Answer:
T = 5.66 s
Explanation:
The system formed by the bar plus ball forms a physical pendulum
w = [tex]\sqrt{mgd/I}[/tex]
the moment of inertia of a rod held at one end is
I = [tex]\frac{1}{3}[/tex] m L²
we substitute
w = [tex]\sqrt{\frac{d \ d}{ 3 L^2 } }[/tex]
in this case the turning distance and the length of the rod are equal
d = L
w = [tex]\sqrt{\frac{g}{3L} }[/tex]
angular velocity and period are related
w = 2π / T
2π / T = [tex]\sqrt{\frac{g}{3L} }[/tex]
T = 2π [tex]\sqrt{3L/g}[/tex]
let's calculate
T = 2π [tex]\sqrt{3 \ 2.65 / 9.8}[/tex]
T = 5.66 s
How many loops are in this circuit?
I see six (6) loops.
I attached a drawing to show where I get six loops from.
The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is:__________.
A) 0 m/s
B) 1 m/s
C) 2 m/s
D) 3 m/s
E) Not enough info
Answer:
The correct option is (E).
Explanation:
Given that,
Mass of object 1, m₁ = 1 kg
Mass of object 2, m₂ = 2 kg
They collides after the collision. We need to find the speed of the two boxes after the collision.
The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.
So, the information is not enough.
What is the escape speed on a spherical asteroid whose radius is 517 km and whose gravitational acceleration at the surface is 0.636 m/s2
Answer:
810.94 m/s
Explanation:
Applying,
v = √(2gR)............. Equation 1
Where v = escape velocity of the spherical asteroid, g = acceleration due to gravity, R = radius of the earth
From the question,
Given: g = 0.636 m/s², R = 517 km = 517000 m
Substitute these values into equation 1
v = √(2×0.636×517000)
v = √(657624)
v = 810.94 m/s
Hence, the escape velocity is 810.94 m/s
I need help with this problem can anybody help me please , it’s physics 2 course
Answer:
ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|
Explanation:
The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:
[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]
Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is
[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]
or
[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]
The cannon on a battleship can fire a shell a maximum distance of 33.0 km.
(a) Calculate the initial velocity of the shell.
Answer:
v = 804.23 m/s
Explanation:
Given that,
The maximum distance covered by a cannon, d = 33 km = 33000 m
We need to find the initial velocity of the shell. Let it is v. It can be calculated using the conservation of energy such that,
[tex]v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 33000} \\\\v=804.23\ m/s[/tex]
So, the initial velocity of the shell is 804.23 m/s.
A child is outside his home playing with a metal hoop and stick. He uses the stick to keep the hoop of radius 45.0 cm rotating along the road surface. At one point the hoop coasts downhill and picks up speed. (a) If the hoop starts from rest at the top of the hill and reaches a linear speed of 6.35 m/s in 11.0 s, what is the angular acceleration, in rad/s2, of the hoop? rad/s2 (b) If the radius of the hoop were smaller, how would this affect the angular acceleration of the hoop? i. The angular acceleration would decrease. ii. The angular acceleration would increase. iii. There would be no change to the angular acceleration.
Answer:
a) [tex] \alpha = 1.28 rad/s^{2} [/tex]
b) Option ii. The angular acceleration would increase
Explanation:
a) The angular acceleration is given by:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular speed = v/r
v: is the tangential speed = 6.35 m/s
r: is the radius = 45.0 cm = 0.45 m
[tex]\omega_{0}[/tex]: is the initial angular speed = 0 (the hoop starts from rest)
t: is the time = 11.0 s
α: is the angular acceleration
Hence, the angular acceleration is:
[tex] \alpha = \frac{\omega}{t} = \frac{v}{r*t} = \frac{6.35 m/s}{0.45 m*11.0 s} = 1.28 rad/s^{2} [/tex]
b) If the radius were smaller, the angular acceleration would increase since we can see in the equation that the radius is in the denominator ([tex] \alpha = \frac{v}{r*t} [/tex]).
Therefore, the correct option is ii. The angular acceleration would increase.
I hope it helps you!
A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.
What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.
Answer:
sup qwertyasdfghjk
Explanation:
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas. During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles, while during the adiabatic expansion the volume increases by a factor of 5.7. The work output of the engine is 940 J in each cycle. Compute the temperatures of the two reservoirs between which this engine
operates.
Answer:
The hot temperature is 157.5 K
The cold temperature is 48.8 K
Explanation:
Step 1: Data given
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas.
The volume increases by a factor of 5.7
The work output of the engine is 940 J in each cycle.
During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles. This means V2 = 2*V1 (and V4 = 2*V3)
Step 2:For a carnot engine:
V2/V1 = V4/V3
Work = nR((T1)ln(V2/V1) - (T2)ln(V4/V3))
⇒with Work = the work done in the cycle = 940J
⇒with n = the number of moles = 1.50 moles
⇒with R = the gas constant = 8.314 J/mol*K
⇒with T1 = the hot temperature
⇒With T2⇒ the cold temperature
where R = 8.31 J/mol K Gas Constant
940J = 1.5moles * 8.314 J/mol*K * (T1*ln(2) - T2*ln(2)))
940 = 1.5 * 8.314 ln(2) * (T1-T2)
(T1-T2) = 940 / (1.5*8.314*ln(2))
(T1-T2) = 108.7K
For the reversible adiabatic expansion: T2 = T1*(V1/V2)^(R/Cv). Where V2/V1 = 5.7 (Because during the adiabatic expansion the volume increases by a factor of 5.7)
For a monatomic ideal gas, Cv = 3/2R
When we combine both, we'll have:
T2 = T1*(1/5.7)^(R/3/2R)
T2 = T1*(1/5.7)^(2/3)
T2= T1 * 0.31
Since we know that (T1-T2) = 108.7K
we have:
T1 - 0.31T1= 108.7K
0.69T1 = 108.7K
T1 = 157.5K
T2 = 157.5*0.31 = 48.8K
The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.2 m. The block has a thermal conductivity of 150 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 35 ˚C and that of the cooler surface is 16 ˚C Determine the heat that flows in 6 s for each case.
Answer:
1140 J, 6840 J, 10260 J
Explanation:
Lo x 2 Lo x 3 Lo, Lo = 0.2 m, K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,
time, t = 6 s
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]
A cylindrical tank with radius 7 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing (in m/min)?
Answer:
0.013 m/min
Explanation:
Applying,
dV/dt = (dh/dt)(dV/dh)............. Equation 1
Where
V = πr²h................ Equation 2
Where V = volume of the tank, r = radius, h = height.
dV/dh = πr²............ Equation 3
Substitute equation 3 into equation 1
dV/dt = πr²(dh/dt)
From the question,
Given: dV/dt = 2 m³/min, r = 7 m, π = 3.14
Substitute these values into equation 3
2 = (3.14)(7²)(dh/dt)
dh/dt = 2/(3.14×7²)
dh/dt = 0.013 m/min
A grade 12 Physics student shoots a basketball
from the ground at a hoop which is 2.0 m above
her release. The shot was at a velocity of 10 m/s
and at an angle of 80° to the ground.
a. Determine the vertical velocity of the ball
when it is at the level of the net. You
should get two answers.
Please show ALL steps
Answer:
7.84 m/s
Explanation:
Height, h = 2 m
Initial velocity, u = 10 m/s
Angle, A = 80°
(a) Let the time taken to go to the net is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]
Time cannot be negative.
So, t = 0.2 s
The vertical velocity at t = 0.2 s is
v = u + at
v = 10 sin 80 - 9.8 x0.2
v = 9.8 - 1.96 = 7.84 m/s
A system gains 1500J of heat and 2200J of work is done by the system on its surroundings. Determine the change in internal energy of the system
Answer:
-700
formula is heat gained - work done
The change in internal energy if A system gains 1500J of heat and 2200J of work is done by the system on its surroundings, is 700 joules.
What is Energy?Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc.
Additionally, there is heat and work, which is energy being transferred from one body to another. Energy is always assigned based on its nature once it has been transmitted. Thus, heat transmitted may manifest as thermal energy while work performed may result in mechanical energy.
Given:
A system gains 1500J of heat and 2200J of work is done by the system on its surroundings,
Calculate the change in internal energy as shown below,
The change in internal energy = heat gained - work done
The change in internal energy = 1500 - 2200
The change in internal energy = -700 J
Thus, the change in internal energy is 700 joules.
To know more about Energy:
https://brainly.com/question/8630757
#SPJ5
A beam of light has a wavelength of 549nm in a material of refractive index 1.50. In a different material of refractive index 1.07, its wavelength will be:_________.
Explanation:
someone to check if the answer is correct
You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area of 245 SF. The berm is 300 ft long and is assumed to taper evenly between the two cross-sectional areas, what is the calculated volume of the berm in cubic feet
Find the volume of cuboid of side 4cm. Convert it in SI form
Answer:
0.000064 cubic meters.
Explanation:
Given the following data;
Length of side = 4 centimeters
Conversion:
100 centimeters = 1 meters
4 cm = 4/100 = 0.04 meters
To find the volume of cuboid;
Mathematically, the volume of a cuboid is given by the formula;
Volume of cuboid = length * width * height
However, when all the sides are equal the formula is;
Volume of cuboid = L³
Volume of cuboid = 0.04³
Volume of cuboid = 0.000064 cubic meters.
What is the energy equivalent of an object with a mass of 2.5 kg? 5.5 × 108 J 7.5 × 108 J 3.6 × 1016 J 2.25 × 1017 J
Answer:
E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules
Answer:
The answer is D. 2.25 × 1017 J
Explanation:
got it right on edge 2021
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by
The question is incomplete. The complete question is :
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].
Determine the following:
(a) frequency of the motion
(b) period of the motion
(c) amplitude of the motion
(d) first time after t = 0 that the object reaches the position x = 2.6 cm
Solution :
Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].
Comparing it with the general equation of simple harmonic motion,
x = A sin (ωt + Φ)
A = 4.7 cm
ω = 7.9 π
a). Therefore, frequency, [tex]$f=\frac{\omega}{2 \pi}$[/tex]
[tex]$=\frac{7.9 \pi}{2 \pi}$[/tex]
= 3.95 Hz
b). The period, [tex]$T=\frac{1}{f}$[/tex]
[tex]$T=\frac{1}{3.95}[/tex]
= 0.253 seconds
c). Amplitude is A = 4.7 cm
d). We have,
x = A sin (ωt + Φ)
[tex]$x_t=4.7 \sin (7.9 \pi t)$[/tex]
[tex]$2.6 = 4.7 \sin (7.9 \pi t)$[/tex]
[tex]$\sin (7.9 \pi t) = \frac{26}{47}$[/tex]
[tex]$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$[/tex]
Hence, t = 0.0236 seconds.
helppp!!! what's the answer to this??
when an ideal capacitor is connected across an ac voltage supply of variable frequency, the current flowing
a) is in phase with voltage at all frequencies
b) leads the voltage with a phase independent of frequency
c) leads the voltage with a phase which depends on frequency
d) lags the voltage with a phase independent of frequency
what would be the correct option?
Answer:
(b)
Explanation:
The voltage always lags the current by 90°, regardless of the frequency.
When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?
Answer:
because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.
prove mathematically :
1. v = u + at
2. s = ut+1*2 at
Answer:
a.v=u+v/2
a.v=s/t
combining two equation we get,
u+v/2=s/t
(u+v)t/2=s
(u+v)t/2=s
{u+(u+at)}t/2=s
(u+u+at)t/2=s
(2u+at)t/2=s
2ut+at^2/2=s
2ut/2+at^2/2=s
UT +1/2at^2=s
proved
a=v-u/t
at=v-u
u+at=v
Turning a corner at a typical large intersection in a city means driving your car through a circular arc with a radius of about 25 m. if the maximum advisable acceleration of your vehicle through a turn on wet pavement is 0.40 times the free-fall acceleration, what is the maximum speed at which you should drive through this turn?
Answer:
9.89 m/s.
Explanation:
Given that,
The radius of the circular arc, r = 25 m
The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²
Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{3.92\times 25} \\\\=9.89 m/s[/tex]
So, the maximum speed of the car should be 9.89 m/s.
The force an ideal spring exerts on an object is given by , where measures the displacement of the object from its equilibrium position. If , how much work is done by this force as the object moves from to
Answer:
The correct answer is "1.2 J".
Explanation:
Seems that the given question is incomplete. Find the attachment of the complete query.
According to the question,
x₁ = -0.20 mx₂ = 0 mk = 60 N/mNow,
⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]
⇒ [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]
⇒ [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]
By putting the values, we get
⇒ [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]
⇒ [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]
⇒ [tex]=1.2 \ J[/tex]
In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?
Answer:
59.26°
Explanation:
Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.
Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.
Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.
So, y = ut + 1/2a't²
y = 0 × t + 1/2(acosθ)t²
y = 0 + 1/2(acosθ)t²
y = 1/2(acosθ)t² (1)
Also, both particles must move the same horizontal distance to collide in time, t.
Let x be the horizontal distance,
x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision
Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.
So, x = ut + 1/2a"t²
x = 0 × t + 1/2(ainsθ)t²
x = 0 + 1/2(asinθ)t²
x = 1/2(asinθ)t² (3)
Equating (2) and (3), we have
vt = 1/2(asinθ)t² (4)
From (1) t = √[2y/(acosθ)]
Substituting t into (4), we have
v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²
v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)
v√[2y/(acosθ)] = ytanθ
√[2y/(acosθ)] = ytanθ/v
squaring both sides, we have
(√[2y/(acosθ)])² = (ytanθ/v)²
2y/acosθ = (ytanθ/v)²
2y/acosθ = y²tan²θ/v²
2/acosθ = ytan²θ/v²
1/cosθ = aytan²θ/2v²
Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1
secθ = ay(sec²θ - 1)/2v²
2v²secθ = aysec²θ - ay
aysec²θ - 2v²secθ - ay = 0
Let secθ = p
ayp² - 2v²p - ay = 0
Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have
ayp² - 2v²p - ay = 0
0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0
10.85p² - 15.68p - 10.85 = 0
dividing through by 10.85, we have
p² - 1.445p - 1 = 0
Using the quadratic formula to find p,
[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]
Since p = secθ
secθ = 1.95625 or secθ = -0.51125
cosθ = 1/1.95625 or cosθ = 1/-0.51125
cosθ = 0.5112 or cosθ = -1.9956
Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.
So, cosθ = 0.5112
θ = cos⁻¹(0.5112)
θ = 59.26°
So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.
1. A message signal m(t) has a bandwidth of 5kHz and a peak magnitude of 2V. Estimate the bandwidth of the signal u(t) obtained when m(t) frequency modulates a carrier with a) kf = 10 Hz/V, b) kf = 100 Hz/V, and c) kf = 1000 Hz/V.
Answer:
3v at 5.3 herts
Explanation:
Upon completing an interview, it is important that you send a follow-up thank you
note/letter/e-mail because it will show that you are a person who appreciates an opportunity.
A True
B
False
A particle charge of 2.7 µC is at the center of a Gaussian cube 55 cm on edge. What is the net electric flux through the surface?
Answer:
3.05×10⁵ Nm²C⁻¹
Explanation:
According to Gauss' law,
∅' = q/e₀............... Equation 1
Where ∅' = net flux through the surface, q = net charge, e₀ = electric permittivity of the space
From the question,
Given: q = 2.7 μC = 2.7×10⁻⁶ C,
Constant: e₀ = 8.85×10⁻¹² C²/N.m²
Substituting these values into equation 1
∅' = (2.7×10⁻⁶)/(8.85×10⁻¹²)
∅' = 3.05×10⁵ Nm²C⁻¹
The outer surface of a spacecraft in space has an emissivity of 0.44 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.
Answer:
[tex]T=326.928K[/tex]
Explanation:
From the question we are told that:
Emissivity [tex]e=0.44[/tex]
Absorptivity [tex]\alpha =0.3[/tex]
Rate of solar Radiation [tex]R=0.3[/tex]
Generally the equation for Surface absorbed energy is mathematically given by
[tex]E=\alpha R[/tex]
[tex]E=0.3*950[/tex]
[tex]E=285W/m^2[/tex]
Generally the equation for Emitted Radiation is mathematically given by
[tex]\mu=e(\sigmaT^4)[/tex]
Where
T=Temperature
[tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]
Therefore
[tex]\alpha*E=e \sigma T^4[/tex]
[tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]
[tex]T=326.928K[/tex]
two point charges two point charges are separated by 25 cm in the figure find The Net electric field these charges produced at point a and point b
The image is missing and so i have attached it.
Answer:
A) E = 8740 N/C
B) E = -6536 N/C
Explanation:
The formula for electric field is;
E = kq/r²
Where;
q is charge
k is a constant with value 8.99 x 10^(9) N•m²/C²
A) Now, to find the net electric field at point A, the formula would now be;
E = (kq1/(r1)²) - (kq2/(r2)²)
Where;
r1 is distance from charge q1 to point A
r2 is distance from charge q2 to point A.
q1 = -6.25 nC = -6.25 × 10^(-9) C
q2 = -12.5 nC = -12 5 × 10^(-9) C
From the attached image, r1 = 25 cm - 10 cm = 15 cm = 0.15 m
r2 = 10 cm = 0.1 m
Thus;
E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.15^(2)) - ((-12.5 × 10^(-9))/0.1^(2))
E = 8740 N/C
B) similarly, electric field at point B;
E = (kq1/(r1)²) + (kq2/(r2)²)
Where;
r1 is distance from charge q1 to point B
r2 is distance from charge q2 to point B.
q1 = -6.25 nC = -6.25 × 10^(-9) C
q2 = -12.5 nC = -12 5 × 10^(-9) C
From the attached image, r1 = 10 cm = 0.1 m
r2 = 25cm + 10 cm = 35 cm = 0.35 m
Thus;
E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.1^(2)) + ((-12.5 × 10^(-9))/0.35^(2))
E = -6536 N/C
A ball has a mass of 4.65kg and approximates a ping pong ball of mass 0.060kg that is at rest by striking it in an elastic collision. The initial velocity of the bowling ball is 5.00m/s, determine the final velocities of both masses after the collision.
Answer:
Look at work
Explanation:
Elastic Collision: Ki=Kf
M1=4.65kg
M2: 0.060kg
v1=5m/s
v2=0m/s
4.65*5+0.060*0=4.65*v1'+0.060*v2'
23.25+0=4.65v1'+0.060v2'
Also since it is an elastic collision we can use
v1+v1'=v2+v2'
4.65+v1'=v2'
4.65+v1'=v2'
Substitute into the earlier equation
23.25=4.65v1'+0.060(4.65+v1')
Expand
23.25=4.65v1'+0.279+0.06v1'
Solve for v1'
22.971=4.71v1'
v1'=4.88m/s
v2'=4.65+4.88=9.53m/s
What has a wind speed of 240 kph or greater?
Answer:
SUPER TYPHOON (STY), a tropical cyclone with maximum wind speed exceeding 220 kph or more than 120 knots.
