Answer:
The speed when it strikes the ground below is V= 157.64 m/s < -56.92º .
Explanation:
V= 150m/s
α= 55º
hi= 120m
Vy= V*sinα
Vy= 122.87 m/s
Vx= V * cos α
Vx= 86.03 m/s
h= hi + Vy * t - g*t²/2
clearing t we get the total flying time of the projectile:
t(total fly)= 26.01 sec
0= Vy - g*t
clearing t we get the maximum height time:
t(max height)= 12.53 sec
to get the fall time:
t(fall)= t(total fly) - t(max height)
t(fall)= 13.48 sec
Vy'= g* t(fall)
Vy'= 132.1 m/s
V'= √(Vx² +Vy'²)
V'= 157.64 m/s
α'= tg⁻¹ (Vy'/Vx)
α'= -56.92º
By using the conservation of energy principle, we find the speed of the projectile when it hits the ground to be roughly 165 m/s (rounded to 3 significant figures).
Explanation:
Given that a projectile is fired upward at an angle of 55° from the top of a 120 m cliff at a speed of 150 m/s, it is asked what its speed will be when it strikes the ground. To answer such a question, we employ the principle of conservation of energy, which states that the total mechanical energy (kinetic energy + potential energy) of an isolated system remains constant if non-conservative forces like air resistance are negligible.
Now, the energy of the projectile at the top of the cliff is equal to its kinetic energy (as it's launched) and its potential energy (due to its height). When the projectile hits the ground, all its potential energy will be converted into kinetic energy as the body has descended from the height, which the body will possess as its speed.
Let's calculate:
Initial Energy = Final Energy
1/2 * mass * (initial speed)² + mass * g * height = 1/2 * mass * (final speed)²
Solving for final speed, -(initial speed)² - 2gh = - (final speed)²
finalSpeed = sqrt((initial speed)² + 2*g*h)
Given that the initial speed = 150m/s, g (acceleration due to gravity) = 9.81m/s², and the height = 120m, the final speed would amount to 164.833 m/s. Therefore, when rounding to 3 significant figures, the speed when it hits the ground will be 165 m/s.
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An electron is to be accelerated in a uniform electric field having a strength of 2.00 × 106 V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?
Answer:
a) 800 keV
b) 25 km
Explanation:
[tex]Strength\ of\ Electric\ field=2\times 10^6\ V/m\\a)\ Potential\ Difference=Strength\ of\ Electric\ field\times Distance\\\Rightarrow Potential\ Difference=Kinetic\ Energy\ =2\times 10^6\times 0.4\\\therefore Energy=0.8\times 10^6\ eV=800\ keV\\[/tex]
[tex]b)\ Potential\ difference=50\ GeV=50\times 10^9\ eV\\Distance=\frac{Potential\ difference}{electric\ field}\\\Rightarrow Distance=\frac{50\times 10^9}{2\times 10^6}\\\Rightarrow Distance=25\times 10^3\ m\\\therefore Distance=25\ km[/tex]
Answer:
a) 800 keV
b) 24.996 km.
Explanation:
(a) we have
[tex]\large \Delta K.E=q\Delta V[/tex] .............(1)
where,
[tex]\large \Delta K.E[/tex] = Change in kinetic energy
[tex]q[/tex] = charge of an electron
[tex]\Delta V[/tex] = Potential difference
also
[tex]\large E=\frac{V}{d}[/tex] .......(2)
E = electric field
d = distance traveled
Now from (1) and (2) we have,
[tex]\large \Delta K. E=qV=qEd[/tex]
substituting the values in the above equation, we get
[tex]\large \Delta K. E=(1.6\times 10^{-19}C)(2\times 10^6V/m)(0.400m)(\frac{1eV}{1.6\times 10^{-19}J})(\frac{1keV}{1000eV})[/tex]
[tex]\large \Delta K. E=800keV[/tex]
Thus, the energy gained by the electron is 800 keV if it is accelerated over a distance of 0.400 m.
(b) Using the equation (1), we have
[tex]\large d=\frac{\Delta K.E}{qE}[/tex]
[tex]\large d=\frac{(50\times 10^9eV)}{(1.6\times 10^{-19C})(2\times 10^6V/m)}(\frac{1.6\times 10^{-19}J}{1eV})[/tex]
or
[tex]\large d=2.4996\times 10^4m[/tex]
or
[tex]\large d=24.996\times 10^3m=24.996km[/tex]
Thus, to gain 50.0 GeV of energy the electron must be accelerated over a distance of 24.996 km.
A marble rolls off the edge of a table top with a speed of 2.00 m/s. a.) What is the magnitude of its velocity 0.100 s later? b.) How far from the table does it land? The height of the table is 1.00m.
Answer:
(a) 2.23 m/s
(b) 0.9 m
Explanation:
h = 1 m, t = 0.1 second
horizontal component of initial velocity, ux = 2 m/s
vertical component of initial velocity, uy = 0
(a) Let v be the velocity after 0.1 seconds. Its vertical component is vy and horizontal component is vx.
The horizontal component of velocity remains constant as in this direction, acceleration is zero.
vx = ux = 2 m/s
Use first equation of motion in Y axis direction.
vy = uy + g t
vy = 0 + 9.8 x 0.1 = 0.98 m/s
Resultant velocity after 0.1 second
v^2 = vx^2 + vy^2
v^2 = 2^2 + 0.98^2
v = 2.23 m/s
(b) Let it takes time t to land.
Use second equation of motion along Y axis
h = uy t + 1/2 g t^2
1 = 0 + 1/2 x 9.8 x t^2
t = 0.45 second
Let it lan at a distance x.
so, x = ux x t
x = 2 x 0.45 = 0.9 m
The force between two electrical charges is 0.3 N. What would be the force if the charges are doubled, and the distance between them increased by 100%?
(Please show clear steps/explanation)
Answer:
The new force F' will be same of the original force F.
Explanation:
Given that,
Charges = 0.3
We need to calculate the force between the charges
Suppose that the distance between the charges is r.
The force between the charges
[tex]F =\dfrac{kq_{1}q_{2}}{r^2}[/tex]
Put the value into the formula
[tex]F=\dfrac{k\times0.3\times0.3}{r^2}[/tex]
[tex]F=\dfrac{k\times(0.09)}{r^2}[/tex]
If the charges are doubled, and the distance between them increased by 100%.
So, The charges are 0.6 and the distance is 2r.
Then,
The force between the charges
[tex]F'=\dfrac{k\times0.6\times0.6}{(2r)^2}[/tex]
[tex]F'=\dfrac{k\times0.09}{r^2}[/tex]
[tex]F'=F[/tex]
Hence, The new force F' will be same of the original force F.
Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lightweight material of density 600 kg/m3. (Assume that acceleration due to gravity is g = 9.8 m/s2. Round your answer to one decimal place.)
Answer:
Work done = 35467.278 J
Explanation:
Given:
Height of the cone = 4m
radius (r) of the cone = 1.2m
Density of the cone = 600kg/m³
Acceleration due to gravity, g = 9.8 m/s²
Now,
The total mass of the cone (m) = Density of the cone × volume of the cone
Volume of the cone = [tex]\frac{1}{3}\pi r^2 h[/tex]
thus,
volume of the cone = [tex]\frac{1}{3}\pi 1.2^2\times 4[/tex] = 6.03 m³
therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg
The center of mass for the cone lies at the [tex]\frac{1}{4}[/tex]times the total height
thus,
center of mass lies at, h' = [tex]\frac{1}{4}\times4=1m[/tex]
Now, the work gone (W) against gravity is given as:
W = mgh'
W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J
The work against gravity required to build a cone of height 4 m and base of radius 1.2 m out of a material of density 600 kg/m3 is 35,467.278 J.
Given to us
Height of the cone = 4m
The radius of the cone = 1.2 m
Density of the material = 600 kg/m³
We know that the work gone against the gravity is given as,
[tex]W = mgh'[/tex]
where W is the work, m is the mass, g is the acceleration due to gravity, and h' is the center of mass.
Also, the mass of an object is the product of its volume and density, therefore,
[tex]m = v \times \rho[/tex]
We know the value of the volume of a cone,
[tex]m = \dfrac{1}{3}\pi r^2 h \times \rho[/tex]
The center of mass of a cone lies at the center of its base at 1/4 of the total height from the base,
[tex]h' = \dfrac{h}{4}[/tex]
Substitute all the values we will get,
[tex]W = mgh'\\\\W = (v \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi r^2h \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi (1.2)^2 \times4 \times 600) \times 9.81 \times \dfrac{4}{4}\\\\W = 35467.278\rm\ J[/tex]
Hence, the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lightweight material of density 600 kg/m3 is 35,467.278 J.
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An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s1045 rad/s ). If a particular disk is spun at 646.1 rad/s646.1 rad/s while it is being read, and then is allowed to come to rest over 0.234 seconds0.234 seconds , what is the magnitude of the average angular acceleration of the disk?
Answer:
The magnitude of the average angular acceleration of the disk is 2761.1 rad/s².
Explanation:
Given that,
Angular velocity of optical disk= 1045 rad/s
Angular velocity of particular disk = 646.1 rad/s
Time = 0.234 sec
We need to calculate the average angular acceleration
Using formula of angular acceleration
[tex]\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}[/tex]
Where, [tex]\omega_{f}[/tex] = final angular velocity
[tex]\omega_{i}[/tex] = Initial angular velocity
t = time
Put the value in the equation
[tex]\alpha = \dfrac{0-646.1}{0.234}[/tex]
[tex]\alpha= -2761.1\ rad/s^2[/tex]
Negative sign shows the angular deceleration.
Hence, The magnitude of the average angular acceleration of the disk is 2761.1 rad/s².
A flowerpot falls off a windowsill and passes the win- dow of the story below. Ignore air resistance. It takes the pot 0.380 s to pass from the top to the bottom of this window, which is 1.90 m high. How far is the top of the window below the window- sill from which the flowerpot fell?
Answer:
d = 0.50 m
Explanation:
Let say the speed at the top and bottom of the window is
[tex]v_1 \: and \: v_2[/tex] respectively
now we have
[tex]d = \frac{v_1 + v_2}{2}t[/tex]
[tex]1.90 = \frac{v_1 + v_2}{2} (0.380)[/tex]
[tex]v_1 + v_2 = 10 [/tex]
also we know that
[tex]v_2 - v_1 = 9.8(0.380)[/tex]
[tex]v_2 - v_1 = 3.72[/tex]
now we have from above equations
[tex]v_2 = 6.86 m/s[/tex]
[tex]v_1 = 3.14 m/s[/tex]
now the distance from which it fall down is given as
[tex]v_f^2 - v_i^2 = 2ad[/tex]
[tex]3.14^2 - 0^2 = 2(9.8)d[/tex]
[tex]d = 0.50 m[/tex]
horizontal block–spring system with the block on a frictionless surface has total mechanical energy E 5 47.0 J and a maximum displacement from equilibrium of 0.240 m. (a) What is the spring constant? (b) What is the kinetic energy of the system at the equilibrium point? (c) If the maximum speed of the block is 3.45 m/s, what is its mass? (d) What is the speed of the block when its displacement is 0.160 m? (e) Find the kinetic energy of the block at x 5 0.160 m. (f) Find the potential energy stored in the spring when x 5 0.160 m. (g) Suppose the same system is released from rest at x 5 0.240 m on a rough surface so that it loses
Answer:
Part a)
[tex]k = 1632 J[/tex]
Part b)
[tex]KE = 47 J[/tex]
Part c)
[tex]m = 7.9 kg[/tex]
Part d)
[tex]v = 2.57 m/s[/tex]
Part e)
[tex]KE = 26.1 J[/tex]
Part f)
[tex]PE = 20.9 J[/tex]
Explanation:
Total Mechanical energy is given as
[tex]E = 47.0 J[/tex]
Its maximum displacement from mean position is given as
[tex]A = 0.240 m[/tex]
Part a)
Now from the formula of energy we know that
[tex]E = \frac{1}{2}kA^2[/tex]
[tex]47 = \frac{1}{2}k(0.240)^2[/tex]
[tex]k = 1632 J[/tex]
Part b)
At the mean position of SHM whole mechanical energy will convert into kinetic energy
so it is given as
[tex]KE = 47 J[/tex]
Part c)
As per the formula of kinetic energy we know that
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]47 = \frac{1}{2}m(3.45^2)[/tex]
[tex]m = 7.9 kg[/tex]
Part d)
As we know by the equation of the speed of SHM is given as
[tex]v = \sqrt{\frac{k}{m}(A^2 - x^2)}[/tex]
[tex]v = \sqrt{\frac{1632}{7.9}(0.24^2 - 0.16^2)}[/tex]
[tex]v = 2.57 m/s[/tex]
Part e)
As we know by the formula of kinetic energy
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}(7.9)(2.57^2)[/tex]
[tex]KE = 26.1 J[/tex]
Part f)
As per energy conservation we know
KE + PE = Total energy
[tex]26.1 + PE = 47[/tex]
[tex]PE = 20.9 J[/tex]
(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]
(b) The KE at equilibrium point =47j
(c) The mass =7.9kg
(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]
(e)The KE of block 0.160m will be 26.1
(f) The PE will be 20.9j
What will be the asked values of the spring-mass system in the question?(a) for finding spring constant
KE=47 j
A=0.240m
By using the formula
[tex]E=\dfrac{1}{2} kx^2[/tex]
[tex]47=\dfrac{1}{2} k(0.240)^2[/tex]
[tex]k=1632\frac{N}{m^2}[/tex]
(b) At the mean position the whole mechanical energy will be equal to KE so
KE=47j
(c) The mass of the system
[tex]KE =\dfrac{1}{2} mv^2[/tex]
[tex]47=\dfrac{1}{2} m(3.45^2)[/tex]
[tex]m=7.9kg[/tex]
(d)Now the speed of the block
[tex]v=\sqrt{\dfrac{k}{m} (A^2-x^2)}[/tex]
[tex]v=\sqrt{\dfrac{1632}{7.9} (0.24^2-0.16^2)}[/tex]
[tex]v=2.57\frac{m}{s}[/tex]
(e) The KE of the block
[tex]KE=\dfrac{1}{2} mv^2=\dfrac{1}{2} 7.9(2.57)^2=26.1J[/tex]
(f) The PE of the system
[tex]Total Energy = KE+PE[/tex]
[tex]PE= 47-26.1 =20.9J[/tex]
Thus
(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]
(b) The KE at equilibrium point =47j
(c) The mass =7.9kg
(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]
(e)The KE of block 0.160m will be 26.1
(f) The PE will be 20.9j
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5.90. A force is applied to a block to move it up a 30° incline. The incline is frictionless. If F = 65.0 N and M = 5.00 kg , what is the magnitude of the acceleration of the block? Enter your answer in units of m/s^2m/s 2 , without units, to the nearest hundredth.
Answer:
The acceleration of the block is 6.35 m/s².
Explanation:
It is given that,
A force is applied to a block to move it up a 30° incline. The applied force is, F = 65 N
Mass of the block, m = 5 kg
We need to find the acceleration of the block. From the attached figure, it is clear that.
[tex]F_x=ma_x[/tex]
[tex]F\ cos\theta-mg\ sin\theta=ma_x[/tex]
[tex]a_x=\dfrac{F\ cos\theta-mg\ sin\theta}{m}[/tex]
[tex]a_x=\dfrac{65\ N\ cos(30)-5\ kg\times 9.8\ m/s^2\ sin(30)}{5\ kg}[/tex]
[tex]a_x=6.35\ m/s^2[/tex]
So, the acceleration of the block is 6.35 m/s². Hence, this is the required solution.
Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².The magnitude of the acceleration of the block will be 6.35 m/sec².
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
On resolving the given force and accelertaion in the different components and balancing the equation gets.Components in the x-direction.
The given data in the problem,
F is the applied force =65 N
Θ is the angle of inclined plane=30°
m is the mass of the block= 5 kg
We need to find the acceleration of the block in the x-direction
[tex]\rm F_x=ma_x\\\\\rm F_x=Fcos\theta-mgsin\theta\\\\\rm Fcos\theta-mgsin\theta=ma_x\\\\\rm a_x=\frac{Fcos\theta-mgsin\theta}{m}\\\\ \rm a_x=\frac{65\timescos30^0-mgsin30^0}{5} \\\\\rm a_x=6.35 m/sec^2[/tex]
Hence the magnitude of the acceleration of the block will be 6.35 m/sec².
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A pump is used to empty a 5200 L wading pool. The water exits the 3.0-cm-diameter hose at a speed of 2.2 m/s .How long will it take to empty the pool? Express your answer using two significant figures.
Answer:
55.79 minutes
Explanation:
Volume of water = 5200 L = 5.2 m^3
Diameter = 3 cm
radius, r = 1.5 cm = 0.015 m
v = 2.2 m/s
The volume of water coming out in 1 second = velocity x area
= 2.2 x 3.14 x 0.015 x 0.015
= 1.55 x 10^-3 m^3
1.55 x 10^-3 m^3 water flows = 1 second
5.2 m^3 water flows = 5.2 / (1.55 x 10^-3) = 3345.57 second
= 55.79 minutes
How many centimeters are there in meter? b. 10 c. d 1000 e. 10000 100 2. A centimeter is equal to 1 inch b.½inch C 1/2.54 inch d. 2.54 inches e. 37.39 inches
Answer:
a) There are 100 centimeters in 1 meter.
b) [tex]\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}[/tex]
Explanation:
a) We have the conversion
1 m = 100 cm
So there are 100 centimeters in 1 meter.
b) 1 inch = 2.54 cm
[tex]1cm=\frac{1}{2.54}inch[/tex]
[tex]\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}[/tex]
A 900 kg SUV is traveling at a constant speed of 1.44 m/s due north. What is the total force (in N) on the vehicle? (Assume north is the positive direction. Indicate the direction with the sign of your answer.)
Explanation:
It is given that,
Mass of SUV, m = 900 kg
It is moving with a constant speed of 1.44 m/s due south. We need to find the total force on the vehicle. The second law of motion gives the magnitude of force acting on the object. It is given by :
F = m a
Since, [tex]a=\dfrac{dv}{dt}[/tex] i.e. the rate of change of velocity
As SUV is travelling at a constant speed. This gives acceleration of it as zero.
So, F = 0
So, the total force acting on SUV is 0 N. Hence, this is the required solution.
Answer:
The total force acting on the vehicle is 0 N.
Explanation:
Given that,
Mass of SUV = 900 kg
Velocity = 1.44 m/s
SUV is travelling at a constant speed of 1.44 m/s due north.
We need to calculate the force on the vehicle.
Using newton's second law
[tex]F = ma[/tex]....(I)
We know that,
The acceleration is the first derivative of the velocity of the particle.
[tex]a = \dfrac{dv}{dt}[/tex]
Now,put the value of a in equation (I)
[tex]F=m\dfrac{dv}{dt}[/tex]
SUV is traveling at a constant speed it means acceleration is zero.
So, The force will be zero.
Hence, The total force acting on the vehicle is 0 N.
An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary, the tension in the cable was 7000 N {\rm N}. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N {\rm N} drag force.
Part A
What was the tension in the cable when the craft was being lowered to the seafloor?
Express your answer to two significant figures and include the appropriate units.
Part B
What was the tension in the cable when the craft was being raised from the seafloor?
Express your answer to two significant figures and include the appropriate units.
Answer:
A) 5.2 x 10³ N
B) 8.8 x 10³ N
Explanation:
Part A)
[tex]F_{g}[/tex] = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N
[tex]T[/tex] = Tension force in upward direction
[tex]F_{d}[/tex] = Drag force in upward direction = 1800 N
Force equation for the motion of craft is given as
[tex]F_{g}[/tex] - [tex]F_{d}[/tex] - [tex]T[/tex] = 0
7000 - 1800 - [tex]T[/tex] = 0
[tex]T[/tex] = 5200 N
[tex]T[/tex] = 5.2 x 10³ N
Part B)
[tex]F_{g}[/tex] = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N
[tex]T[/tex] = Tension force in upward direction
[tex]F_{d}[/tex] = Drag force in downward direction = 1800 N
Force equation for the motion of craft is given as
[tex]T[/tex] - [tex]F_{g}[/tex] - [tex]F_{d}[/tex] = 0
[tex]T[/tex] - 7000 - 1800 = 0
[tex]T[/tex] = 8800 N
[tex]T[/tex] = 8.8 x 10³ N
A research-level Van de Graaff generator has a 2.15 m diameter metal sphere with a charge of 5.05 mC on it. What is the potential near its surface in MV? (Assume the potential is equal to zero far away from the surface.)
Answer:
42.3 MV
Explanation:
d = diameter of the metal sphere = 2.15 m
r = radius of the metal sphere
diameter of the metal sphere is given as
d = 2r
2.15 = 2 r
r = 1.075 m
Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C
Potential near the surface is given as
[tex]V = \frac{kQ}{r}[/tex]
[tex]V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}[/tex]
V = 4.23 x 10⁷ volts
V = 42.3 MV
A metal ring 4.20 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.240 T/s . (A) What is the magnitude of the electric field induced in the ring? (B) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?
Answer:
09m
Explanation:
yes the current flow in clockwise direction
Explanation:
Given that,
Diameter of the metal ring, d = 4.2 cm
Radius, r = 2.1 cm
Initial magnetic field, B = 1.12 T
The magnetic field is decreasing at the rate of, [tex]\dfrac{dB}{dt}=0.24\ T/s[/tex]
Due to change in magnetic field, an emf is induced in it. And hence, electric field is induced. It is given by :
[tex]\int\limits {E.dl}=\dfrac{d}{dt}(\int\limits{B.dA)}[/tex]
[tex]E.(2\pi r)=\dfrac{dB}{dt}(\pi r^2)[/tex]
[tex]E=\dfrac{dB}{dt}\times \dfrac{r}{2}[/tex]
[tex]E=0.24\times \dfrac{2.1\times 10^{-2}}{2}[/tex]
[tex]E=0.00252\ N/C[/tex]
So, the magnitude of the electric field induced in the ring has a magnitude of 0.00252 N/C.
The direction of electric field will be counter clock wise direction as viewed by someone on the south pole of the magnet.
A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0; (5.0 m, 5.0 m) at t = 2.0 s; and (14.0 m, 12.0 m) at t = 5.0 s. (a) Find the magnitude of the average velocity from t = 0 to t = 2 s. (b) Find the magnitude of the average velocity from t = 0 to t = 5 s.
Answer:
a) Magnitude of the average velocity from t = 0 to t = 2 s = 2.24 m/s
b) Magnitude of the average velocity from t = 0 to t = 5 s = 3.16 m/s
Explanation:
a) Velocity is rate of change of position.
A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0
A particle's position coordinates (x, y) are (5.0 m, 5.0 m) at t = 2.0 s
Displacement = (5-1)i + (5-3)j = 4i + 2j
Change in time = 2s
Velocity
[tex]v=\frac{4i+2j}{2}=2i+j[/tex]
Magnitude of velocity
[tex]v=\sqrt{2^2+1^2}=2.24m/s[/tex]
b) Velocity is rate of change of position.
A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0
A particle's position coordinates (x, y) are (14.0 m, 12.0 m) at t = 5.0 s
Displacement = (14-1)i + (12-3)j = 13i + 9j
Change in time = 5s
Velocity
[tex]v=\frac{13i+9j}{5}=2.6i+1.8j[/tex]
Magnitude of velocity
[tex]v=\sqrt{2.6^2+1.8^2}=3.16m/s[/tex]
The speed of sound through the ground is about 6.0 km/s while the speed of sound in air is 343 m/s. A very powerful explosion occurs some distance away and you feel the ground vibrate 60 seconds before you hear the sound of the explosion. How far away is the explosion?
Answer:
21828 m
Explanation:
[tex]v_{ground}[/tex] = speed of sound through ground = 6 km/s = 6000 m/s
[tex]v_{air}[/tex] = speed of sound through air = 343 m/s
t = time taken for the vibrations to arrive
t' = time taken for sound to arrive = t + 60
d = distance of the point of explosion
time taken for the vibrations to arrive is given as
[tex]t = \frac{d}{v_{ground}}[/tex] eq-1
time taken for the sound to arrive is given as
[tex]t' = \frac{d}{v_{air}}[/tex]
[tex]t + 60 = \frac{d}{v_{air}}[/tex]
using eq-1
[tex]\frac{d}{v_{ground}} + 60 = \frac{d}{v_{air}}[/tex]
[tex]\frac{d}{6000} + 60 = \frac{d}{343}[/tex]
d = 21828 m
An alpha particle travels at a velocity of magnitude 760 m/s through a uniform magnetic field of magnitude 0.034 T. (An alpha particle has a charge of charge of 3.2 × 10-19 C and a mass 6.6 × 10-27 kg) The angle between the particle's direction of motion and the magnetic field is 51°. What is the magnitude of (a) the force acting on the particle due to the field, and (b) the acceleration of the particle due to this force
Answer:
(a) 6.42 x 10^-18 N
(b) 9.73 x 10^8 m/s^2
Explanation:
v = 760 m/s, B = 0.034 T, m = 6.6 x 10^-27 kg, q = 3.2 x 10^-19 C, theta = 51 degree
(a) F = q v B Sin theta
F = 3.2 x 10^-19 x 760 x 0.034 x Sin 51
F = 6.42 x 10^-18 N
(b) Acceleration, a = Force / mass
a = (6.42 x 10^-18) / (6.6 x 10^-27)
a = 0.973 x 10^9
a = 9.73 x 10^8 m/s^2
The earth orbits the sun once per year at the distance of 1.50 x 1011 m. Venus orbits the sun at a distance of 1.08 x 1011 m. These distances are between the centers of the planets and the sun. How long (in earth days) does it take for Venus to make one orbit around the sun
It takes 225 Earth days for Venus to orbit the Sun. Interestingly, Venus spins on its axis so slowly that its day (243 Earth days) is longer than its year. The Sun takes 117 Earth days to return to the same place in Venus' sky.
Explanation:The time it takes for Venus to make one orbit around the sun, also known as its orbital period, is actually 225 Earth days. This is quite different compared to Earth which takes 365.25 days to orbit the Sun. Additionally, Venus spins on its axis very slowly, with its rotational period being 243 Earth days. As a result, a day on Venus - considering its rotation - is longer than its year! Also, this leads to an unusual phenomenon where the sun takes 117 Earth days to return to the same place in the Venusian sky. This suggests that Venus' rotation and orbit display unique characteristics compared to other planets in our solar system, likely due to factors from its formation.
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Venus completes one orbit around the Sun in approximately 225 Earth days. This is shorter than Earth's orbital period due to Venus's closer distance to the Sun.
Orbit of Venus Around the Sun-
The distance between the Earth and the Sun is approximately 1.50 x 1011 meters, while Venus is closer, at 1.08 x 1011 meters. This difference in distance influences the orbital period of each planet. Venus takes about 225 Earth days to complete one full orbit around the Sun.
Venus has a nearly circular orbit and, being closer to the Sun than Earth, receives almost twice as much light and heat. The elliptical orbits mean that different planets have different orbital periods, with those closer to the Sun having shorter years.
When a comet comes close to the Sun, what happens to it that makes it bright and easier to see?
Answer:
Explanation:
Comet is made by dust particles, icy particles, gases etc.
A comet has a fixed time to complete a revolution around the sun.
As a comet comes nearer to the sun, due to the heat of the sun the vapour and the icy particles becomes gases and due to the radial pressure of energy od sun, we observe a tail of comet which has vapours mainly. SI the comet is visible easily.
A satellite orbits the earth in a circle with constant speed. Which of the following is true? The net force on the satellite is zero because the satellite is not accelerating. The net force on the satellite is directed forward, in the direction of travel. The net force on the satellite is directed straight down, toward the Earth. The net force on the satellite is directed outward, away from the Earth.
Explanation:
A satellite orbiting the Earth is a characteristic example of the uniform circular motion, where the direction of the velocity vector [tex]\vec{V}[/tex] is perpendicular to the radius [tex]r[/tex] of the trajectory. Hence, the velocity is changing although the speed is constant.
On the other hand, acceleration [tex]\vec{a}[/tex] is directed toward the center of the circumference (that is why it is called centripetal acceleration).
Now, according to Newton's 2nd law, the force [tex]\vec{F}[/tex] is directly proportional and in the same direction as the acceleration:
[tex]\vec{F}=m.\vec{a}[/tex]
Therefore, the net force on the satellite resulting from its circular motion points towards the center of the circle (where the Earth is in the Earth-satellite system).
When 1.50 ✕ 10^5 J of heat transfer occurs into a meat pie initially at 20.0°C, its entropy increases by 465 J/K. What is its final temperature (in degrees)?
Answer:
The final temperature is 79.16°C.
Explanation:
Given that,
Heat [tex]Q=1.50\times10^{5}\ J[/tex]
Temperature = 20.0°C
Entropy = 465 J/k
We need to calculate the average temperature
Using relation between entropy and heat
[tex]\Delta S=\dfrac{\Delta Q}{T}[/tex]
[tex]T=\dfrac{\Delta Q}{\Delta S}[/tex]
Where, T = average temperature
[tex]\Delta Q[/tex]= transfer heat
[tex]\Delta S[/tex]= entropy
Put the value into the formula
[tex]T=\dfrac{1.50\times10^{5}}{465}[/tex]
[tex]T=322.58\ K[/tex]
We need to calculate the final temperature
Using formula of average temperature
[tex]T = \dfrac{T_{i}+T_{f}}{2}[/tex]
[tex]T_{f}=2T-T_{i}[/tex]....(I)
Put the value in the equation (I)
[tex]T_{f}=2\times322.58-293[/tex]
[tex]T_{f}=352.16\ K[/tex]
We convert the temperature K to degrees
[tex]T_{f}=352.16-273[/tex]
[tex]T_{f}=79.16^{\circ}\ C[/tex]
Hence, The final temperature is 79.16°C.
A 2.0 kg ball and a 3.5 kg ball, each moving at 0.90 m/s, undergo a head-on collision. The lighter ball rebounds opposite its initial direction, with speed 0.90 m/s.Find the post-collision velocity of the heavier ball. Assume the initial direction of the lighter ball as positive.How much mechanical energy was lost in this collision? Express your answer in J.How much mechanical energy was lost in this collision? Express your answer as a fraction of the system's initial mechanical energy.
Explanation:
It is given that,
Mass of first ball, m₁ = 2 kg
Mass of other ball, m₂ = 3.5 kg
Velocities of both balls, u = 0.9 m/s
(1) The lighter ball rebounds opposite its initial direction, with speed 0.90 m/s. We need to find the final velocity of second ball. Applying the conservation of momentum as :
[tex]2\ kg\times 0.9-3.5\ kg\times 0.9\ m/s=-2\ kg\times 0.9\ m/s+3.5v[/tex]
v is the final velocity of heavier ball.
v = 0.128 m/s
or
v = 0.13 m/s
Initial kinetic energy, [tex]E_i=\dfrac{1}{2}\times (2\ kg+3.5\ kg)\times (0.9\ m/s)^2=2.23\ J[/tex]
Final kinetic energy, [tex]E_f=\dfrac{1}{2}\times 2\ kg\times (0.9\ m/s)^2+\dfrac{1}{2}\times 3.5\ kg\times (0.13\ m/s)^2=0.84\ J[/tex]
Lost in kinetic energy, [tex]\Delta KE=0.84-2.23=-1.39\ J[/tex]
Hence, this is the required solution.
slope of stress-strain curve in elastic deformation region is a. Plastic modulus b. Poisson's ratio c. Elastic modulus d. None the these a) all of these
Answer:
c) Elastic Modulus
Explanation
As we know that when deformation is under elastic limit then stress applied to the given material is proportional to the strain developed in it
So here we can say that since they both are directly proportional to each other so the proportionality constant here is known as Modulus of elasticity.
So we can say it is given as
[tex]stress = E (strain)[/tex]
so now if we draw a graph between between stress and strain then it must be a straight line and the the slope of this straight line is given as
[tex]Slope = \frac{Stress}{Strain}[/tex]
So here correct answer will be
c) Elastic Modulus
Final answer:
The slope of the stress-strain curve in the elastic deformation region is known as the Elastic Modulus or Young's Modulus, which measures the stiffness of the material. The correct answer to the given question is 'c. Elastic Modulus'.
Explanation:
The slope of the stress-strain curve in the elastic deformation region represents the relationship between stress and strain under elastic conditions. This slope is in fact the Elastic Modulus, which is also known as Young's Modulus when it's in tension or compression. It serves as a measure of the stiffness or rigidity of the material, indicating how much stress is required to achieve a certain amount of strain.
In the context of a stress-strain curve, the plastic modulus is associated with plastic deformation, not the elastic region. Poisson's ratio is another material property that describes the ratio of transverse strain to axial strain, and is not the slope of the curve. Hence, the correct answer to the question is 'c. Elastic Modulus'.
A motoris st enters a freeway at a speed of 35 mi/h and accelerates uniformly to a speed of 55 mi h. The motorist travels 500 ft while accelerating. I Determine: ) The acceleration of the car 2) The time required to reach 55 mi/h.
Answer:
1) The acceleration of the car = 1.17 m/s²
2) The time required to reach 55 mi/h = 7.59 seconds.
Explanation:
1) Initial speed of motorist, u = 35 mph = 15.56 m/s
Final speed of motorist, v = 55 mph = 24.44 m/s
Distance traveled, s = 500ft = 152.4 m
We have v² = u² + 2as
24.44² = 15.56² + 2 x a x 152.4
a = 1.17 m/s²
The acceleration of the car = 1.17 m/s²
2) We have v = u + at
24.44 = 15.56 + 1.17 x t
t = 7.59 s
The time required to reach 55 mi/h = 7.59 seconds.
Consider a simple tension member that carries an axial load of P=22.44N. Find the total elongation in the member due to the load. Assume that the member is made of steel, which has a modulus of elasticity of E=204.00 N/mm2. Also assume that the member is 3048 mm long and has a cross-sectional area of 1290 mm2
Answer:
The total elongation for the tension member is of 0.25mm
Explanation:
Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:
[tex]\sigma=E*\epsilon[/tex] (1)
Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:
[tex]\delta L=L*\epsilon[/tex] (2)
Here L is the member extension and δL the change total longitudinal elongation.
Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:
[tex]\sigma=P/A[/tex]
[tex]\sigma=22.44N / 1290 mm^2[/tex]
[tex]\sigma=0.0174 N/mm^2[/tex]
Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:
[tex]=\sigma=E*\epsilon[/tex]
[tex]\epsilon=\sigma/E[/tex]
[tex]\epsilon=0.0174 \frac{N}{mm^2}/\ 204 \frac{N}{mm^2} [/tex]
[tex]\epsilon=8.53*10^-{5}[/tex]
Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):
[tex]\delta L=3048 mm * 8.53*10^{-5} [/tex]
[tex]\delta L= 0.25 mm [/tex]
A stone with a mass of 10 kg is sitting on the ground, not moving. a) What is the weight of the stone? b) What is the normal force acting on the stone?
Answer:
A) weight=mass X gravitational force
10Kg X 10 M/S 2
100 KgM/S2
100Newton
B) Force =Mass X gravitational force
10Kg X 10 M/S2
100 Newton
The weight of a 10kg stone sitting at rest on the ground is 98 Newtons, and the normal force acting on it is also 98 Newtons.
Explanation:The weight of an object is calculated by multiplying its mass by the acceleration due to gravity. In this case, the mass of the stone is 10kg and The standard acceleration due to gravity on Earth is approximately 9.8 m/s2. Therefore, the weight of the stone would be 10 kg x 9.8 m/s^2 = 98 Newtons.
b) The normal force acting on the stone is the force exerted by the surface of the ground on the stone in the upward direction. When the stone is sitting on the ground, the normal force is equal in magnitude and opposite in direction to the weight of the stone. Therefore, the normal force is also 98 N.
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A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.715 m0.715 m and 2.51 kg2.51 kg , respectively. When the propellor rotates at 527 rpm527 rpm (revolutions per minute), what is its rotational kinetic energy?
Answer:
3260.33 J
Explanation:
[tex]n [/tex] = number of rods = 5
[tex]L [/tex] = length of each rod = 0.715 m
[tex]m [/tex] = mass of rod = 2.51 kg
[tex]I [/tex] = total moment of inertia
Total moment of inertia is given as
[tex]I = \frac{nmL^{2}}{3}[/tex]
[tex]I = \frac{(5)(2.51)(0.715)^{2}}{3}[/tex]
[tex]I [/tex] = 2.14 kgm²
[tex]w [/tex] = angular speed = 527 rpm = 55.2 rad/s
Rotational kinetic energy is given as
E = (0.5) [tex]I [/tex] ([tex]w [/tex] )²
E = (0.5) (2.14) (55.2)²
E = 3260.33 J
The temperature of air decreases as it is compressed by an adiabatic compressor. a)-True b)-False
Answer: false
Explanation: the air will increase if it is compressed by an adiabatic compressor
A 62.0-kg woman runs up a flight of stairs having a rise of 4.28 m in a time of 4.20 s. What average power did she supply? a) 63.2 W b) 619 W c) 596 W d) 629 W e) 609 W
Answer:
The average power is 619 W.
(B) is correct option
Explanation:
Given that,
Weight = 62.0 kg
Height = 4.28 m
Time t = 4.20 s
We need to calculate the work done
Work done by woman
[tex]W=mgh[/tex]
Where,
m = mass
g = acceleration due to gravity
t = time
Put the value into the formula
[tex]W=62\times9.8\times 4.28[/tex]
[tex]W=2600.528 J[/tex]
We need to calculate the power
Using formula of power
[tex]P=\dfrac{W}{t}[/tex]
Where,
W = work done
t = time
Put the value in to the formula
[tex]P=\dfrac{2600.528}{4.20}[/tex]
[tex]P=619\ W[/tex]
Hence, The average power is 619 W.
The average power supplied by the woman is calculated by first determining the work done against gravity to move up the stairs, and then dividing this by the time taken. After performing these calculations, the most accurate answer is approximately 629 W.
Explanation:The question is asking to calculate the average power supplied by a woman while running up the stairs. Power is defined as the rate at which work is done. In this scenario, the work is the woman's movement against the gravitational force, which is calculated using the formula W = mgh, where m is mass, g is the acceleration due to gravity and h is the height of the stairs. Substituting the given values, we get W = 62.0 Kg * 9.8 m/s² * 4.28 m = 2650.368 Joules. Power is calculated by dividing the work done by the time taken, so P = W/t = 2650.368 J / 4.20 s = 630.8 Watts, which is closest to option (d) 629 W.
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The small piston of a hydraulic lift has a diameter of 8.0 cm, and its large piston has a diameter of 40 cm. The lift raises a load of 15,000 N. Assume the pistons each have negligible weight. (a) Determine the force that must be applied to the small piston. (b) Determine the pressure applied to the fluid in the lift.
Answer:
a)
600 N
b)
1.2 x 10⁵ Pa
Explanation:
(a)
d₁ = diameter of small piston = 8 cm = 0.08 m
d₂ = diameter of large piston = 40 cm = 0.40 m
F₂ = force applied to large piston = 15000 N
F₁ = force applied to small piston = ?
Using pascal's law
[tex]\frac{F_{1}}{(0.25)\pi d_{1}^{2}} = \frac{F_{2}}{(0.25)\pi d_{2}^{2}}[/tex]
Inserting the values
[tex]\frac{F_{1}}{(0.08)^{2}} = \frac{15000}{(0.40)^{2}}[/tex]
F₁ = 600 N
b)
Pressure applied is given as
[tex]P = \frac{F_{1}}{(0.25)\pi d_{1}^{2}}[/tex]
[tex]P = \frac{(600)}{(0.25)(3.14) (0.08)^{2}}[/tex]
P = 1.2 x 10⁵ Pa
To determine the force needed on the small piston of a hydraulic lift, the ratios of the pistons' areas are used in combination with the load on the large piston, applying Pascal's Principle. The pressure in the hydraulic fluid is then found by dividing the force by the area of the piston where the force is applied.
Force applied to small piston = (Force on large piston) × (Area of small piston / Area of large piston)Pressure in the hydraulic fluid = Force / AreaGiven the diameters of the pistons, we can find the areas by using the formula for the area of a circle, A = πr², where r is the radius of the piston.
For the provided diameters of 8.0 cm and 40 cm, the areas of the pistons are:
Area of small piston = π × (4 cm)²Area of large piston = π × (20 cm)²With a load of 15,000 N on the large piston:
Force applied to small piston = 15,000 N × (Area of small piston / Area of large piston)Pressure = 15,000 N / (Area of large piston)Once we perform these calculations, we can determine the required force to apply to the small piston and the pressure applied to the fluid within the lift.