Answer:
[tex]f = +42.9 cm[/tex]
[tex]P =+2.33Dioptre[/tex]
Explanation:
As we know that Far sighted person has near point shifted to 60 cm distance
so he is able to see the object 60 cm
and the person want to see the objects at distance 25 cm
so here the image distance from lens is 60 cm and the object distance from lens is 25 cm
now from lens formula we have
[tex]\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}[/tex]
[tex]-\frac{1}{60} + \frac{1}{25} = \frac{1}{f}[/tex]
[tex]f = +42.9 cm[/tex]
Now we know that power of lens is given as
[tex]P = \frac{1}{f}[/tex]
[tex]P = \frac{1}{0.429} = +2.33Dioptre[/tex]
Be sure to answer all parts. Indicate the number of protons, neutrons and electrons in each of these species: (a) 13 N 7 (b) 35 S 16 (c) 63 Cu 29 (d) 89 Sr 38 How many protons are in each isotope? How many neutrons are in each isotope? How many electrons are in each isotope?
Answer:
(a) No. of protons = atomic number = 7
No. of electrons = no. of protons = 7
No. of neutrons = mass no. - atomic no. = 6
(b) No. of protons = atomic number = 16
No. of electrons = no. of protons = 16
No. of neutrons = mass no. - atomic no. = 19
(c) No. of protons = atomic number = 29
No. of electrons = no. of protons = 29
No. of neutrons = mass no. - atomic no. = 34
(d) No. of protons = atomic number = 38
No. of electrons = no. of protons = 38
No. of neutrons = mass no. - atomic no. = 51
Explanation:
No. of protons = No. of electrons = atomic number
No. of neutrons = Mass no. - atomic no.
(a) 13 N 7
Here atomic number is 7, mass number is 13
No. of protons = atomic number = 7
No. of electrons = no. of protons = 7
No. of neutrons = mass no. - atomic no. = 13 - 7 = 6
(b) 35 S 16
Here atomic number is 16, mass number is 35
No. of protons = atomic number = 16
No. of electrons = no. of protons = 16
No. of neutrons = mass no. - atomic no. = 35 - 16 = 19
(c) 63 Cu 29
Here atomic number is 29, mass number is 63
No. of protons = atomic number = 29
No. of electrons = no. of protons = 29
No. of neutrons = mass no. - atomic no. = 63 - 29 = 34
(d) 89 Sr 38
Here atomic number is 38, mass number is 89
No. of protons = atomic number = 38
No. of electrons = no. of protons = 38
No. of neutrons = mass no. - atomic no. = 89 - 38 = 51
Final answer:
The isotopes ¹N⁷, ³⁵S¹⁶, ⁶³Cu₂₉, and ₈₉Sr₃₈ have 7, 16, 29, and 38 protons, respectively, with neutrons calculated by subtracting the number of protons from the mass number. Lastly, the number of electrons is equal to the number of protons in a neutral atom.
Explanation:
The number of protons, neutrons, and electrons for each isotope can be determined using the atomic number, mass number, and charge. The atomic number represents the number of protons (and also the number of electrons in a neutral atom), the mass number represents the sum of protons and neutrons, and the charge indicates how many electrons have been lost or gained.
(a) ¹N⁷: 7 protons, 6 neutrons, 7 electrons(b) ³⁵S¹⁶: 16 protons, 19 neutrons, 16 electrons(c) ⁶³Cu₂₉: 29 protons, 34 neutrons, 29 electrons(d) ₈₉Sr₃₈: 38 protons, 51 neutrons, 38 electronsEach isotope comprises an equal number of protons and electrons in their neutral states. Neutrons are the difference between the mass number and the number of protons.
A pendulum clock was moved from a location where g = 9.8135 m/s 2 to another location where g = 9.7943 m/s 2 . During the move, the length of the clock’s pendulum did not change; nevertheless, the clock lost accuracy. Assuming the clock was perfectly accurate at its previous location, how many seconds a day does it lose at the new location?
Answer:
The pendulum clock does it loses 83.0304 seconds a day at the new location.
Explanation:
T1= 1 s
T2= ?
g1= 9.8135 m/s
g2= 9.7943 m/s
L=?
L= (T1/2π)² * g1=
L= 0.24857m
T2= 2π * √(L/g2)
T2= 1.000961 s
ΔT= T2 - T1
ΔT = 0.000961 s
seconds lost a day = 24 * 3600 * ΔT
seconds lost a day= 24 * 3600 * 0.000961 s
seconds lost a day= 83.0304 s
A pendulum clock loses accuracy when it is moved from one location to another due to changes in the acceleration due to gravity. The period of a pendulum is dependent on its length and the acceleration due to gravity. To calculate how many seconds a day the clock loses at the new location, we can subtract the new period from the original period and convert it to seconds.
Explanation:A pendulum clock loses accuracy when it is moved from one location to another due to changes in the acceleration due to gravity, not changes in the length of the pendulum. In this case, the clock was moved from a location where g = 9.8135 m/s2 to a new location where g = 9.7943 m/s2. The clock loses accuracy because the period of a pendulum is dependent on its length and the acceleration due to gravity.
The period of a pendulum can be calculated using the equation: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
So, to calculate how many seconds a day the clock loses at the new location, we can subtract the new period from the original period and convert it to seconds. Since the original period is 2.00000 s, we have:
ΔT = Toriginal - Tnew
ΔT = 2.00000 s - Tnew
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In his explanation of the threshold frequency in the photoelectric effect, Einstein reasoned that the absorbed photon must have a minimum energy to dislodge an electron from the metal surface. This energy is called the work function (φ) of the metal. Many calculators use photocells to provide their energy. Find the maximum wavelength needed to remove an electron from silver (φ = 7.59 × 10 −19 J).
Answer:
[tex]\lambda = 260.9nm[/tex]
Explanation:
As we know that work function is defined as the minimum energy of photons required to produce the photo electric effect
so here we have
[tex]\phi = \frac{hc}[\lambda}[/tex]
now we know that
[tex]h = 6.63\times 10^{-34}[/tex]
[tex]c = 3\times 10^8 m/s[/tex]
[tex]\lambda[/tex] = wavelength of photons
now from above formula
[tex]7.59 \times 10^{-19} = \frac{(6.36 \times 10^{-34})(3\times 10^8)}{\lambda}[/tex]
[tex]\lambda = \frac{(6.36 \times 10^{-34})(3\times 10^8)}{7.59 \times 10^{-19}}[/tex]
[tex]\lambda = 260.9nm[/tex]
Suppose all the mass of the Earth were compacted into a small spherical ball. Part A What radius must the sphere have so that the acceleration due to gravity at the Earth's new surface was equal to the acceleration due to gravity at the surface of the Moon?
Answer:
0.4 times the radius of moon
Explanation:
gravity on moon is equal to the one sixth of gravity on earth.
g' = g / 6
where, g' is the gravity on moon and g be the gravity on earth.
As the earth shrinks, the mass of earth remains same.
The acceleration due to gravity is inversely proportional to the square of radius of planet.
g' ∝ 1/R'² .....(1)
Where, R' is the radius of moon.
g ∝ 1/R² ..... (2)
Where, R be the radius of earth.
Divide equation (1) by (2)
g / g' = R'² / R²
Put g' = g / 6
6 = R'² / R²
2.5 = R' / R
R = R' / 2.5 = 0.4 R'
Thus, the radius of earth should be 0.4 times the radius of moon.
The motion of a particle is defined by the relation x = 2t3 – 9t2 +12t +10, where x and t are expressed in feet and seconds, respectively. Determine the time, position, and acceleration of the particle when v = 2.00 ft/s. (Round the final answer to two decimal places.) The time, the position, and the acceleration of the particle when v = 2.00 are:
Explanation:
The motion of a particle is defined by the relation as:
[tex]x=2t^3-9t^2+12t+10[/tex]........(1)
Differentiating equation (1) wrt t we get:
[tex]v=\dfrac{dx}{dt}[/tex]
[tex]v=\dfrac{d(2t^3-9t^2+12t+10)}{dt}[/tex]
[tex]v=6t^2-18t+12[/tex]............(2)
When v = 2 ft/s
[tex]2=6t^2-18t+12[/tex]
[tex]6t^2-18t+10=0[/tex]
t₁ = 2.26 s
and t₂ = 0.73 s
Put the value of t₁ in equation (1) as :
[tex]x=2(2.26)^3-9(2.26)^2+12(2.26)+10[/tex]
x₁ = 14.23 ft
Put the value of t₂ in equation (1) as :
[tex]x=2(0.73)^3-9(0.73)^2+12(0.73)+10[/tex]
x₁ = 14.74 ft
For acceleration differentiate equation (2) wrt t as :
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]a=\dfrac{d(6t^2-18t+12)}{dt}[/tex]
a = 12 t - 18.........(3)
Put t₁ and t₂ in equation 3 one by one as :
[tex]a_1=12(2.26)-18=9.12\ ft/s^2[/tex]
[tex]a_2=12(0.73)-18=-9.24\ ft/s^2[/tex]
Hence, this is the required solution.
The time taken by the particle is 2.26 s and the magnitude of acceleration and position of the particle at the obtained instant are [tex]9.12 \;\rm ft/s^{2}[/tex] and 14.23 ft respectively.
What is linear motion?When an object is observed to move in a straight line, then the motion of the object is known as linear motion. It can be referenced as motion in one dimension.
Given data:
The position of the particle is, [tex]x = 2t^{3}-9t^{2}+12t+10[/tex].
The velocity of the particle is, [tex]v = 2.00 \;\rm ft/s[/tex].
We know that velocity can be obtained by differentiating the position. Then,
[tex]v =\dfrac{dx}{dt}\\\\v = \dfrac{d(2t^{3}-9t^{2}+12t+10)}{dt}\\\\v=6t^{2}-18t+12[/tex]
Now when v = 2.00 ft/s. The time is,
[tex]2 = 6t^{2}-18t+12\\\\6t^{2}-18t+10=0\\\\t = 2.26 \;\rm s[/tex]
Now, the position of the particle is,
[tex]x = 2t^{3}-9t^{2}+12t+10\\\\x = (2 \times 2.26^{3})-(9 \times 2.26^{2})+(12 \times 2.26)+10\\\\x =14.23\;\rm ft[/tex]
And the magnitude of the acceleration is calculated by differentiating the velocity as,
[tex]a = \dfrac{dv}{dt}\\\\a = \dfrac{d(6t^{2}-18t+12)}{dt}\\\\a=12t-18[/tex]
Substituting t = 2.26 s as,
[tex]a = 12(2.26)-18\\\\a=9.12 \;\rm ft/s^{2}[/tex]
Thus, we can conclude that the time taken by the particle is 2.26 s and the magnitude of acceleration and position of the particle at the obtained instant are [tex]9.12 \;\rm ft/s^{2}[/tex] and 14.23 ft respectively.
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If a copper wire has a resistance of 23.7 Ω at a temperature of 20.3 oC, what resistance does it have at 79.0 oC? (Neglect any change in length or cross-sectional area resulting from the change in temperature.)
Answer:
[tex]R_{79} = 28.91 OHM[/tex]
Explanation:
Resistance ca be determine by using following formula
[tex]R = R_{ref}\left [ 1+ \alpha (T - T_{ref}) \right ][/tex]
where
R = Conductor resistance = 23.7Ω
Rref = conductor reistance at reference temperature,
α = temperature coefficient of resistance for material, for copper 40.41*10^{-4}
T = Conductor temperature in Celcius = 20.3°C
Tref = reference temperature at which α is specified.
[tex]23.7 = R_{ref}\left [ 1+ 40.41*10^{-4} (20.3) \right ][/tex][tex]R_{ref} =21.92 OHM[/tex]
now for 79 degree celcius
[tex]R_{79} = R_{ref}\left [ 1+ \alpha (T - T_{ref}) \right ][/tex]
[tex]R_{79} =21.92\left [ 1+ 40.41*10^{-4} (79) \right ][/tex]
[tex]R_{79} = 28.91 OHM[/tex]
In a ballistic pendulum experiment, a small marble is fired into a cup attached to the end of a pendulum. If the mass of the marble is 0.0265 kg and the mass of the pendulum is 0.250 kg, how high will the pendulum swing if the marble has an initial speed of 5.05 m/s? Assume that the mass of the pendulum is concentrated at its end so that linear momentum is conserved during this collision.
Answer:
0.012 m
Explanation:
m = mass of the marble = 0.0265 kg
M = mass of the pendulum = 0.250 kg
v = initial velocity of the marble before collision = 5.05 m/s
V = final velocity of marble-pendulum combination after the collision = ?
using conservation of momentum
m v = (m + M) V
(0.0265) (5.05) = (0.0265 + 0.250) V
V = 0.484 m/s
h = height gained by the marble-pendulum combination
Using conservation of energy
Potential energy gained by the combination = Kinetic energy of the combination just after collision
(m + M) gh = (0.5) (m + M) V²
gh = (0.5) V²
(9.8) h = (0.5) (0.484)²
h = 0.012 m
Using the concept of inelastic collison and principle of conservation of momentum, the height of swing made by the pendulum would be 0.012 meters
Collison is inelastic :
m1v1 = (m1 + m2)vv = final velocity after the collision(0.0265 × 5.05) = (0.0265 + 0.250)v
0.133825 = 0.2765v
v = (0.133825 ÷ 0.2765)
v = 0.484 m/s
Final velocity after collision = 0.484 m/s
Assuming linear momentum is conserved :
Kinetic energy = Potential Energy0.5mv² = mgh
Mass cancels out
0.5v² = gh
0.5(0.484)² = 9.8h
9.8h = 0.1171
h = (0.1171) ÷ 9.8
h = 0.0119
Therefore, height of the swing made by the pendulum would be 0.012 meters.
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A parachutist bails out and freely falls 50 m. Then the parachute opens, and thereafter she deceler- ates at 2.0 m/s2. She reaches the ground with a speed of 3.0 m/s. (a) How long is the parachutist in the air? (b) At what height does the fall begin?
Answer:
a)The parachutist in the air for 12.63 seconds.
b)The parachutist falls from a height of 293 meter.
Explanation:
Vertical motion of parachutist:
Initial speed, u = 0m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 50 m
We have equation of motion, v² = u² + 2as
Substituting
v² = 0² + 2 x 9.81 x 50
v = 31.32 m/s
Time taken for this
31.32 = 0 + 9.81 x t
t = 3.19 s
After 50m we have
Initial speed, u = 31.32m/s
Acceleration, a = -2 m/s²
Final speed , v = 3 m/s
We have equation of motion, v² = u² + 2as
Substituting
3² = 31.32² - 2 x 2 x s
s = 243 m
Time taken for this
3 = 31.32 - 2 x t
t = 9.44 s
a) Total time = 3.19 + 9.44 = 12.63 s
The parachutist in the air for 12.63 seconds.
b) Total height = 50 + 243 = 293 m
The parachutist falls from a height of 293 meter.
A solenoid has 332 turns and a length of 14 cm. If a current of 0.88 A produces a magnetic flux density of 0.28 T in the core of the solenoid, what is the relative permeability of the core material?
Answer:
106.83
Explanation:
N = 332, l = 14 cm = 0.14 m, i = 0.88 A, B = 0.28 T
Let ur be the relative permeability
B = u0 x ur x n x i
0.28 = 4 x 3.14 x 10^-7 x ur x 332 x 0.88 / 0.14 ( n = N / l)
ur = 106.83
A ball is thrown straight up. Taking the drag force of air into account, does it take longer for the ball to travel to the top of its motion or for it to fall back down again?
Taking the drag force of air into account, it takes longer for the ball to travel to the top of its motion compared to the time it takes to fall back down.
Explanation:When a ball is thrown straight up, the drag force of air acts against its motion. As the ball ascends, the drag force opposes its upward velocity, causing a reduction in the net force acting on the ball. This results in a slower upward acceleration and a longer time to reach the top of its motion.
On the descent, the gravitational force aids the motion, overcoming the drag force. The downward velocity increases, and the time taken to fall back down is less than the time taken to ascend.
In the absence of air resistance, the ascent and descent times would be equal. However, with air resistance considered, the ascent time is prolonged, impacting the overall motion of the ball.
What is the volumetric flow rate if water is flowing at 3m/s through a 10cm diameter pipe?
Answer:
Volumetric flow rate = 0.024 m³/s
Explanation:
Volumetric flow rate = Discharge = Q
We have expression for discharge, Q = Area ( A) x Velocity (v)
Velocity , v = 3 m/s
[tex]\texttt{Area, A}=\frac{\pi d^2}{4}=\frac{\pi \times (10\times 10^{-2})^2}{4}=7.85\times 10^{-3}m^2[/tex]
Substituting
Discharge, Q =Av = 7.85 x 10⁻³ x 3 = 0.024 m³/s
Volumetric flow rate = 0.024 m³/s
A uniform steel plate has an area of 0.819 m2. When subjected to a temperature difference between its sides, a heat current* of 31700 W is found to flow through it. What is the temperature gradient? What is the temperature difference when the plate is 0.0475 m thick? The thermal conductivity of steel is 50.2 W/(m·K).
Answer:
ΔT / Δx = 771 K/m
ΔT = 771 x 0.0475 = 36.62 k
Explanation:
P = 31700 W, A = 0.819 m^2, Δx = 0.0475 m, K = 50.2 W /m k
Use the formula of conduction of heat
H / t = K A x ΔT / Δx
So, ΔT / Δx = P / K A
ΔT / Δx = 31700 / (50.2 x 0.819)
ΔT / Δx = 771 K/m
Now
ΔT = 771 x 0.0475 = 36.62 k
A particle with an initial velocity of 50 m slows at a constant acceleration to 20 ms-1 over a distance of 105 m. How long does it take for the particle to slow down? (a) 2 s (c) 3 s (b)4 s (d)5 s
Answer:
Time taken, t = 3 s
Explanation:
It is given that,
Initial velocity of the particle, u = 50 m/s
Final velocity, v = 20 m/s
Distance covered, s = 105 m
Firstly we will find the acceleration of the particle. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2as[/tex]
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
[tex]a=\dfrac{(20\ m/s)^2-(50\ m/s)^2}{2\times 105\ m}[/tex]
[tex]a=-10\ m/s^2[/tex]
So, the particle is decelerating at the rate of 10 m/s². Let t is the time taken for the particle to slow down. Using first equation of motion as :
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{20\ m/s-50\ m/s}{-10\ m/s^2}[/tex]
t = 3 s
So, the time taken for the particle to slow down is 3 s. Hence, this is the required solution.
A parallel plate capacitor of area A = 30 cm2 and separation d = 5 mm is charged by a battery of 60-V. If the air between the plates is replaced by a dielectric of κ = 4 with the battery still connected, then what is the ratio of the initial charge on the plates divided by the final charge on the plates?
Answer:
0.25
Explanation:
A = area of each plate = 30 cm² = 30 x 10⁻⁴ m²
d = separation between the plates = 5 mm = 5 x 10⁻³ m
[tex]C_{air}[/tex] = Capacitance of capacitor when there is air between the plates
k = dielectric constant = 4
[tex]C_{dielectric}[/tex] = Capacitance of capacitor when there is dielectric between the plates
Capacitance of capacitor when there is air between the plates is given as
[tex]C_{air} = \frac{\epsilon _{o}A}{d}[/tex] eq-1
Capacitance of capacitor when there is dielectric between the plates is given as
[tex]C_{dielectric} = \frac{k \epsilon _{o}A}{d}[/tex] eq-2
Dividing eq-1 by eq-2
[tex]\frac{C_{air}}{C_{dielectric}}=\frac{\frac{\epsilon _{o}A}{d}}{\frac{k \epsilon _{o}A}{d}}[/tex]
[tex]\frac{C_{air}}{C_{dielectric}}=\frac{1}{k}[/tex]
[tex]\frac{C_{air}}{C_{dielectric}}=\frac{1}{4}[/tex]
[tex]\frac{C_{air}}{C_{dielectric}}=0.25[/tex]
Charge stored in the capacitor when there is air is given as
[tex]Q_{air}=C_{air}V[/tex] eq-3
Charge stored in the capacitor when there is dielectric is given as
[tex]Q_{dielectric}=C_{dielectric}V[/tex] eq-4
Dividing eq-3 by eq-4
[tex]\frac{Q_{air}}{Q_{dielectric}}=\frac{C_{air}V}{C_{dielectric} V}[/tex]
[tex]\frac{Q_{air}}{Q_{dielectric}}=\frac{C_{air}}{C_{dielectric}}[/tex]
[tex]\frac{Q_{air}}{Q_{dielectric}}=0.25[/tex]
A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed 14.0 m/s, the total horizontal force on the driver has magnitude 130 N. What is the total horizontal force on the driver if the speed on the same curve is 18.0 m/s instead?
Explanation:
It is given that,
Initial speed, v₁ = 14 m/s
Initial force, F₁ = 130 N
We need to find the total horizontal force (F₂) on the driver if the speed on the same curve is 18.0 m/s instead, v₂ = 18 m/s
The centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]\dfrac{F_1}{F_2}=\dfrac{mv_1^2/r}{mv_2^2/r}[/tex]
[tex]\dfrac{F_1}{F_2}=\dfrac{v_1^2}{v_2^2}[/tex]
[tex]F_2=\dfrac{v_2^2\times F_1}{v_1^2}[/tex]
[tex]F_2=\dfrac{(18\ m/s)^2\times 130\ N}{(14\ m/s)^2}[/tex]
[tex]F_2=214.8\ N[/tex]
So, if the speed is 18 m/s, then the horizontal force acting on the car is 214.8 N. Hence, this is the required solution.
Final answer:
The total horizontal force on the driver when the car's speed increases to 18.0 m/s can be found using the proportionality between the centripetal force and the square of the velocity. By comparing the initial and final conditions, we can calculate the new total horizontal force required for the increased speed.
Explanation:
The question involves finding the total horizontal force on a driver when the speed of the car increases on the same curve. The horizontal force experiences by the driver in a curved path, which is part of a horizontal circle, can be determined by using the formula for centripetal force (Fc = mv²/r). We know from the initial condition that the horizontal force is 130 N when the car is traveling at 14.0 m/s. Since the force depends on the square of the velocity, we can use proportions to find the new force when the speed increases to 18.0 m/s.
Thus, (F2/F1) = (v₂²/v₁²). Substituting the given values, we get (F2/130 N) = (18.0 m/s)² / (14.0 m/s)². Calculating this provides us with the new force F2 which represents the total horizontal force on the driver at the higher speed of 18.0 m/s.
In the Olympic shotput event, an athlete throws the shot with an initial speed of 13.0 m/s at a 47.0° angle from the horizontal. The shot leaves her hand at a height of 1.80 m above the ground. How far does the shot travel?
Answer:
18.7 m
Explanation:
[tex]v_{o}[/tex] = initial speed of the shot = 13 m/s
θ = angle of launch from the horizontal = 47 deg
Consider the motion along the vertical direction
[tex]v_{oy}[/tex] = initial velocity along vertical direction = 13 Sin47 = 9.5 m/s
[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²
y = vertical displacement = - 1.80 m
t = time of travel
Using the kinematics equation
[tex]y=v_{oy} t+(0.5)a_{y} t^{2}[/tex]
- 1.80 = (9.5) t + (0.5) (- 9.8) t²
t = 2.11 s
Consider the motion along the horizontal direction
x = horizontal displacement of the shot
[tex]v_{ox}[/tex] = initial velocity along horizontal direction = 13 Cos47 = 8.87 m/s
[tex]a_{x}[/tex] = acceleration along the horizontal direction = 0 m/s²
t = time of travel = 2.11 s
Using the kinematics equation
[tex]x=v_{ox} t+(0.5)a_{x} t^{2}[/tex]
x = (8.87) (2.11) + (0.5) (0) (2.11)²
x = 18.7 m
The horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.
What is projectile motion?A projectile motion is a curved motion that is projected near the surface of the earth such that it falls under the influence of gravity only.
Given to us
The initial velocity of the shotput, u = 13 m/s
The angle from the horizontal, θ = 47.0°
The height from which shotput was thrown h = 1.80 m
As we know that when the shotput will be thrown it will be in a projectile motion, therefore, the distance that will be covered by the shot put will be the range of the projectile motion,
[tex]R = \dfrac{u\ cos\theta}{g} [u\cdot sin\theta + \sqrt{u^2sin^2\theta+2gh}][/tex]
Substitute the values,
[tex]R = \dfrac{13\ cos47^o}{9.81} [13\cdot sin47 + \sqrt{13^2(sin47)^2+2(9.81)(1.80)}][/tex]
R = 18.7 m
Hence, the horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.
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A vertical straight wire carrying an upward 28-A current exerts an attractive force per unit length of 7.83 X 10 N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?
Answer:
[tex]i_2 = 978750 A[/tex]
Since the force between wires is attraction type of force so current must be flowing in upward direction
Explanation:
Force per unit length between two current carrying wires is given by the formula
[tex]F = \frac{\mu_0 i_1 i_2}{2 \pi d}[/tex]
here we know that
[tex]F = 7.83 \times 10 N/m[/tex]
[tex]d = 7.0 cm = 0.07 m[/tex]
[tex]i_1 = 28 A[/tex]
now we will have
[tex]F = \frac{4\pi \times 10^{-7} (28.0)(i_2)}{2\pi (0.07)}[/tex]
[tex]7.83 \times 10 = \frac{2\times 10^{-7} (28 A)(i_2)}{0.07}[/tex]
[tex]i_2 = 978750 A[/tex]
Since the force between wires is attraction type of force so current must be flowing in upward direction
An adventurous archeologist (m = 80.5 kg) tries to cross a river by swinging from a vine. The vine is 11.0 m long, and his speed at the bottom of the swing is 7.80 m/s. The archeologist doesn't know that the vine has a breaking strength of 1,000 N. Does he make it safely across the river without falling in?
Answer:
He made it, he across safely the river without falling ing.
Explanation:
Vt= 7.8 m/s
r= 11m
m= 80.5 kg
Vt= ω * r
ω= 0.71 rad/s
ac= ω² * r
ac= 5.54 m/s²
F= m * ac
F= 445.97 N < 1,000 N
Convert: Thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C). Surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m20C)
To convert the thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C), multiply by 694.7. To convert the surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m2 °C), multiply by 11,545.
Explanation:To convert the thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C), we can use the following conversion factors:
1 Btu = 1055.06 J1 ft = 0.3048 m1 °F = 0.5556 °CUsing these conversion factors, we can convert the thermal conductivity value as follows:
0.3 Btu/(h ft°F) = (0.3 * 1055.06 J)/(h * 0.3048 m * 0.5556 °C) = 694.7 W/(m °C)
Similarly, to convert the surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m2 °C), we can use the following conversion factors:
1 ft2 = 0.0929 m21 °F = 0.5556 °CUsing these conversion factors, we can convert the surface heat transfer coefficient value as follows:
105 Btu/(h ft2 oF) = (105 * 1055.06 J)/(h * 0.0929 m2 * 0.5556 °C) = 11,545 W/(m2 °C)
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The conversion of the thermal conductivity value yields approximately 0.52 W/(m°C), and the surface heat transfer coefficient conversion yields approximately 596.15 W/(m²°C). Conversion factors for BTU to J, ft to m, and °F to °C are critical in these calculations.
To convert thermal conductivity from BTU/(h ft°F) to W/(m°C):
Given value: 0.3 BTU/(h ft°F).Use conversion factors: 1 BTU = 1055.06 J, 1 ft = 0.3048 m, 1 °F = (5/9) °C.Calculation: 0.3 BTU/(h ft°F) * (1055.06 J/BTU) * (1/3600 s/h) * (1/0.3048 m) * (9/5 °C/°F).Simplify: 0.3 * (1055.06/3600) * (1/0.3048) * (9/5) ≈ 0.52 W/(m°C).To convert surface heat transfer coefficient from BTU/(h ft²°F) to W/(m²°C):
Given value: 105 BTU/(h ft²°F).Use conversion factors: 1 BTU = 1055.06 J, 1 ft = 0.3048 m, 1 °F = (5/9) °C.Calculation: 105 BTU/(h ft²°F) * (1055.06 J/BTU) * (1/3600 s/h) * (1/0.3048² m²) * (9/5 °C/°F).Simplify: 105 * (1055.06/3600) * (1/0.3048²) * (9/5) ≈ 596.15 W/(m²°C).At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constant speed. At time t2 = 5.00 s, its acceleration is 4.00 i - 6.00 j m/s2 . What is the radius of the path taken by the particle if t2-t1 is less than one period?
The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.
At time [tex]t_1[/tex], the acceleration vector has direction [tex]\theta_1[/tex] such that
[tex]\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ[/tex]
which indicates the particle is situated at a point on the lower left half of the circle, while at time [tex]t_2[/tex] the acceleration has direction [tex]\theta_2[/tex] such that
[tex]\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ[/tex]
which indicates the particle lies on the upper left half of the circle.
Notice that [tex]\theta_1-\theta_2=90^\circ[/tex]. That is, the measure of the major arc between the particle's positions at [tex]t_1[/tex] and [tex]t_2[/tex] is 270 degrees, which means that [tex]t_2-t_1[/tex] is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.
Recall that
[tex]\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}[/tex]
where [tex]R[/tex] is the radius of the circle and [tex]T[/tex] is the period. We have
[tex]t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s[/tex]
and the magnitude of the particle's acceleration toward the center of the circle is
[tex]\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}[/tex]
So we find that the path has a radius [tex]R[/tex] of
[tex]7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}[/tex]
The acceleration of a particle moving in a circular path with constant speed is given by the centripetal acceleration. Since the velocity is not given, the radius cannot be determined with this information alone.
Explanation:In circular motion, the acceleration of a particle is given by the centripetal acceleration, which is equal to the square of the speed divided by the radius of the circular path (v^2/r). As the particle moves at a constant speed, we know that the magnitude of acceleration (which gives us the centripetal acceleration) remains the same at both times t1 and t2. Therefore, we can calculate the magnitude of acceleration at both points and set them equal to each other. This will allow us to solve for the radius of the path.
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A point charge Q is located a distance d away from the center of a very long charged wire. The wire has length L >> d and total charge q. What force does the wire experience?
Answer:
[tex]F = \frac{Qq}{2\pi \epsilon_0 L d}[/tex]
Explanation:
As we know that if a charge q is distributed uniformly on the line then its linear charge density is given by
[tex]\lambda = \frac{q}{L}[/tex]
now the electric field due to long line charge at a distance d from it is given as
[tex]E = \frac{2k\lambda}{d}[/tex]
[tex]E = \frac{q}{2\pi \epsilon_0 d}[/tex]
now the force on the other charge in this electric field is given as
[tex]F = QE[/tex]
[tex]F = \frac{Qq}{2\pi \epsilon_0 L d}[/tex]
If a 2 x 10^-4C test charge is given 6.5J of energy, determine the electric potential difference.
Answer:
The electric potential difference is 32500 volt.
Explanation:
Given that,
Charge[tex]q=2\times10^{-4}C[/tex]
Energy = 6.5 J
We need to calculate the electric potential difference
Potential difference :
Potential difference is equal to the energy divide by charge.
Using formula of potential difference
[tex]V=\dfrac{E}{Q}[/tex]
[tex]V=\dfrac{6.5}{2\times10^{-4}}[/tex]
[tex]V=32500\ volt[/tex]
Hence, The electric potential difference is 32500 volt.
The electric potential difference is 32,500 volts.
Explanation:To determine the electric potential difference, we can use the formula:
V = ΔE / q
where V is the electric potential difference, ΔE is the energy and q is the test charge. In this case, the energy given is 6.5J and the test charge is 2 x 10-4C. Plugging these values into the formula, we get:
V = 6.5J / (2 x 10-4C)
V = 32,500V
Therefore, the electric potential difference is 32,500 volts.
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In the pair of supply and demand equations below, where x represents the quantity demanded in units of a thousand and p the unit price in dollars, find the equilibrium quantity and the equilibrium price. p = −x2 − 3x + 80 and p = 7x + 5
In this exercise we have to use the knowledge of finance to calculate the equilibrium value and the quantity, so we have:
Equilibrium quantity = 5Equilibrium price = 40Organizing the information given in the statement we have:
p = -x²-3x+80p = 7x+5So equating the two given equations we have:
[tex]-x^2 - 3x + 80 = 7x+5\\x^2 +3x + 7x + 5 - 80 = 0\\x^2 + 10x - 75 = 0\\x^2- 5x + 15x -75 = 0\\x(x-5) + 15(x-5) = 0[/tex]
So we can see that the roots will be x = 5 and x = -15 since the quantity cannot be in negative therefore, the equilibrium quantity will be = 5 So replace that value at:
[tex]p = -(5)^2-3(5) + 80 = 40\\p = 7(5) + 5 = 40[/tex]
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A certain part of an aircraft engine has a volume of 1.4 ft^3. (a) Find the weight of the piece when it is made of lead. (b) If the same piece is made of aluminum, what is its weight? Determine how much weight is saved by using aluminum instead of lead.
Final answer:
The weight of the piece when made of lead is 992.6 lb. The weight of the piece when made of aluminum is 236.6 lb. By using aluminum instead of lead, a weight of 756 lb is saved.
Explanation:
(a) Weight of the piece when it is made of lead:
To find the weight of the piece when it is made of lead, we need to know the density of lead. The density of lead is approximately 709 lb/ft³. We can use the formula:
Weight = Density x Volume
Given that the volume of the piece is 1.4 ft³, the weight of the piece when it is made of lead can be calculated as:
Weight = 709 lb/ft³ x 1.4 ft³ = 992.6 lb
(b) Weight of the piece when made of aluminum, and weight saved:
To find the weight of the piece when it is made of aluminum, we need to know the density of aluminum. The density of aluminum is approximately 169 lb/ft³. Using the same formula as before:
Weight = Density x Volume
Given that the volume of the piece is still 1.4 ft³, the weight of the piece when it is made of aluminum can be calculated as:
Weight = 169 lb/ft³ x 1.4 ft³ = 236.6 lb
The weight saved by using aluminum instead of lead can be determined by subtracting the weight of the aluminum piece from the weight of the lead piece:
Weight saved = Weight of lead piece - Weight of aluminum piece
Weight saved = 992.6 lb - 236.6 lb = 756 lb
Two resistors, the first 12 ? and the second 6 2, are connected in parallel to a 48 V battery What is the power dissipated by each resistor? A) P1 144 W, P2-288 W B) P1-162 W, P2-324 W C) P1-180 W, P2-360 W D) P1-192 W, P2 384 W
Answer:
P₁ = 192 W and P₂ = 384 W
Explanation:
It is given that,
Resistor 1, [tex]R_1=12\ \Omega[/tex]
Resistor 2, [tex]R_1=6\ \Omega[/tex]
Voltage, V = 48 V
Power dissipated by resistor 1 is given by :
[tex]P_1=\dfrac{V^2}{R_1}[/tex]
[tex]P_1=\dfrac{(48)^2}{12}[/tex]
P₁ = 192 watts
Power dissipated by resistor 2 is given by :
[tex]P_2=\dfrac{V^2}{R_1}[/tex]
[tex]P_2=\dfrac{(48)^2}{6}[/tex]
P₂ = 384 watts
So, the power dissipated by both the resistors is 192 watts and 384 watts respectively. Hence, this is the required solution.
A rock is dropped from the top of a tower. When it is 40 meters above the ground velocity of 17 m/s. When its velocity is 24 m/s, how far is it above the ground? Select one: O a. 29.4 meters o b. 25.4 meters o c. 33.3 meters d. 10.7 meters O e. 147 meters
Answer:
Option B is the correct answer.
Explanation:
Let us consider 40 meter above ground as origin.
Initial velocity = 17 m/s
Final velocity = 24 m/s
Acceleration = 9.81 m/s
We have equation of motion v² = u² + 2as
Substituting
24² = 17² + 2 x 9.81 x s
s = 14.63 m
Distance traveled by rock = 14.63 m down.
Height of rock from ground = 40 - 14.63 = 25.37 m = 25.4 m
Option B is the correct answer.
ver shines light up to the surface of a flat glass-bottomed boat at an angle of 30 relative to the normal. If the index of refraction of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its n
Answer:
[tex]\beta = 41.68°[/tex]
Explanation:
according to snell's law
[tex]\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }[/tex]
refractive index of water n_w is 1.33
refractive index of glass n_g is 1.5
[tex]sin\alpha = \frac{n_w}{n_g}* sin30[/tex]
[tex]sin\alpha = 0.443[/tex]
now applying snell's law between air and glass, so we have
[tex]\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}[/tex]
[tex]sin\beta = \frac{n_g}{n_a} sin\alpha[/tex]
[tex]\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha][/tex]
we know that [tex]sin\alpha = 0.443[/tex]
[tex]\beta = 41.68°[/tex]
A 10-g bullet moving horizontally with a speed of 1.8 km/s strikes and passes through a 5.0-kg block initially at rest on a horizontal frictionless surface. The bullet emerges from the block with a speed of 1.0 km/s. What is the kinetic energy of the block immediately after the bullet emerges? a. 8.0.J 6.4 J 5.3 J 9.4 J 10 J e.
Answer:
6.4 J
Explanation:
m = mass of the bullet = 10 g = 0.010 kg
v = initial velocity of bullet before collision = 1.8 km/s = 1800 m/s
v' = final velocity of the bullet after collision = 1 km/s = 1000 m/s
M = mass of the block = 5 kg
V = initial velocity of block before collision = 0 m/s
V' = final velocity of the block after collision = ?
Using conservation of momentum
mv + MV = mv' + MV'
(0.010) (1800) + (5) (0) = (0.010) (1000) + (5) V'
V' = 1.6 m/s
Kinetic energy of the block after the collision is given as
KE = (0.5) M V'²
KE = (0.5) (5) (1.6)²
KE = 6.4 J
Final answer:
Using the conservation of momentum, and given that the bullet exits the wooden block with a lower speed, we calculate the block's final speed to be 7.5 m/s after the bullet exits.
Explanation:
To solve this problem, we need to apply the law of conservation of momentum. The law states that if no external forces act on a system, the total momentum of the system remains constant, even though individual objects within the system might change velocities. In this case, the system is made up of the bullet and the wooden block. Before the collision, the block is at rest, so its initial momentum is 0. The bullet’s initial momentum is its mass times its velocity (momentum = mass × velocity). After the collision, the bullet exits the block with a lower speed, indicating that some of its momentum has been transferred to the block.
Step 1: Calculate the initial momentum of the bullet:
Initial momentum of the bullet = mass of bullet × initial velocity of bullet
= 0.05 kg × 500 m/s
= 25 kg·m/s
Step 2: Calculate the final momentum of the bullet:
Final momentum of the bullet = mass of bullet × final velocity of bullet
= 0.05 kg × 200 m/s
= 10 kg·m/s
Step 3: Calculate the change in momentum of the bullet:
Change in momentum = Initial momentum - Final momentum
Change in momentum = 25 kg·m/s - 10 kg·m/s
= 15 kg·m/s
Step 4: Calculate the final momentum of the block:
Since the block was initially at rest, its final momentum is equal to the change in momentum of the bullet (due to conservation of momentum). Therefore, the final momentum of the block is 15 kg·m/s.
Step 5: Calculate the final velocity of the block:
Using the final momentum and the mass of the block, we can find the final velocity:
Final velocity of the block = Final momentum of block / mass of block
= 15 kg·m/s / 2 kg
= 7.5 m/s
Thus, the speed of the block after the bullet has come out the other side is 7.5 m/s.
A baseball approaches home plate at a speed of 43.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 56.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 1.90 ms. What is the average vector force the ball exerts on the bat during their interaction? (Let the +x-direction be in the initial direction of motion, and the +y-direction be up.)
Answer:
(4273.7 [tex]\hat{j}[/tex] - 3281.6 [tex]\hat{i}[/tex])
Explanation:
[tex]\underset{v_{i}}{\rightarrow}[/tex] = initial velocity of the baseball before collision = 43 [tex]\hat{i}[/tex] m/s
[tex]\underset{v_{f}}{\rightarrow}[/tex] = final velocity of the baseball after collision = 56 [tex]\hat{j}[/tex] m/s
m = mass of the ball = 145 g = 0.145 kg
t = time of contact of the ball with the bat = 1.90 ms = 0.0019 s
[tex]\underset{F_{avg}}{\rightarrow}[/tex] = Average force vector
Using Impulse-change in momentum equation
[tex]\underset{F_{avg}}{\rightarrow}[/tex] t = m ([tex]\underset{v_{f}}{\rightarrow}[/tex] - [tex]\underset{v_{i}}{\rightarrow}[/tex] )
[tex]\underset{F_{avg}}{\rightarrow}[/tex] (0.0019) = (0.145) (56 [tex]\hat{j}[/tex] - 43 [tex]\hat{i}[/tex])
[tex]\underset{F_{avg}}{\rightarrow}[/tex] = (4273.7 [tex]\hat{j}[/tex] - 3281.6 [tex]\hat{i}[/tex])
A diver can reduce her moment of inertia by a factor of about 3.5 when changing from the straight position to the tuck position. If she makes two rotations in 1.5 seconds when in the tuck position, what is her angular speed (rad/sec) when in the straight position?
Answer:
29.3 rad/s
Explanation:
Moment of inertia in straight position, I1 = I
He takes 2 rotations in 1.5 second
So, Time period, T = 1.5 / 2 = 0.75 second
w1 = 2 x 3.14 / T = 2 x 3.14 / 0.75 = 8.37 rad/s
Moment of inertia in tucked position, I2 = I / 3.5
Let the new angular speed is w2.
By the use of conservation of angular momentum, if no external torque is applied, then angular momentum is constant.
L1 = L2
I1 w1 = I2 w2
I x 8.37 = I / 3.5 x w2
w2 = 29.3 rad/s
To find the angular speed of the diver in the straight position, we can use the conservation of angular momentum. The moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position. By applying the conservation of angular momentum equation, we can calculate the angular speed in the straight position.
Explanation:To find the angular speed of the diver in the straight position, we can use the conservation of angular momentum. According to the conservation of angular momentum, the product of the initial moment of inertia and initial angular speed is equal to the product of the final moment of inertia and final angular speed.
Let's assume the initial angular speed in the tuck position is denoted by w' and the final angular speed in the straight position is denoted by w. We are given that the moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position. Therefore, the final moment of inertia (I) is 3.5 times greater than the initial moment of inertia (I').
The conservation of angular momentum equation can be written as:
I' * w' = I * w
Since the moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position, we have:
w = w' * (I' / I)
Using the given values, we can calculate the angular speed in the straight position.
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