T
With the above formula, we can obtain the frequency of the wave as follow:
Period (T) = 0.0202 s
Frequency (f) =?
f = 1/T
f = 1/0.0202
f = 49.5 Hz
Therefore the frequency of the wave is 49.5 Hz.
5. ssm How long does it take a plane, traveling at a constant speed of 110 m/s, to fl y once around a circle whose radius is 2850 m
Answer:
162.71 s
Explanation:
Applying,
v = 2πr/t................... Equation 1
Where v = Velocity of the plane, r = radius of the plane, t = time, π = pie.
make t the subject of the equation
t = 2πr/v................ Equation 2
From the question,
Given: r = 2850 m, v = 110 m/s
Constant: π = 3.14
Substitute these values into equation 2
t = (2×3.14×2850)/(110)
t = 17898/110
t = 162.71 s
Find the acceleration due to gravity on the surface of a planet with a mass of 3.5 * 10^24 kg and an average radius of 4.5 * 10^6 m.
A 9.5 V battery supplies a 3.5 mA current to a circuit for 7.0 h. Part A How much charge has been transferred from the negative to the positive terminal
Answer:
88.2 C
Explanation:
The current can be defined as the rate of flow of charge in a conductor.
The relation between charge current and time is given as
I = Q/T
I = current, Q= charge and T = time
that is ampere = coulomb / second
The amount of charge passed is from the negative to the positive terminal
shall be given by:
Q = I * t = 3.5mA * 7h * 3600s/h = 88.2 C
Note: take care of the units.
A small airplane flies at 60 m/s relative to the air. The wind is blowing at 20 m/s to the south. The pilot heads this plane east. a) What is the speed, v, that the plane will travel relative to the ground
Answer:
[tex]V_g=63.3m/s[/tex]
Explanation:
From the question we are told that:
Speed of Plane [tex]v_s=60m/s[/tex]
Wind speed [tex]V_w=20m/s[/tex]
Generally the equation for Speed of plane relative to the ground V_g is mathematically given by
[tex]V_g=\sqrt{V_s^2+V_w^2}[/tex]
[tex]V_g=\sqrt{60^2+20^2}[/tex]
[tex]V_g=\sqrt{4000}[/tex]
[tex]V_g=63.3m/s[/tex]
A 4.38 kg sphere makes a perfectly inelastic collision with a second sphere that is initially at rest. The composite system moves with a speed equal to one third the original speed of the 4.38 kg sphere. What is the mass of the second sphere?
Answer: The mass of the second sphere is 8.76 kg
Explanation:
The equation for a perfectly inelastic collision follows:
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
where,
[tex]m_1\text{ and }u_1[/tex] are the mass and initial velocity of first sphere
[tex]m_2\text{ and }u_2[/tex] are the mass and initial velocity of second sphere
v = final velocity of the system
We are given:
[tex]m_1=4.38kg\\u_2=0m/s\\v=\frac{u_1}{3}[/tex]
Rearranging the above equation, we get:
[tex]m_1u_1-m_1v=m_2v\\\\m_2=m_1\frac{u_1-v}{v}\\\\m_2=m_1(\frac{u_1}{v}-1)[/tex]
Plugging values in the above equation, we get:
[tex]m_2=4.38(\frac{3u_1}{u_1}-1)\\\\m_2=(4.38\times 2)=8.76kg[/tex]
Hence, the mass of the second sphere is 8.76 kg
40ml of Liquid A are poured into a beaker, and 40.0ml of Liquid B are poured into an identical beaker. Stirrers in each beaker are connected to motors, and the forces FA and FB needed to stir each liquid at a constant rate are measured.
a. FA will be greater than F B
b. FA will be less than FB
c. FA will be equal to FB
d. It's impossible to predict whether FA or FB will be greater without more information.
Answer:
d
Explanation:
there is not enough information about the liquid to know the force required for each. ie. stirring a cup of water is different than stirring a cup of pudding.