Answer:
8.85 x 10⁴ Nm²/C
Explanation:
d = diameter of the conducting sphere = 0.10 m
r = radius of the conducting sphere = (0.5) d = (0.5) (0.10) = 0.05 m
Area of the sphere is given as
A = 4πr²
A = 4 (3.14) (0.05)²
A = 0.0314 m²
σ = Surface charge density = 150 x 10⁻⁶ C/m²
Q = total charge enclosed
Total charge enclosed is given as
Q = σA
Q = (150 x 10⁻⁶) (0.0314)
Q = 4.7 x 10⁻⁶ C
Electric flux through one of the side is given as
[tex]\phi = \frac{Q}{6\epsilon _{o}}[/tex]
[tex]\phi = \frac{4.7\times 10^{-6}}{6(8.85\times 10^{-12})}[/tex]
[tex]\phi [/tex] = 8.85 x 10⁴ Nm²/C
Two flat rectangular mirrors are set edge to edge and placed perpendicular to a flat nonreflecting surface. The edges of the two mirrors meet at a 30° angle. A light ray that approaches mirror 1 is parallel to mirror 2. The angle of reflection of that ray from mirror 1 is:
Answer:
Angle of refelction is 60°
Explanation:
Let two mirrors XY and ZY meet at Y such that ∠XYZ = 30°
Let an incident ray PO incident on the 1st mirror XY and is parallel to the 2nd mirror YZ.
Let ON be the normal.
Now we know that , angle of incidence is equal to the angle of reflection
∴∠PON = ∠NOQ =60°
Therefore, the angle of reflection of the ray incident on the 1st mirror is 60 degree.
The angle of reflection of the light ray from Mirror 1 is 30°. This is in accordance with the Law of Reflection which states that the angle of incidence equals the angle of reflection. Examples of this principle are corner reflectors on bikes or cars and binoculars.
Explanation:According to The Law of Reflection, the angle of incidence is equal to the angle of reflection. When a light ray approaches a mirror parallel to another, it will reflect off at the same angle it hit, but in the opposite direction. Therefore, if the light ray approaches Mirror 1 parallel to Mirror 2, and considering that the two mirrors meet at a 30° angle, the light should reflect off at a 30° angle to the normal (perpendicular line) of Mirror 1.
This behavior of light can be observed in corner reflectors used on bikes and cars, where light is reflected back exactly parallel to the direction from which it came. It's also a principle used in binoculars and periscopes, where light is made to reflect multiple times in the system at respective angles of incidence and reflection.
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A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the normal force to the weight of the car when θ is equal to the following. (a) θ = 13°
Explanation:
The weight of the car is equal to, [tex]W_c=m\times g[/tex]...........(1)
Where
m is the mass of car
g is the acceleration due to gravity
The normal or vertical component of the force is, [tex]F_N=mg\ cos\theta[/tex]
or
[tex]F_N=mg\ cos(13)[/tex].............(2)
The horizontal component of the force is, [tex]F_H=mg\ sin\theta[/tex]
Taking ratio of equation (1) and (2) as :
[tex]\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}[/tex]
[tex]\dfrac{F_N}{W_c}=cos(13)[/tex]
[tex]\dfrac{F_N}{W_c}=0.97[/tex]
or
[tex]\dfrac{F_N}{W_c}=\dfrac{97}{100}[/tex]
Hence, this is the required solution.
The ratio of the magnitude of the normal force to the weight of the car can be determined by taking the cosine of the angle of inclination.
Explanation:When a car is traveling up a hill inclined at an angle θ above the horizontal, the ratio of the magnitude of the normal force to the weight of the car can be determined. The vertical component of the normal force is N cos θ, while the weight of the car is mg. Since the net vertical force must be zero, these two forces must be equal in magnitude:
N cos θ = mg
To find the ratio, divide both sides of the equation by mg:
N/mg = cos θ
So, the ratio of the magnitude of the normal force to the weight of the car is cos θ.
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After an initial test run John determines that his cooling system generates 45 W of heat loss. Calculate the amount of heat loss (H2), in W, that Mike expects his pump to do if its fan speed were 3.5 times greater and the coolant density was 9.5 times smaller.
Given:
heat generated by John's cooling system, [tex]H = \rho A v^{3}[/tex] = 45 W (1)
If ρ, A, and v corresponds to John's cooling system then let [tex]\rho_{1}, A_{1}, v_{1}[/tex] be the variables for Mike's system then:
[tex]\rho = 9.5\rho_{1}[/tex]
[tex]\rho_{1} = \frac{\rho}{9.5}[/tex]
[tex]v_{1} =3.5 v[/tex]
Formula use:
Heat generated, [tex]H = \rho A v^{3}[/tex]
where,
[tex]\rho[/tex] = density
A = area
v = velocity
Solution:
for Mike's cooling system:
[tex]H_{2}[/tex] = [tex]v_{1}^{3}{1}A_{1}\rho_{1}[/tex]
⇒ [tex]H_{2}[/tex] = [tex](3.5v)^{3}[/tex] × A × [tex]\frac{\rho}{9.5}[/tex]
[tex]H_{2}[/tex] = 4.513[tex]v^{3}[/tex] A [tex]\rho[/tex]
Using eqn (1) in the above eqn, we get:
[tex]H_{2}[/tex] = 4.513 × 45 = 203.09 W
We have that The Heat loss H2 is given as
H2=1347.5wFrom the question we are told
After an initial test run John determines that his cooling system generates 45 W of heat loss. Calculate the amount of heat loss (H2), in W, that Mike expects his pump to do if its fan speed were 3.5 times greater and the coolant density was 9.5 times smaller. heat loss (H2)Generally the equation for the is mathematically given as
H=PAV^3
Therefore
[tex]\frac{110}{h_2}=\frac{P_1}{p1/3.5}*(\frac{V_1}{3.5*v_1})^3[/tex]
H_2=1347.5w
Therefore
The Heat loss H2 is given as
H2=1347.5wFor more information on this visit
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Three charges are located at 100-m intervals along a horizontal line: a charge of –3.0 C on the left, 2.0 C in the middle, and 1.0 C on the right. What is the electric field on the horizontal line halfway between the –3.0 C and 2.0 C charges?
Answer:
1.84 x 10^7 N/C
Explanation:
Let q 1 = - 3 C, q2 = 2 C, q3 = 1 C
Electric field at P due to q1 is E1, due to q2 is E2 and due to q3 is E3.
E1 = k (3) / (50)^2 = 9 x 10^9 x 3 / 2500 = 108 x 10^5 N/C
E2 = k (2) / (50)^2 = 9 x 10^9 x 2 / 2500 = 72 x 10^5 N/C
E3 = k (1) / (150)^2 = 9 x 10^9 x 1 / 22500 = 4 x 10^5 N/C
Resultant electric field at P is given by
E = E1 + E2 + E3 = (108 + 72 + 4) x 10^5 = 184 x 10^5 = 1.84 x 10^7 N/C
Describe two ways of detecting black holes in space.
Answer and Explanation:
Since black holes deal with dark matter and dark energy and are dark enough not to allow even light to escape them, so their detection is much more difficult.
There are two basic methods to detect a black hole in space and these are listed below:
One way is to determine on the basis of strong influence of gravity that the black holes have due to their no bound dense massesAnother way to detect a black hole is through observation of a falling matter into the black hole. Matter, after falling into the black hole settles in a disk around it. The temperature of the disc can sometimes hike to extreme hot temperatures. Some amount of energy due to trap matter inside is liberated and turns into light which can be seen as in X-rays.A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in amperes and t is in seconds. What is the electric field induced at a radius r = 1.00 cm from the axis of the solenoid? (Use the following as necessary: t. Let E be measured in millivolts/meter and t be measured in seconds.)
To find the electric field induced at a radius of 1.00 cm from the axis of the solenoid, we can use the formula for the magnetic field inside a solenoid:
B = μ₀nI,
where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current.
First, let's find the magnetic field at a radius of 2.00 cm from the axis of the solenoid using the given values:
B = μ₀nI,
B = (4π×10⁻⁷ T·m/A)(1.65×10³ turns/m)(6.00 sin 90πt A),
B = (4π×10⁻⁷)(1.65×10³)(6.00 sin 90πt) T.
Now, we can use Ampere's law to find the electric field at a radius of 1.00 cm from the axis of the solenoid. Ampere's law states that the line integral of the magnetic field around a closed loop is equal to the permeability times the current enclosed by the loop.
∮B·dl = μ₀I_enclosed.
For a solenoid, the magnetic field is constant along any circular loop inside the solenoid. Therefore, the left side of the equation simplifies to B multiplied by the circumference of the loop, 2πr.
B(2πr) = μ₀I_enclosed.
Since the electric field is induced by a changing magnetic field, we can use Faraday's law of electromagnetic induction to relate the electric field to the time derivative of the magnetic field:
E = -d(B·A)/dt,
where E is the electric field, B is the magnetic field, A is the cross-sectional area of the loop, and dt is the change in time.
To find the electric field at a radius of 1.00 cm, we need to differentiate the magnetic field with respect to time and multiply by the cross-sectional area of the loop. Since the cross-sectional area of the loop is proportional to the square of the radius, A = πr², we have:
E = -d(B·A)/dt,
E = -(d(B·πr²)/dt,
E = -πr²(dB/dt).
Taking the time derivative of the magnetic field B, we get:
dB/dt = (4π×10⁻⁷)(1.65×10³)(6.00 cos 90πt) T/s.
Substituting this expression back into the equation for the electric field, we have:
E = -πr²(dB/dt),
E = -π(1.00×10⁻² m)²[(4π×10⁻⁷)(1.65×10³)(6.00 cos 90πt)] T/s,
E = -6.27×10⁻⁸π cos 90πt mV/m.
So, the electric field induced at a radius of 1.00 cm from the axis of the solenoid is -6.27×10⁻⁸π cos 90πt mV/m.
The induced electric field at ( r = 1.00 ) cm from the solenoid axis is approximately [tex]\( -12\pi^2 \cos(90\pi t) \) mV/m.[/tex]
Electric field induced: [tex]\( 0.111 \cos(90\pi t) \) mV/m at \( r = 1.00[/tex] \) cm from solenoid axis.
To find the electric field induced at a radius of ( r = 1.00 ) cm from the axis of the solenoid, we can use Faraday's law of electromagnetic induction.
The induced electric field [tex](\( E \))[/tex] is given by:
[tex]\[ E = -\frac{d\Phi}{dt} \][/tex]
Where [tex]\( \Phi \)[/tex] is the magnetic flux through the solenoid's cross-sectional area.
For a solenoid, the magnetic flux [tex]\( \Phi \)[/tex] is given by:
[tex]\[ \Phi = BA \][/tex]
Where:
- B is the magnetic field inside the solenoid,
- A is the cross-sectional area of the solenoid.
Given the current [tex]\( I(t) = 6.00 \sin(90\pi t) \)[/tex] A, and using Ampere's law for the magnetic field inside a solenoid:
[tex]\[ B = \mu_0 n I \][/tex]
Where:
[tex]- \( \mu_0 \)[/tex] is the permeability of free space [tex](\( 4\pi \times 10^{-7} \) T m/A),[/tex]
- n is the number of turns per meter.
Substitute [tex]\( I(t) \)[/tex] into the equation for ( B ) and then calculate [tex]\( \frac{d\Phi}{dt} \) to find \( E \).[/tex]
Given:
- [tex]\( n = 1.65 \times 10^3 \)[/tex] turns/m
- [tex]\( I(t) = 6.00 \sin(90\pi t) \)[/tex] A
- [tex]\( A = \pi r^2 \)[/tex] (for a circular cross-section)
We calculate the magnetic field [tex]\( B \):[/tex]
[tex]\[ B = \mu_0 n I(t) = 4\pi \times 10^{-7} \times 1.65 \times 10^3 \times 6.00 \sin(90\pi t) \][/tex]
Next, we calculate the magnetic flux [tex]\( \Phi \):[/tex]
[tex]\[ \Phi = BA = 4\pi \times 10^{-7} \times 1.65 \times 10^3 \times 6.00 \sin(90\pi t) \times \pi \times (0.01)^2 \][/tex]
Now, we take the time derivative of [tex]\( \Phi \) to find \( \frac{d\Phi}{dt} \):[/tex]
[tex]\[ \frac{d\Phi}{dt} = 4\pi \times 10^{-7} \times 1.65 \times 10^3 \times 6.00 \times 90\pi \cos(90\pi t) \times \pi \times (0.01)^2 \][/tex]
Finally, we plug in the values and calculate:
[tex]\[ \frac{d\Phi}{dt} = 12\pi^2 \cos(90\pi t) \times 4.95 \times 10^{-10} \][/tex]
Now, we have [tex]\( \frac{d\Phi}{dt} \),[/tex] the induced electric field [tex]\( E \) is given by \( E = -\frac{d\Phi}{dt} \).[/tex]
[tex]\[ E = -12\pi^2 \cos(90\pi t) \times 4.95 \times 10^{-10} \][/tex]
Therefore, the induced electric field at ( r = 1.00 ) cm from the solenoid axis is approximately [tex]\( -12\pi^2 \cos(90\pi t) \) mV/m.[/tex]
A car going initially with a velocity 13.5 m/s accelerates at a rate of 1.9 m/s for 6.2 s. It then accelerates at a rate of-1.2 m/s until it stops. a)Find the car's maximum speed. b) Find the total time from the start of the first acceleration until the car is stopped. c) What is the total distance the car travelled?
Answer:
a) Maximum speed = 25.28 m/s
b) Total time = 27.27 s
c) Total distance traveled = 402.43 m
Explanation:
a) Maximum speed is obtained after the end of acceleration
v = u + at
v = 13.5 + 1.9 x 6.2 = 25.28 m/s
Maximum speed = 25.28 m/s
b) We have maximum speed = 25.28 m/s, then it decelerates 1.2 m/s² until it stops.
v = u + at
0 = 25.28 - 1.2 t
t = 21.07 s
Total time = 6.2 + 21.07 = 27.27 s
c) Distance traveled for the first 6.2 s
s = ut + 0.5 at²
s = 13.5 x 6.2 + 0.5 x 1.9 x 6.2² = 120.22 m
Distance traveled for the second 21.07 s
s = ut + 0.5 at²
s = 25.28 x 21.07 - 0.5 x 1.2 x 21.07² = 282.21 m
Total distance traveled = 120.22 + 282.21 = 402.43 m
Answer:
a) Maximum speed = 25.28 m/s
b) Total time = 27.26 s
c) Total distance traveled = 390,5537
Explanation:
In order to solve the first proble we just have to use the next formula:
[tex]Vf= Vo+ Acc-t\\Vf= 13,5 + 6,2*1,9\\Vf=25,28 m/s\\[/tex]
So the maximum speed would be 25,28 m/s.
THe total time of the trip will be given by adding the inital time plus the velocity divided by the acceleration rate:
[tex]Time= 6,2 s+ \frac{25,28}{-1,2} \\Time= 6,2 +21,06\\Time= 27,26[/tex]
Remember that when dealing with time in physics you will always use positive numbers since there is no negative time.
To calculate the total distance covered we use the next formula:
[tex]D=Vo*t+ 1/2a*t^2\\D= 13,5*6,2 + 1/2(1,9)(6,2)^2\\D=87,75+36,518\\D=124,268m[/tex]
This is the first part now we calculate it with the stopping of the car:
[tex]D=Vo*t+ 1/2a*t^2\\D=25,28*21,06+ 1/2(-1,2)(21,06)^2\\D=532,3968-266,1141\\D= 266,2857meters[/tex]
No we just add the two distances to discover the whole distance:
Total distance= 124,268+266,2857= 390,5537
A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60° = 0.87; cos 60° = 0.50. Ignore friction and the weights of the pulleys.)
Answer:
Answer:196 Joules
Explanation:
Hello
Note: I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem
the work is the product of a force applied to a body and the displacement of the body in the direction of this force
assuming that the force goes in the same direction of the displacement, that is upwards
W=F*D (work, force,displacement)
the force necessary to move the object will be
[tex]F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules[/tex]
Answer:196 Joules
I hope it helps
A sample of N2 gas is added to a mixture of other gases originally at 0.85 atm. When the nitrogen is added, the pressure of the gases increases to 988 mmHg. Explain why the pressure increased and give the partial pressure of nitrogen in atm.
Answer:
Partial pressure of nitrogen gas,[tex]p^o_{N_2] =0.44 atm[/tex]
Explanation:
Pressure of the mixture of gases before adding nitrogen gas = 0.85 atm
Pressure of the mixture of gases after adding nitrogen gas = 988 mmHg
1 mmHg = 0.001315 atm
988 mmHg=[tex]988 mmHg\times 0.001315 atm = 1.29 atm[/tex]
Partial pressure of nitrogen gas,[tex]p^o_{N_2] = 1.29 atm - 0.85 atm = 0.44 atm[/tex]
According to Dalton's law of partial pressure , the total pressure of the mixture of gases is equal to sum of all the partial pressures of each gas present in the mixture.
[tex]P_{total}=\sum p^o_{i}[/tex]
So, on addition of nitrogen gas to the mixture the pressure of the mixture increases.
A car that was initially moving at 10 m/s is accelerated until its velocity reached 61 m/s. The change of the velocity took 6 s. What is the acceleration of the car in m/s^2? Please round your answer to two decimal places.
Answer:
The acceleration of the car is 8.50 m/s²
Explanation:
Given that,
Initial velocity = 10 m/s
Final velocity = 61 m/s
Time = 6 s
We need to calculate the acceleration of the car
Using equation of motion
[tex]v=u+at[/tex].....(I)
[tex]a=\dfrac{v-u}{t}[/tex]
Where, u = initial velocity
v = final velocity
t = time
Put the value in the equation (I)
[tex]a=\dfrac{61-10}{6}[/tex]
[tex]a=8.50\ m/s^2[/tex]
Hence, The acceleration of the car is 8.50 m/s²
When atom A loses an electron to atom B: a) atom A becomes a negative ion and atom B becomes a positive ion. b) atom A acquires more neutrons than atom B. c) atom A becomes more negative than atom B. d) atom A acquires less neutrons than atom B. e) atom A becomes a positive ion and atom B becomes a negative ion.
The correct answer is e) atom A becomes a positive ion and atom B becomes a negative ion. Ions are charged particles that result when a neutral atom gains or loses an electron.
Further Explanation:
In a neutral atom, the number of protons is equal to the number of electrons.In a positive ion, the number of protons is greater than the number of electrons.In a negative ion, the number of protons is less than the number of electrons.In the problem, atom A loses an electron to atom B.
Suppose atom A has 11 protons and 11 electrons and atom B has 9 protons and 9 electrons.
BEFORE TRANSFER OF ELECTRONS
atom A: 11 protons, 11 electrons
atom B: 9 protons, 9 electrons
AFTER TRANSFER OF ELECTRONS
atom A: 11 protons, 10 electrons
atom B: 9 protons, 10 electrons
a) atom A becomes a negative ion and atom B becomes a positive ion FALSE because atom A has more protons than electrons turning it into a positive ion, and atom B now has more electrons making a negative ion.
b) atom A acquires more neutrons than atom B FALSE because only the number of electrons changed. The number of neutrons are unchanged in the formation of ions.
c) atom A becomes more negative than atom B FALSE because atom A becomes more positive since there is more protons (positively charged) than electrons (negatively charged) while atom B, with its extra electron becomes negatively charged
d) atom A acquires less neutrons than atom B FALSE because the number of neutrons does not change. Only an electron was transferred.
e) atom A becomes a positive ion and atom B becomes a negative ion TRUE because there are more protons than electrons in A after the electron transfer and there are more electrons in B than protons after receiving the electron.
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An unknown metal alloy with a mass of 390 g is taken from boiling water and dropped into an insulated cup that contains 553 g of water at an initial temperature of 18.7°C. The final temperature of the system is 26.8°C. What is the metal's specific heat capacity?
Answer:
so specific heat capacity of unknown metal is 656.8 J/degree C kg
Explanation:
Here by energy balance we can say that energy given by metal alloy at 100 degree = heat absorbed by water at 18.7 degree C
now we have
[tex]Q_{in} = Q_{out}[/tex]
[tex]390 s (100 - 26.8) = 553 (4186 ) ( 26.8 - 18.7)[/tex]
[tex] 28548 s = 18750349.8[/tex]
now we have
[tex]s = \frac{18750349.8}{28548}[/tex]
[tex]s = 656.8 J/kg ^o C[/tex]
so specific heat capacity of unknown metal is 656.8 J/degree C kg
What is an S0 type galaxy?
A disk-shaped galaxy with no evidence of spiral arms
An irregular
A large elliptical
A large spiral
Answer:
A large elliptical
Explanation:
A S0 type galaxy is a large elliptical.
A S0 type galaxy is a large elliptical.
Answer is C.
A stone is thrown vertically upward with a speed of 35.0 m/s a) Howfast sit moving when it reaches a height of 13.0m How much time is required to reach this height? Wht s the maximum height it will reach? Explain why there are two answers for part 3
Answer:
a)
v = 31.15 m/s
t = 0.393 sec
h = 62.5 m
Explanation:
a)
v₀ = initially speed of the stone = 35.0 m/s
v = final speed of the stone = ?
y = vertical displacement of the stone = 13.0 m
a = acceleration due to gravity = - 9.8 m/s²
using the equation
v² = v₀² + 2 a y
v² = 35² + 2 (- 9.8) (13.0)
v = 31.15 m/s
t = time taken
using the equation
v = v₀ + a t
31.15 = 35 + (- 9.8) t
t = 0.393 sec
h = maximum height
v' = final speed at the maximum height = 0 m/s
using the equation
v'² = v₀² + 2 a h
0² = 35² + 2 (- 9.8) h
h = 62.5 m
A hot-water stream at 80 ℃ enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold water at 20 ℃. If it is desired that the mixture leave the chamber at 42 ℃, determine the mass flow rate of the cold-water stream. Assume all the streams are at a pressure of 250 kPa
Answer:
[tex]\dot{m_{2}}=0.865 kg/s[/tex]
Explanation:
[tex]\dot{m_1}= 0.5kg/s[/tex]
from steam tables , at 250 kPa, and at
T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg
T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg
T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg
we know
[tex]\dot{m_{in}}=\dot{m_{out}}[/tex]
[tex]\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}[/tex]
according to energy balance equation
[tex]\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}[/tex]
[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}[/tex]
[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s[/tex]
The mass flow rate of the cold-water stream ( m₂ ) = 0.865 kg/s
Given data :
m₁ = 0.5 kg/s
m₂ = ?
From steam tables
At 250 kPa
at T1 = 80℃ , h₁ = 335.02 kJ/kg
at T2 = 20℃, h₂ = 83.915 kJ/kg
at T3 = 42℃, h₃ = 175.90 kJ/kg
Determine the mass flow rate of the cold water streamGiven that :
Min = Mout also m₁ + m₂ = m₃
applying the principle of energy balance
M₁h₁ + M₂ h₂ = ( m₁ + m₂ ) h₃ ---- ( 1 )
insert values into equation ( 1 )
m₂ = 0.865 kg/s
Hence In conclusion The mass flow rate of the cold-water stream ( m₂ ) = 0.865 kg/s
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Block A, with a mass of 4.0 kg, is moving with a speed of 3.0 m/s while block B, with a mass of 6.0 kg, is moving in the opposite direction with a speed of 5.0 m/s. What is the momentum of the two-block system
Answer:
The momentum of the two-block system is 18 kg-m/s
Explanation:
It is given that,
Mass of block A, [tex]m_A=4\ kg[/tex]
Mass of block B, [tex]m_B=6\ kg[/tex]
Velocity of block A, [tex]v_A=3\ m/s[/tex]
Velocity of block B, [tex]v_B=-5\ m/s[/tex] (it is moving in opposite direction)
We need to find the momentum of two block system. It is given by the product of mass and velocity for both blocks i.e.
[tex]p=m_Av_A+m_Bv_B[/tex]
[tex]p=4\ kg\times 3\ m/s+6\ kg\times (-5\ m/s)[/tex]
p = -18 kg-m/s
So, the momentum of two block system is 18 kg-m/s. Hence, this is the required solution.
Final answer:
The total momentum of the two-block system, involving block A with a mass of 4.0 kg moving at 3.0 m/s and block B with a mass of 6.0 kg moving at 5.0 m/s in the opposite direction, is computed as -18 kg·m/s, which is in the -x-direction.
Explanation:
The question involves finding the momentum of a two-block system where the blocks are moving in opposite directions. To compute the total momentum of the system, we apply the principle of conservation of momentum, which states that the momentum of a system remains constant if no external forces act on it.
Momentum is defined as the product of an object's mass and its velocity (p = mv). For block A, which has a mass of 4.0 kg and is moving at 3.0 m/s, its momentum is calculated as:
pA = (4.0 kg) × (3.0 m/s) = 12 kg·m/s in the +x-direction.
For block B, which has a mass of 6.0 kg and is moving at 5.0 m/s in the opposite direction, its momentum is calculated as:
pB = (6.0 kg) × (-5.0 m/s) = -30 kg·m/s in the -x-direction (note the negative sign because it is moving in the opposite direction).
To find the total momentum of the system, we sum the momenta of both blocks:
ptotal = pA + pB = 12 kg·m/s + (-30 kg·m/s) = -18 kg·m/s.
The total momentum of the two-block system is -18 kg·m/s, indicating that there is a net momentum in the -x-direction.
A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground and experience negligible air resistance. (a) How far from where the child is standing does the ball hit the ground?
The distance from where the ball hits the ground is approximately 7.53 m.
Explanation:To find the distance from where the ball hits the ground, we need to analyze the horizontal and vertical motion separately. First, we can find the time it takes for the ball to hit the ground using the vertical motion equation. The final position in the y-axis is 0, the initial position is 1 m, the initial vertical velocity is 8*sin(40°) m/s, and the acceleration is -9.8 m/s² (due to gravity). By solving the equation, we find that the time of flight is 1.15 s. Using this time, we can find the horizontal distance traveled by the ball using the horizontal motion equation. The initial horizontal velocity is 8*cos(40°) m/s, and multiplying it by the time of flight gives us the answer: approximately 7.53 m.
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A ray of light in air strikes the flat surface of a liquid, resulting in a reflected ray and a refracted ray. If the angle of reflection is known, what additional information is needed in order to determine the relative refractive index of the liquid compared to air?
Answer:
Angle of refraction.
Explanation:
Refractive index is given as the ratio of the angle of incidence to the angle of refraction.
If angle of incidence is i and the refractive index of the incident medium is n₁ , and if angle of refraction is r and refractive index of the refracting medium is n₂,
according to Snell's law, n₁ sin i = n₂ sin r,
relative refractive index = sin i / sin r
Two spherical point charges each carrying a charge of 40 C are attached to the two ends of a spring of length 20 cm. If its spring constant is 120 Nm-1, what is the length of the spring when the charges are in equilibrium?
Answer:
[tex]L = 20 + 37 = 57 cm[/tex]
Explanation:
As we know that two charges connected with spring is at equilibrium
so here force due to repulsion between two charges is counter balanced by the spring force between them
so here we have
[tex]F_e = F_{spring}[/tex]
here we have
[tex]\frac{kq_1q_2}{r^2} = kx[/tex]
[tex]\frac{(9 \times 10^9)(40 \mu C)(40 \mu C)}{(0.20 + x)^2} = 120 x[/tex]
[tex]14.4 = (0.20 + x)^2 ( 120 x)[/tex]
by solving above equation we have
[tex]x = 0.37 m[/tex]
so the distance between two charges is
[tex]L = 20 + 37 = 57 cm[/tex]
A 20 kg mass with an initial velocity of 10 m/s slides to a stop in 2 meters. What is the net work down on the mass? a.) 250 J b.) 500 J c.) 120 J d.) 20 J e.) None of the above
Answer:
12500 J
Explanation:
m = 20 kg, u = 10 m/s, s = 2 m, v = 0
Use third equation of motion
v^2 = u^2 = 2 a s
0 = 100 - 2 a x 2
100 / 4 = - a
a = - 25 m/s^2
Force, F = m a = 20 x 25 = 500 N
Work done = F x s = 500 x 25 = 12500 J
Suppose that a steel bridge, 1000 m long, were built without any expansion joints. Suppose that only one end of the bridge was held fixed. What would the difference in the length of the bridge be between winter and summer, taking a typical winter temperature as 0°C, and a typical summer temperature as 40°C? The coefficient of thermal expansion of steel is 10.5 × 10-6 K-1.
Answer:
The difference in the length of the bridge is 0.42 m.
Explanation:
Given that,
Length = 1000 m
Winter temperature = 0°C
Summer temperature = 40°C
Coefficient of thermal expansion [tex]\alpha= 10.5\times10^{-6}\ K^{-1}[/tex]
We need to calculate the difference in the length of the bridge
Using formula of the difference in the length
[tex]\Delta L=L\alpha\Delta T[/tex]
Where, [tex]\Delta T [/tex]= temperature difference
[tex]\alpha[/tex]=Coefficient of thermal expansion
L= length
Put the value into the formula
[tex]\Delta L=1000\times10.5\times10^{-6}(40^{\circ}-0^{\circ})[/tex]
[tex]\Delta L=0.42\ m[/tex]
Hence, The difference in the length of the bridge is 0.42 m.
A 50-kg student gets in a 1000-kg elevator at rest. As the elevator begins to move, she has an apparent weight of 600 N for the first 3 s. How far has the elevator moved, and in which direction, at the end of 3 s?
Answer:
The elevator has moved 54 meters upwards.
Explanation:
m= 50kg
F= 600 N
t= 3 sec
a= [F-(m*g)]/m
a= (600N-490N)/50kg
a= 2.2 m/s²
h= (a*t²)/2
h= (2.2m/s²*(3 sec)²)/2
h= 9.9 m
Final answer:
The elevator has moved upward a distance of 7 meters at the end of 3 seconds by considering the motion and acceleration of the elevator and the student.
Explanation:
The elevator has moved upward a distance of 7 meters at the end of 3 seconds.
To calculate this, we need to consider the motion of the elevator and the student. The apparent weight of the student is the normal force exerted by the elevator floor on the student, which is 600 N when moving up. This value is different from the student's actual weight due to the acceleration of the elevator.
Using the equation Net Force = Mass x Acceleration, and knowing that the student's weight (mg) - normal force = ma, we can find the acceleration, and subsequently, the distance the elevator has moved 7 meters after 3 seconds.
Wanda exerts a constant tension force of 12 N on an essentially massless string to keep a tennis ball (m=60g) attached to the end of the string traveling in uniform circular motion above her head at a constant speed of 9.0 m/s. What is the length of the string between her hand and the tennis ball? You may ignore gravity in this problem (assume the motion of the tennis ball and string happen in a purely horizontal plane). a. 22 m
b. 2.5 m
c. 2.2 cm
d. 10 m
Final answer:
The length of the string between Wanda's hand and the tennis ball is 4.5 cm.
Explanation:
To find the length of the string between Wanda's hand and the tennis ball, we need to use the formula for centripetal force:
F = (m * v^2) / r
where F is the tension force, m is the mass of the ball, v is the velocity, and r is the radius of the circular motion.
In this case, the tension force is 12 N, the mass of the ball is 60 g (0.06 kg), and the velocity is 9.0 m/s. We can rearrange the formula to solve for r:
r = (m * v^2) / F
Substituting the given values, we get:
r = (0.06 * 9.0^2) / 12
r = 0.045 m = 4.5 cm
Therefore, the length of the string between Wanda's hand and the tennis ball is 4.5 cm.
The average speed of the space shuttle is 19800 mi/h. Calculate the altitude of the shuttle's orbit.
Answer:
[tex]r = 5.13 \times 10^6[/tex]
Explanation:
As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed
so here we will have
[tex]\frac{mv^2}{r} = \frac{GMm}{r^2}[/tex]
here we know that
v = 19800 mi/h
[tex]v = 19800 \frac{1602 m}{3600 s}[/tex]
[tex]v = 8811 m/s[/tex]
also we know that
[tex]M = 5.97 \times 10^{24} kg[/tex]
now we will have
[tex]r = \frac{GM}{v^2}[/tex]
[tex]r = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(8811)^2}[/tex]
[tex]r = 5.13 \times 10^6[/tex]
A meteoroid, heading straight for Earth, has a speed of 14.8 km/s relative to the center of Earth as it crosses our moon's orbit, a distance of 3.84 × 108 m from the earth's center. What is the meteroid's speed as it hits the earth
Answer:
The meteoroid's speed is 18.5 km/s
Explanation:
Given that,
Speed = 14.8 km/s
Distance [tex]d= 3.84\times10^{8}[/tex]
We need to calculate the meteoroid's speed
The total initial energy
[tex]E_{i}=K_{i}+U_{i}[/tex]
[tex]E_{i}=\dfrac{1}{2}mv_{i}^2-\dfrac{GM_{e}m}{r}[/tex]
Where, m = mass of meteoroid
G = gravitational constant
[tex]M_{e}[/tex]=mass of earth
r = distance from earth center
Now, The meteoroid hits the earth then the distance of meteoroid from the earth 's center will be equal to the radius of earth
The total final energy
[tex]E_{f}=K_{f}+U_{f}[/tex]
[tex]E_{f}=\dfrac{1}{2}mv_{f}^2-\dfrac{GM_{e}m}{r_{e}}[/tex]
Where,
[tex]r_{e}[/tex]=radius of earth
Using conservation of energy
[tex]E_{i}=E_{j}[/tex]
Put the value of initial and final energy
[tex]\dfrac{1}{2}mv_{i}^2-\dfrac{GM_{e}m}{r}=\dfrac{1}{2}mv_{f}^2-\dfrac{GM_{e}m}{r_{e}}[/tex]
[tex]v_{f}^2=v_{i}^2+2GM_{e}(\dfrac{1}{r_{e}}-\dfrac{1}{r})[/tex]
Put the value in the equation
[tex]v_{f}^2=(14.8\times10^{3})^2+2\times6.67\times10^{-11}\times5.97\times10^{24}(\dfrac{1}{6.37\times10^{6}}-\dfrac{1}{3.84\times10^{8}})[/tex]
[tex]v_{f}=\sqrt{(14.8\times10^{3})^2+2\times6.67\times10^{-11}\times5.97\times10^{24}(\dfrac{1}{6.37\times10^{6}}-\dfrac{1}{3.84\times10^{8}})}[/tex]
[tex]v_{f}=18492.95\ m/s[/tex]
[tex]v_{f}=18.5\ km/s[/tex]
Hence, The meteoroid's speed is 18.5 km/s
To find the meteoroid's speed as it hits the Earth, we can use the principle of conservation of mechanical energy. The final velocity of the meteoroid is approximately 13.4 km/s.
Explanation:To find the meteoroid's speed as it hits the Earth, we can use the principle of conservation of mechanical energy. Since there is no air friction, the mechanical energy of the meteoroid is conserved as it falls towards Earth. The initial kinetic energy of the meteoroid is equal to the final kinetic energy plus the gravitational potential energy.
First, we find the initial kinetic energy of the meteoroid using the formula KE = (1/2)mv^2, where m is the mass of the meteoroid and v is its initial velocity relative to the center of the Earth. Since the mass is not given, we can assume it cancels out in the equation.
Next, we calculate the gravitational potential energy of the meteoroid using the formula PE = mgh, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height from which the meteoroid fell. The height can be calculated by subtracting the radius of the Earth from the distance from the center of the Earth to the moon's orbit (h = 3.84 × 10^8 m - 6.37 x 10^6 m).
Solving for the final velocity, we equate the initial kinetic energy and the sum of the final kinetic energy and gravitational potential energy. Rearranging the equation, we find that the final velocity is the square root of (initial velocity squared minus 2 times g times h).
Plugging in the given values, the final velocity of the meteoroid as it hits the Earth is approximately 13.4 km/s.
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The size of the picture of a nanparticle is measured to be 5.2 cm by a ruler. If the scale bar size is 3 cm and is labeled 40 nm, find the actual size of the particle.
Answer:
The actual size is
69.33 nm.
Explanation:
It means that
3 cm is equivalent to 40 nm
So, 1 cm is equivalent to 40 / 3 nm
Thus, 5.2 cm is equivalent to
40 × 5.2 / 3 = 69.33 nm
A very long wire generates a magnetic field of 0.0020x 10^-4 T at a distance of 10 mm. What is the magnitude of the current? A) 2.0 mA B) 3100 mA C) 4000 mA D) 1.0 mA
Answer:
2*10^-5
Explanation:
B=IL
I=B/L
I=0.0020*10^-4/10
I=2*10^5
The magnitude of the current in a magnetic field is 2 x10⁵ Amperes.
What is magnetic field?The magnetic field is the region of space where a charged object experiences magnetic force when it is moving.
The formula for magnetic field is given as;
B=IL
I=B/L
A very long wire generates a magnetic field of 0.0020 x 10⁻⁴ T at a distance of 10 mm.
Plug the values, we get
I=0.0020 x 10⁻⁴/(10 x 10⁻³)
I=2 x10⁵
Thus, the magnitude of the current is 2 x10⁵ Amperes.
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A 6.0 Ω and a 12. Ω resistor are connected in series to a 36. V battery, what power is dissipated by the 12.0 Ω resistor? A) 24. W B) 486. W C)216. W D) 12. W E) 48. W
Answer:
Is dissipated E) P= 48 W in the 12 Ω resistor.
Explanation:
V= 36v
Req= (12 + 6)Ω
Req= 18Ω
I = V/ Req
I= 2 A
P= I² * R(12Ω)
P= 48 W
How much heat is required to change 0.500 kg of water from a liquid at 50. °C to vapor at 110. °C?
Answer:
Heat energy needed = 1243.45 kJ
Explanation:
We have
heat of fusion of water = 334 J/g
heat of vaporization of water = 2257 J/g
specific heat of ice = 2.09 J/g·°C
specific heat of water = 4.18 J/g·°C
specific heat of steam = 2.09 J/g·°C
Here wee need to convert 0.500 kg water from 50°C to vapor at 110°C
First the water changes to 100°C from 50°C , then it changes to steam and then its temperature increases from 100°C to 110°C.
Mass of water = 500 g
Heat energy required to change water temperature from 50°C to 100°C
[tex]H_1=mC\Delta T=500\times 4.18\times (100-50)=104.5kJ[/tex]
Heat energy required to change water from 100°C to steam at 100°C
[tex]H_2=mL=500\times 2257=1128.5kJ[/tex]
Heat energy required to change steam temperature from 100°C to 110°C
[tex]H_3=mC\Delta T=500\times 2.09\times (110-100)=10.45kJ[/tex]
Total heat energy required
[tex]H=H_1+H_2+H_3=104.5+1128.5+10.45=1243.45kJ[/tex]
Heat energy needed = 1243.45 kJ
Consider a vector 4.08 + 3.0 , wherex, are the unit vectors in x-, y-directions, respectively. (a) What is the magnitude of the vector A? (b) What are the angles vector A makes with the x and y axes, respectively?
Answer:
Part a)
Magnitude = 5.06 unit
Part b)
[tex]\theta = 36.2 ^0[/tex]
Explanation:
Part a)
Vector is given as
[tex]\vec A = 4.08 \hat x + 3.0 \hat y[/tex]
now from above we can say that
x component of the vector is 4.08
y component of the vector is given as 3.0
so the magnitude of the vector is given as
[tex]|A| = \sqrt{4.08^2 + 3^2}[/tex]
[tex]|A| = 5.06 unit[/tex]
Part b)
Now the angle made by the vector is given as
[tex]\theta = tan^{-1}(\frac{y}{x})[/tex]
[tex]\theta = tan^{-1}(\frac{3}{4.08})[/tex]
[tex]\theta = 36.3 degree[/tex]