To solve this question, we need to use the concept of the conversion between potential and kinetic energy.
The roller coaster's initial potential energy before it goes down the hill can be calculated using the formula for gravitational potential energy: `PE = m * g * h`,
where `m` is the mass,
`g` is the acceleration due to gravity, and
`h` is the height.
Substituting the given values, where `m` is 1200 kg, `g` is 9.81 m/s² and the initial height `h` is 19 m,
`PE_initial = 1200 kg * 9.81 m/s² * 19 m = 223668 J`
This is the energy the roller coaster has due to its position at the top of the 19m high hill before it starts to move.
When the roller coaster reaches the hill that is 13 m high, we can calculate its potential energy at this point the same way we calculated the initial potential energy, with `h` being now the final height of 13 m,
`PE_final = 1200 kg * 9.81 m/s² * 13 m = 153036 J`
This is the energy the coaster has due to its position at the top of the 13m high hill.
The kinetic energy (KE) of the roller coaster at this point is gained by the conversion of some of the initial potential energy into kinetic energy. This conversion is equal to the difference between the initial potential energy and the final potential energy:
`KE = PE_initial - PE_final = 223668 J - 153036 J = 70632 J`
So, when the roller coaster goes over the hill that is 13 m high, its kinetic energy is 70632 J. This is the energy the roller coaster has due to its speed at this point.
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A rhinoceros beetle rides the rín of a small disk that rotates like a merry-go-round. If the beetle crawls toward the center of the disk, do the following (each relative to the central axis) increase, decrease, or remain the same for the beetle-disk system: (a) rotational inertia, (b) angular momentum, and (c) angular speed?
Answer:
a) Rotational Inertia = Decreases
Because the distance from the axis is decreasing
b) Angular momentum = Remains the same
because there is no external torque
c) Angular speed = increases
because here rotational inertia decreases due to which angular speed will increase
Explanation:
Here the Beetle is initially moving along the rim of the disc
So here during the motion of beetle there is no external force on the system of beetle and the disc.
So here we can also say that there is no torque acting on the system
so angular momentum of the disc + beetle system will remain conserved
so here we have
[tex]I_1\omega_1 = I_2 \omega_2[/tex]
here as the beetle crawls towards the centre of the disk then
a) Rotational Inertia = Decreases
Because the distance from the axis is decreasing
b) Angular momentum = Remains the same
because there is no external torque
c) Angular speed = increases
because here rotational inertia decreases due to which angular speed will increase
A car starts from the origin and is driven 1.88 km south, then 9.05 km in a direction 47° north of east. Relative to the origin, what is the car's final location? a) Express your answer in terms of an angle (in degree)
b) and a distance.
Answer:
(a) θ = 55.85 degree
(b) 7.89 km
Explanation:
Using vector notations
A = 1.88 km south = 1.88 (- j) km = - 1.88 j km
B = 9.05 km 47 degree north of east
B = 9.05 ( Cos 47 i + Sin 47 j) km
B = (6.17 i + 6.62 j) km
Net displacement is
D = A + B
D = - 1.88 i + 6.17 i + 6.62 j = 4.29 i + 6.62 j
(a) Angle made with positive X axis
tanθ = 6.62 / 4.29 = 1.474
θ = 55.85 degree
(b) distance = [tex]Distance = \sqrt{(4.29)^{2} + (6.62)^{2}}[/tex]
distance = 7.89 km
To find the car's final location, add the displacements in the north and east directions. The car's final location is 6.4 km north and 6.0 km east relative to the origin.
Explanation:To find the car's final location, we need to add the displacements in both the north and east directions.
First, let's calculate the north displacement by using the distance formula:
North Displacement = 9.05 km * sin(47°) = 6.4 km
Next, let's calculate the east displacement by using the distance formula:
East Displacement = 9.05 km * cos(47°) = 6.0 km
Therefore, the car's final location relative to the origin is 6.4 km north and 6.0 km east.
A batter hits a 0.140-kg baseball that was approaching him at 50.0 m/s and, as a result, the ball leaves the bat at 35.0 m/s in the direction of the pitcher. What is the magnitude of the impulse delivered to the baseball?
Answer:
Impulse, J = 2.1 kg-m/s
Explanation:
Given that,
Mass of baseball, m = 0.14 kg
It was approaching him at 50.0 m/s and, as a result, the ball leaves the bat at 35.0 m/s in the direction of the pitcher. We need to find the magnitude of Impulse delivered to the baseball.
The change in momentum is equal to the Impulse imparted to the ball i.e.
[tex]J=m(v-u)[/tex]
[tex]J=0.14(-35-50)[/tex]
J = -2.1 kg-m/s
So, the Impulse delivered to the baseball is 2.1 kg-m/s
A 10 kg mass starts from rest at the top of a frictionless incline and slides down the ramp. The ramp makes an angle of 0=30" . Take the positive x axis to be down the ramp. What is the acceleration of the mass and it moves down the ramp? m a) a 0 -4.9 m b) a=-4.9- m S m c) a+4.9- a, 0 2. 2 d) None of the above.
Answer:
Explanation:
When a body starts sliding on an inclined plane, the acceleration of body is due to its impotent 9f weight. The component of weight is mg Sin theta along the plane.
Thus, the acceleration is g Sin theta
= 9.8 × Sin 30 = 4.9 m/s^2
A current of 0.2 A flows through a 3 m long wire that is perpendicular to a 0.3 T magnetic field. What is the magnitude of the force on the wire in units of newtons?
Answer:
Magnetic force, F = 0.18 N
Explanation:
It is given that,
Current flowing in the wire, I = 0.2 A
Length of the wire, L = 3 m
Magnetic field, B = 0.3 T
It is placed perpendicular to the magnetic field. We need to find the magnitude of force on the wire. It is given by :
[tex]F=ILB\ sin\theta[/tex]
[tex]F=0.2\ A\times 3\ m\times 0.3\ T\ sin(90)[/tex]
F = 0.18 N
So, the magnitude of force on the wire is 0.18 N. Hence, this is the required solution.
Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 250 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 30 centimeters to 60 centimeters? N-cm
Answer:
112.5 J
Explanation:
F = 250 N, x = 30 cm = 0.3 m
Let the spring constant be K.
By using the Hooke's law
F = k x
250 = k x 0.3
k = 833.3 N / m
xi = 30 cm = 0.3 m, xf 60 cm = 0.6 m
Work done = 1/2 k (xf^2 - xi^2)
Work done = 0.5 x 833.33 x (0.6^2 - 0.3^2)
Work done = 112.5 J
Capacitors, C1 = 1.0 F and C2 = 1.0 F, are connected in parallel to a 6.0 volt battery (ΔV = 6.0V). If the battery is disconnected and the capacitors are connected to a 33 ohm resistor, how long should it take for the voltage to cross the capacitors to drop to 2.2 volts (36.8% of the original 6.0 volts)?
Answer:
66.2 sec
Explanation:
C₁ = 1.0 F
C₂ = 1.0 F
ΔV = Potential difference across the capacitor = 6.0 V
C = parallel combination of capacitors
Parallel combination of capacitors is given as
C = C₁ + C₂
C = 1.0 + 1.0
C = 2.0 F
R = resistance = 33 Ω
Time constant is given as
T = RC
T = 33 x 2
T = 66 sec
V₀ = initial potential difference across the combination = 6.0 Volts
V = final potential difference = 2.2 volts
Using the equation
[tex]V = V_{o} e^{\frac{-t}{T}}[/tex]
[tex]2.2 = 6 e^{\frac{-t}{66}}[/tex]
t = 66.2 sec
The total solar irradiance (TSI) at Earth orbit is 1400 watts/m2 . Assuming this value can be represented as a single Poynting flux. Find the corresponding flux at the solar visible-light surface. Explain your methods.
(Distance between Sun & Earth is 150 million km)
Answer:
3.958 × 10²⁶ watt
Explanation:
Given:
Distance between earth and sun, [tex]R_e[/tex] = 150 ×10⁸ m
Total solar irradiance at earth orbit = 1400 watt/m²
Now,
Area irradiated ([tex]A_e[/tex]) will be = [tex]4\pi R_e^2[/tex]
⇒ [tex]A_e[/tex] = [tex]4\pi \times (150\times 10^{8})^2[/tex]
⇒ [tex]A_e[/tex] = [tex]2.827\times 10^{23}m^2[/tex]
Therefore, the flux = Total solar irradiance at earth orbit × [tex]A_e[/tex]
the flux = [tex]1400watt/m^2\times 2.827\times 10^{23}m^2[/tex]
⇒the flux = 3.958 × 10²⁶ watt
lectric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an electric potential difference of 12 V is placed across its leads. What is the resistance of the heater wire?
Answer:
The resistance of the heater wire is of R= 0.68 Ω.
Explanation:
1 cal/s = 4.184 W
P= 50 cal/s = 209.2 W
V= 12V
P= V* I
I= P/V
I= 17.43 A
P= I² * R
R= P / I²
R= 0.68 Ω
A.
Calculate the specific weight, density and specific gravity of one litre of a liquid, which weighs 7N.
Select one:
1. 7000 N/m3, 713.5 kg/m3, 0.7135
2. 700 N/m3, 71.35 kg/m3, 0.07135
3. 70 N/m3, 7.135 kg/m3, 0.007135
4. None of the above.
B.
The multiplying factor for converting one stoke into m2/s is
Select one:
1. 102
2. 104
3. 10-2
4. 10-4
Answer:
A) Option 1 is the correct answer.
B) Option 4 is the correct answer.
Explanation:
A) Weight of liquid = 7 N
Volume of liquid = 1 L = 0.001 m³
Specific weight = [tex]\frac{7}{0.001}=7000N/m^3[/tex]
Density = [tex]\frac{7000}{9.81}=713.5kg/m^3[/tex]
Specific gravity = [tex]\frac{713.5}{1000}=0.7135[/tex]
Option 1 is the correct answer.
B) The Stokes(St) is the cgs physical unit for kinematic viscosity, named after George Gabriel Stokes.
We have
1 St = 10⁻⁴ m²/s
Option 4 is the correct answer.
After evaluating all the options we have:
A. From all of the options of specific weight, density, and specific gravity of 1 liter of liquid, the correct is option 1: 7000 N/m³, 713.5 kg/m³, 0.7135.
B. The correct option of the multiplying factor for converting one stoke into m²/s is 4: 10⁻⁴.
A. Let's calculate the specific weight, density, and specific gravity of 1 L of the liquid that weighs 7 N.
The specific weight is given by:[tex] \gamma = dg [/tex] (1)
Where:
γ: is the specific weight
d: is the density
g: is the gravity = 9.81 m/s²
We need to find the density which is:
[tex] d = \frac{m}{V} [/tex] (2)
Where:
m: is the mass
V: is the volume = 1 L = 0.001 m³
The mass can be found knowing that the liquid weighs (W) 7 N, so:
[tex] W = mg [/tex]
[tex] m = \frac{W}{g} [/tex] (3)
By entering equations (3) and (2) into (1) we have:
[tex] \gamma = dg = \frac{mg}{V} = \frac{W}{V} = \frac{7 N}{0.001 m^{3}} = 7000 N/m^{3} [/tex]
Hence, the specific weight is 7000 N/m³.
The density can be found as follows:[tex] d = \frac{m}{V} = \frac{W}{gV} = \frac{7 N}{9.81 m/s^{2}*0.001 m^{3}} = 713.5 kg/m^{3} [/tex]
Then, the density is 713.5 kg/m³.
The specific gravity (SG) of a liquid can be calculated with the following equation:[tex] SG = \frac{d}{d_{H_{2}O}} = \frac{713.5 kg/m^{3}}{1000 kg/m^{3}} = 0.7135 [/tex]
Hence, the specific gravity is 0.7135.
Therefore, the correct option is 1: 7000 N/m³, 713.5 kg/m³, 0.7135.
B. A Stokes is a measurement unit of kinematic viscosity.
One m²/s is equal to 10⁴ stokes, so to convert 1 stokes to m²/s we need to multiply for 10⁻⁴.
Hence, the correct option is 4: 10⁻⁴.
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An undiscovered planet, many light-years from Earth, has one moon which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.31500 x 10^5 km and the planet has a radius of 4.150 x10^3 km and a mass of 7.15 x10^22 kg, how long (in days) does it take the moon to make one revolution around the planet. The gravitational constant is 6.67 x10^-11 N m^2/kg^2.
Answer:
118.06 days
Explanation:
d = distance of the center of moon from surface of planet = 2.315 x 10⁵ km = 2.315 x 10⁸ m
R = radius of the planet = 4.15 x 10³ km = 4.15 x 10⁵ m
r = center to center distance between the planet and moon = R + d
M = mass of the planet = 7.15 x 10²² kg
T = Time period of revolution around the planet
Using Kepler's third law
[tex]T^{2}=\frac{4\pi ^{2}r^{3}}{GM}[/tex]
[tex]T^{2}=\frac{4\pi ^{2}(R + d)^{3}}{GM}[/tex]
[tex]T^{2}=\frac{4(3.14)^{2}((4.15\times 10^{5}) + (2.315\times 10^{8}))^{3}}{(6.67\times 10^{-11})(7.15\times 10^{22})}[/tex]
T = 1.02 x 10⁷ sec
we know that , 1 day = 24 h = 24 x 3600 sec = 86400 sec
T = [tex](1.02 \times 10^{7} sec)\frac{1 day}{86400 sec}[/tex]
T = 118.06 days
What is the velocity of a proton that has been accelerated by a potential difference of 15 kV? (i)\:\:\:\:\:9.5\:\times\:10^5\:m.s^{-1} ( i ) 9.5 × 10 5 m . s − 1 (ii)\:\:\:\:2.2\:\times\:10^9\:m.s^{-1} ( i i ) 2.2 × 10 9 m . s − 1 (iii)\:\:\:3.9\:\times\:10^8\:m.s^{-1} ( i i i ) 3.9 × 10 8 m . s − 1 (iv)\:\:\:\:1.7\:\times\:10^6\:m.s^{-1}
Answer:
Velocity of a proton, [tex]v=1.7\times 10^6\ m/s[/tex]
Explanation:
It is given that,
Potential difference, [tex]V=15\ kV=15\times 10^3\ V[/tex]
Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.
Using the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=qV[/tex]
q is the charge of proton
m is the mass of proton
[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}[/tex]
[tex]v=1695361.75\ m/s[/tex]
[tex]v=1.69\times 10^6\ m/s[/tex]
or
[tex]v=1.7\times 10^6\ m/s[/tex]
So, the velocity of a proton is [tex]1.7\times 10^6\ m/s[/tex]. Hence, this is the required solution.
A 15-uF capacitor is connected to a 50-V battery and becomes fullycharged. The battery is
removed and a slab of dielectric that completely fills the spacebetween the plates is inserted.
If the dielectric has a dielectric constant of 5.0:
A. what is the capacitance of the capacitor after the
slab is inserted?
B. what is the voltage across the capacitor's plates
after the slab is inserted?
A. C = 75μF and B. V = 10V.
We have to use the equation k = C/C₀ and k = V₀/V which both are the dielectric constant.
A. The capacitance after the slab is inserted.
With C₀ = 15μF and k = 5.0. Clear k for the equation k = C/C₀:
C = k*C₀
C = (5.0)(15x10⁻⁶F) = 0.000075F
C = 75μF
B. The voltage across the capacitor's plates after the slab is inserted.
With V₀ = 50V and k = 5.0. Clear V from the equation k = V₀/V:
V = V₀/k
V = 50V/5.0
V = 10V
Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.
A. The capacitance of the capacitor after the slab is inserted is 75-uF.B. The voltage across the capacitor's plates after the slab is inserted is 10 volts.What is capacitance of capacitor?The capacitance of capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference.
The capacitance of the capacitor is 15-uF and it is connected to 50-V battery. The value of the dielectric constant is 5.0.
A. The capacitance of the capacitor after the slab is insertedThe dielectric constant of the slab is the ratio of capacitance of the capacitor after the slab and the capacitance of the before the slab inserted.
As the value of the dielectric constant is 5.0. Thus the capacitance of the capacitor after the slab is inserted given as,
[tex]0.5=\dfrac{C}{15\times10^{-6}}\\C=75\rm \mu F[/tex]
Thus, the capacitance of the capacitor after the slab is inserted is 75-uF.
B. The voltage across the capacitor's plates after the slab is insertedThe dielectric constant of the slab is the ratio of voltage across the capacitor's of the capacitor before the slab inserted and the voltage across the capacitor after the slab inserted.
As the value of the dielectric constant is 5.0. Thus the voltage across the capacitor's plates after the slab is inserted can be given as,
[tex]V=\dfrac{50}{5.0}\\C=10\rm V[/tex]
Thus, the voltage across the capacitor's plates after the slab is inserted is 10 volts.
Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.
A. The capacitance of the capacitor after the slab is inserted is 75-uF.B. The voltage across the capacitor's plates after the slab is inserted is 10 volts.Learn more about the capacitance of capacitor here;
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A string of length L= 1.2 m and mass m = 20g is under 400 N of tension, its two ends are fixed. a. How many nodes will you see in the 5th harmonic.
i. 4
ii. 5
iii. 6
iv. 7
v. None of the above.
Answer:
(iii) 6
Explanation:
Part a)
Since it is given that both ends are fixed and it is vibrating in 5th harmonic
So here it will have 5 number of loops in it
so we can draw it in following way
each loop will have 1 antinode and two nodes which means number of nodes is one more than number of anti-nodes
So there are 5 loops which means it will have 5 antinodes
and hence there will be 6 nodes in it
so correct answer will be
(iii) 6
wo kids are on a seesaw. The one on the left has a mass of 75 kg and is sitting 1.5 m from the pivot point. The one on the right has a mass of 60 kg and is sitting 1.8 m from the pivot point. What is the net torque on the system?
Answer:
4.5 Nm (Anticlockwise)
Explanation:
Let the 75 kg kid is sitting at the left end and the 60 kg kid is sitting on the right end.
Anticlockwise Torque = 75 x 1.5 = 112.5 Nm
clockwise Torque = 60 x 1.8 = 108 Nm
Net torque = Anticlockwise torque - clockwise torque
Net Torque = 112.5 - 108 = 4.5 Nm (Anticlockwise)
Identify the least dense and most dense planets and provide an explanation as to why they are the least and most dense.
Answer:
Explained
Explanation:
The most dense planet in our solar system is Earth. Earth is most dense planet because
Earth is made of dense material like rocks, core is made of metals, soil and water. The average density of earth is 5.5 gram/cm^3.The overall mass of Earth is higher than those of other rocky planets. This heaviness accounts for higher density of earth.The least dense planet of the solar system is Saturn because Saturn is mostly made of gases and its size is smaller than the Jupiter. Jupiter has more gravity hence its density is higher to Saturn. Moreover, Uranus and Neptune are ice giants. Although they are also made of gases but due their distance from sun most of these gases have solidified. Hence making them more dense than Saturn and Jupiter.
Saturn is the least dense planet, while Earth is the densest planet in our solar system.
Explanation:The least dense planet in our solar system is Saturn, while the densest planet is Earth. Saturn is a gas giant composed mostly of hydrogen and helium, which have lower densities compared to the rocks and metals found on rocky planets like Earth. Earth, on the other hand, has a higher density due to its solid composition and a core made of iron and nickel.
Rearrange each of the following combinations of units to obtain a units that involves Joule ) Show kg m s intermediate steps, ( ).
Answer:
Explanation:
Joule is SI unit of work
Work = force x distance
Work = mass x acceleration x distance
Unit of mass is kg
Unit of acceleration is m/s^2
Unit of distance is m
So, unit of work = kg x m x m /s^2
So, Joule = kg m^2 / s^2
Part A: A charge +Q is located at the origin and a second charge, +9Q, is located at x= 15.8 cm . Where should a third charge q be placed so that the net force on q is zero? Find q 's position on x -axis.
Part B: A charge +Q is located at the origin and a negative charge, -7Q, is located at a distance x= 19.6 cm . Where should a third charge q be placed so that the net force on q is zero? Find q 's position on x-axis.
Answer:
Part a)
x = 3.95 cm
Part b)
x = - 11.9 cm
Explanation:
Part a)
Since both charges are of same sign
so the position at which net force is zero between two charges is given as
[tex]\frac{kq_1}{r_1^2} = \frac{kq_2}{(15.8 - r)^2}[/tex]
here we know that
[tex]q_1 = Q[/tex]
[tex]q_2 = 9Q[/tex]
[tex]\frac{Q}{r^2} = \frac{9Q}{(15.8 - r)^2}[/tex]
square root both sides
[tex](15.8 - r) = 3r[/tex]
[tex]r = 3.95 cm[/tex]
Part b)
Since both charges are of opposite sign
so the position at which net force is zero will lie on the other side of smaller charges is given as
[tex]\frac{kq_1}{r_1^2} = \frac{kq_2}{(19.6 + r)^2}[/tex]
here we know that
[tex]q_1 = Q[/tex]
[tex]q_2 = -7Q[/tex]
[tex]\frac{Q}{r^2} = \frac{7Q}{(19.6 + r)^2}[/tex]
square root both sides
[tex](19.6 + r) = 2.64r[/tex]
[tex]r = 11.9 cm[/tex]
so on x axis it will be at x = - 11.9 cm
A 3.9 kg block is pushed along a horizontal floor by a force ModifyingAbove Upper F With right-arrow of magnitude 27 N at a downward angle θ = 40°. The coefficient of kinetic friction between the block and the floor is 0.22. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block’s acceleration.
Answer:
a) 12.23 N
b) 2.2 m/s²
Explanation:
m = mass of the block = 3.9 kg
F = applied force = 27 N
θ = angle of the applied force with the horizontal = 40°
μ = Coefficient of kinetic friction = 0.22
[tex]F_{n}[/tex] = normal force
[tex]F_{g}[/tex] = weight of the block = mg
Along the vertical direction, force equation is given as
[tex]F_{n}[/tex] = F Sinθ + [tex]F_{g}[/tex]
[tex]F_{n}[/tex] = F Sinθ + mg
Kinetic frictional force is given as
f = μ [tex]F_{n}[/tex]
f = μ (F Sinθ + mg)
f = (0.22) (27 Sin40 + (3.9)(9.8))
f = 12.23 N
b)
Force equation along the horizontal direction is given as
F Cosθ - f = ma
27 Cos40 - 12.23 = 3.9 a
a = 2.2 m/s²
At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values of L and C. larger than R equal to R Which of the following does the quality factor of the circuit depend on? (Select all that apply.)
Answer:
at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all
Explanation:
we know that for series RLC circuit impedance is given by
[tex]Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2[/tex]
but we know that at resonance [tex]X_L=X_C[/tex]
putting [tex]X_L=X_C[/tex] in impedance formula , impedance will become
Z=R so at resonance impedance of series RLC is equal to resistance only
now quality factor of series resonance is given by
[tex]Q=\frac{\omega L}{R}=\frac{1}{\omega CR}=\frac{1}{R}\sqrt{\frac{L}{C}}[/tex] so from given expression it is clear that quality factor depends on R L and C
A bolt comes loose from underneath an elevator that is moving upward at a speed of 5 m/s. The bolt reaches the bottom of the elevator shaft in 3.1 s. (a) How high up was the elevator when the bolt came loose? (In m)
(b) What is the speed of the bolt when it hits the bottom of the shaft? (In m/s)
Answer:
a) The elevator was 31.64 m high up when the bolt came loose.
b) Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s
Explanation:
a) Considering motion of bolt:-
Initial velocity, u = 5 m/s
Acceleration , a = -9.81 m/s²
Time = 3.1 s
We have equation of motion s= ut + 0.5 at²
Substituting
s= 5 x 3.1 - 0.5 x 9.81 x 3.1²
s = 0 x t + 0.5 x 9.81 x t²
s = -31.64 m
The elevator was 31.64 m high up when the bolt came loose.
b) We have equation of motion v = u + at
Initial velocity, u = 5 m/s
Acceleration , a = -9.81 m/s²
Time = 3.1 s
Substituting
v = u + at
v = 5 - 9.81 x 3.1 = -25.41 m/s
Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s
What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body?
Answer:
Weight, w = mg
Mass is an intrinsic property.
Explanation:
Mass is the measure of amount of matter in an object. It is an intrinsic that is unchanging property of a body. Mass cannot be destroyed nor be created.
Weight is the product of mass and acceleration due to gravity.
It changes with value of acceleration due to gravity. That is weight in Earth not equal to weight in moon since the value of acceleration due to gravity is different. So, weight is not an intrinsic property. It is a changing property for a body.
Weight, w = mg, where m is the mass and g is the acceleration due to gravity value.
Mass is an intrinsic and constant property of a body representing its matter content, while weight, the force of gravity on a body, varies with the gravitational environment. Despite common misuse in everyday language, they are distinct concepts in physics.
Explanation:Mass and weight are two different but closely related concepts. Mass is an intrinsic property of a body, representing the amount of matter it contains, and it remains constant regardless of the body's location, be it on Earth, the moon, or in orbit.
On the other hand, weight is the force exerted on a body due to gravity. It's a product of the body's mass and the acceleration due to gravity, represented by the formula 'Weight = Mass x Gravity'. Unlike mass, weight changes with the gravitational environment. For example, a person's weight on the moon is only one-sixth of their weight on Earth due to the moon's lower gravity, while their mass remains the same.
In everyday language, mass and weight are often used interchangeably, but in the field of physics, it is important to distinguish between them. For example, when we refer to our 'weight' in kilograms, we're technically referring to our mass. The proper unit of weight, consistent with its definition as a force, is the newton in the International System of Units (SI).
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The return-air ventilation duct in a home has a cross-sectional area of 900 cm^2. The air in a room that has dimensions 7.0 m ×× 11.0 m ×× 2.4 m is to be completely circulated in a 40-min cycle. What is the speed of the air in the duct? (Express your answer to two significant figures.) (m/s)
Answer:
0.86 m/s
Explanation:
A = cross-sectional area of the duct = 900 cm² = 900 x 10⁻⁴ m²
v = speed of air in the duct
t = time period of circulation = 40 min = 40 x 60 sec = 2400 sec
V = Volume of the air in the room = volume of room = 7 x 11 x 2.4 = 184.8 m³
Volume of air in the room is given as
V = A v t
inserting the values
184.8 = (900 x 10⁻⁴) (2400) v
v = 0.86 m/s
The output of a generator is 440 V at 20 A. It is to be transmitted on a line with resistance of 0.60 Ω. To what voltage must the generator output be stepped up with a transformer if the power loss in transmission is not to exceed 0.010% of the original power?
Answer:
The voltage of the generator is 7.27 kV.
Explanation:
Given that,
Output of generator = 440 V
Current = 20 A
Resistance = 0.60 Ω
Power loss =0.010%
We need to calculate the total power of the generator
Using formula of power
[tex]P=VI[/tex]
Where, V = voltage
I = current
Put the value into the formula
[tex]P=440\times20[/tex]
[tex]P=8800\ W[/tex]
Th power lost on the transmission lines
[tex]P_{L}=0.010\% P[/tex]
[tex]P_{L} = 0.010\%\times8800[/tex]
[tex]P_{L}=0.88\ W[/tex]
The current passing through the transmission line
[tex]I'=\sqrt{\dfrac{P_{L}}{R}}[/tex]
[tex]I'=\sqrt{\dfrac{0.88}{0.60}}[/tex]
[tex]I'=1.211\ A[/tex]
We need to calculate the voltage of the generator
Using formula of voltage
[tex]V_{g}=\dfrac{P}{I'}[/tex]
Put the value into the formula
[tex]V_{g}=\dfrac{8800}{1.211}[/tex]
[tex]V_{g}=7.27\times10^{3}\ V[/tex]
[tex]V_{g}=7.27\ kV[/tex]
Hence, The voltage of the generator is 7.27 kV.
The voltage of the generator output needs to be stepped up with a transformer to limit power losses in transmission.
Explanation:To limit power losses in transmission, the voltage of the generator output needs to be stepped up with a transformer. The formula to calculate power loss is P_loss = I^2 x R. Given that the power loss should not exceed 0.010% of the original power, we can calculate the maximum allowable power loss. By rearranging the formula, we can find the voltage required for the generator output with a transformer.
First, we calculate the original power using P = V x I. Then, we calculate the maximum allowable power loss as a percentage of the original power. Next, we substitute the given values into the formula and solve for the new voltage using the rearranged formula.
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A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time interval?
Answer:
Number of electrons, [tex]n=5.62\times 10^{21}[/tex]
Explanation:
It is given that,
Resistance, R = 4 ohms
Current, I = 3 A
Time, t = 5 min = 300 s
We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.
i.e. q = n e ...............(1)
And current, [tex]I=\dfrac{q}{t}[/tex]
[tex]I\times t=n\times e[/tex]
[tex]n=\dfrac{It}{e}[/tex]
e is the charge of an electron
[tex]n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}[/tex]
[tex]n=5.62\times 10^{21}[/tex]
So, the number of electrons pass through the resistor is [tex]5.62\times 10^{21}[/tex]. Hence, this is the required solution.
A 0.42 kg football is thrown with a velocity of 17 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.017 S. What is the force exerted on the receiver? Answer in units of N
Answer:
420 N
Explanation:
m = 0.42 Kg, u = 17 m/s, v = 0 m/s, t = 0.017 s
By first law of Newtons' laws of motion, the rate of change in momentum is force, F = m (v - u) / t
F = 0.42 x ( 0 - 17) / 0.017
F = - 420 N
Negative sign shows hat the force is resistive that means the ball finally comes to rest.
The 400 kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F= (3200) N, where t is in seconds. If the car has an initial velocity V1= 2m/s at s 0 and t= 0, determine the distance it moves the plane when (a) t 1 s and (b) f-5 s.
Answer:
(a) 110 m/s
(b) 42 m/s
Explanation:
mass, m = 400 kg, F = 3200 N, V1 = 2 m/s,
acceleration, a = Force / mass = 3200 / 400 = 8 m/s^2
(a) Use first equation of motion
v = V1 + a t
v = 2 + 8 x 1 = 10 m/s
(b) Again using first equation of motion
v = V1 + a t
v = 2 + 8 x 5 = 42 m/s
Thus, the velocity of plane after 1 second is 10 m/s and after 5 second the velocity is 42 m/s.
The 400 kg mine car moves a distance of 6 m after one second and 110 m after five seconds given an initial velocity of 2 m/s and a force of 3200 N.
Explanation:The problem describes a physics scenario where a 400 kg mine car is pulled up an incline by a cable and motor. The force on the cable is given as F = 3200N and the initial velocity, V1, is given as 2 m/s. We can calculate the distance the car moves on the plane at different times using the physics equations of motion.
Let's use the equation of motion: s = ut + 1/2 at², where 's' is the distance moved, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time.
Given that the net force is equal to mass times acceleration (F = ma), we can calculate the acceleration, 'a', as F/m. So, a = 3200N/400kg = 8 m/s².
(a) t = 1s: The distance moved is s = 2m/s * 1s + 1/2 * 8 m/s² * (1s)² = 2m + 4m = 6m.(b) t = 5s: The distance moved is s = 2m/s * 5s + 1/2 * 8 m/s² * (5s)² = 10m + 100m = 110m.Learn more about Physics of Motion here:https://brainly.com/question/13966796
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13. A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V and the secondary is connected to a floodlight that draws 5A, what is the power output? Please show ALL of your work.
Answer:
The output power is 2 kW
Explanation:
It is given that,
Number of turns in primary coil, [tex]N_p=250[/tex]
Number of turns in secondary coil, [tex]N_s=500[/tex]
Voltage of primary coil, [tex]V_p=200\ V[/tex]
Current drawn from secondary coil, [tex]I_s=5\ A[/tex]
We need to find the power output. It is equal to the product of voltage and current. Firstly, we will find the voltage of secondary coil as :
[tex]\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}[/tex]
[tex]\dfrac{250}{500}=\dfrac{200}{V_s}[/tex]
[tex]V_s=400\ V[/tex]
So, the power output is :
[tex]P_s=V_s\times I_s[/tex]
[tex]P_s=400\ V\times 5\ A[/tex]
[tex]P_s=2000\ watts[/tex]
or
[tex]P_s=2\ kW[/tex]
So, the output power is 2 kW. Hence, this is the required solution.
Gaussian surfaces A and B enclose the same positive charge+Q. The area of Gaussian surface A is three times larger than that of Gaussian surface B. The flux of electric field through Gaussian surface A is A) nine times larger than the flux of electric field through Gaussian surface B. B) three times smaller than the flux of electric field through Gaussian surface B. C) unrelated to the flux of electric field through Gaussian surface B. D) equal to the flux of electric field through Gaussian surface B. E) three times larger than the flux of electric field through Gaussian surface B
The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Therefore, the flux through surface A is three times larger than the flux through surface B.
The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Since the flux of electric field is proportional to the area of the surface, the flux through surface A is three times larger than the flux through surface B. Therefore, the correct answer is E) three times larger than the flux of electric field through Gaussian surface B.
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Final answer:
The correct answer is D) equal to the flux of the electric field through Gaussian surface B. The electric flux depends on the charge enclosed by the Gaussian surface, not on its size or shape, according to Gauss's law.
Explanation:
The question asks whether Gaussian surfaces A and B, enclosing the same positive charge +Q, with surface A having three times the area of surface B, have different electric fluxes. According to Gauss's law, the electric flux (Φ) through a closed surface is proportional to the charge enclosed (Φ = Q/ε0). Since both surfaces enclose the same charge, the flux through each surface must be the same, regardless of their respective areas.
Thus, the correct answer is D) equal to the flux of the electric field through Gaussian surface B because the flux depends only on the amount of enclosed charge, not on the size or shape of the Gaussian surface.
At a point 1.2 m out from the hinge, 14.0 N force is exerted at an angle of 27 degrees to the moment arm in a plane which is perpendicular to the door. What is the magnitude of the torque?
Answer:
[tex]\tau = 7.63 Nm[/tex]
Explanation:
As we know that moment of force is given as
[tex]\tau = \vec r \times \vec F[/tex]
now we have
[tex]\vec r = 1.2 m[/tex]
[tex]\vec F = 14 N[/tex]
now from above formula we have
[tex]\tau = r F sin\theta[/tex]
here we know that
[tex]\theta = 27 degree[/tex]
so we have
[tex]\tau = (1.2)(14) sin27[/tex]
[tex]\tau = 7.63 Nm[/tex]