Answer:
a)13.33s
b)at highest point, the centripetal acceleration has its direction at downward path towards the center of the circular path, and the radius vector has its direction upward. Then acceleration=3.555m/s^2
c)c)at lowest point the centripetal acceleration has its direction upward towards the center of the circular path, and the radius vector has its direction downward Then acceleration=3.555m/s^2
Explanation
a)number of turns= 4.5
radius= 16m
We know that the period is the time taken by the wheel to complete one turn which can be calculated using below expresion
T= t/n
Where T= period of motion
t= Time taken by the wheel to finish n turns where our n= 4.5
T= (1×60)/4.5= 13.33s
Hence the period is 13.33s
Then the speed of the woman v= 2πr/T
v= (2×π×16)/13.33
v=7.5417m/s
b)at highest point, the centripetal acceleration has its direction at downward path towards the center of the circular path, and the radius vector has its direction upward.
a= v^2/r
Where r= radius
v= speed of the woman= 7.5417m/s
a=(7.5417m/s)^2/16
a=3.555m/s^2
The centripetal acceleration and radius vector are in opposite direction
c)at lowest point the centripetal acceleration has its direction upward towards the center of the circular path, and the radius vector has its direction downward
The magnitude of the acceleration is calculated below
a= v^2/r
Where r= radius
v= speed of the woman= 7.5417m/s
a=(7.5417m/s)^2/16
a=3.555m/s^2
(II) A baseball pitcher throws a baseball with a speed of 43 m????s. Estimate the average acceleration of the ball during the throwing motion. In throwing the baseball, the pitcher accelerates it through a displacement of about 3.5 m, from behind the body to the point where it is released
Answer:
a = 264.14 m/s²
Explanation:
From the question;
Initial velocity; u will be 0 m/s since the ball will start from rest.
Final velocity; v = 43 m/s
distance covered by the motion; s = 3.5m
To get the acceleration, we will make use of Newton's third equation of motion which is;
v² = u² + 2as
Making a the subject, we have;
a = (v² - u²)/2s
Plugging in the relevant values to give;
a = (43² - 0)/(2 × 3.5)
a = 264.14 m/s²
The average acceleration of the ball during the throwing motion is 265.14m/s².
In order to get the acceleration, the Newton's third law of motion will be used. This will be:
v² = u² + 2as
We'll make a to be the subject of the formula and this will be:
a = (v² - u²) / 2s
We'll plug in the value into the equation and this will be:
a = (43² - 0) / (2 × 3.5)
a = 1849 / 7
= 264.14 m/s²
Therefore, the acceleration is 265.14m/s.
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(A) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other.
(B) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.
Answer:
A) -40° C and -40° F
B) 574.25° K and 574.25° F
Explanation:
see attachment for calculation and explanation
A turntable of radius R1 is turned by a circular rubberroller of radius R2 in contact with it at their outeredges. What is the ratio of their angular velocities,ω1 / ω2 ?
Answer:
The ratio is [tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]
Explanation:
From the question we are told that
The first radius is [tex]R_1[/tex]
The second radius is [tex]R_2[/tex]
Generally the angular speed of the turntable is mathematically represented as
[tex]w_1 = \frac{ v_k }{R_1 }[/tex]
Generally the angular speed of the rubber roller is mathematically represented as
[tex]w_2 = \frac{ v_k }{R_2 }[/tex]
Where [tex]v_k[/tex] is the velocity of both turntable and rubber roller
So
[tex]\frac{w_1}{w_2} = \frac{\frac{v_k}{R_1} }{\frac{v_k}{R_2} }[/tex]
[tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]
A 18-kg hammer strikes a nail at a velocity of 7.6 m/s and comes to rest in a time interval of 8.3 ms .
Required:
a. What is the impulse given to the nail?
b. What is the average force acting on the nail?
Answer:
(a) -136.8 Ns.
(b) -1.135 N
Explanation:
(a)
Impulse: This can be defined as the change in momentum.
From the question,
I = mv-mu.................. Equation 1
Where I = impulse, m = mass of the hammer, v = final velocity, u = initial velocity.
Given: m = 18 kg, u = 7.6 m/s, v = 0 m/s (to rest)
Substitute these values into equation 1
I = 18(0)-18(7.6)
I = -136.8 Ns.
(b)
Average force = It.............. Equation 2
Where t = time.
Given: t = 8.3 ms = 0.0083 s.
Average force = -136.8(0.0083)
Average force = -1.135 N
A ball bearing of radius of 1.5 mm made of iron of density
7.85 g cm is allowed to fall through a long column of
glycerine of density 1.25 g cm. It is found to attain a
terminal velocity of 2.25 cm s-'. The viscosity of glycerine is
Answer:
[tex] \boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise} [/tex]
Given:
Radius of ball bearing (r) = 1.5 mm = 0.15 cm
Density of iron (ρ) = 7.85 g/cm³
Density of glycerine (σ) = 1.25 g/cm³
Terminal velocity (v) = 2.25 cm/s
Acceleration due to gravity (g) = 980.6 cm/s²
To Find:
Viscosity of glycerine ([tex] \sf \eta [/tex])
Explanation:
[tex] \boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}[/tex]
[tex] \sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v} [/tex]
Substituting values of r, ρ, σ, v & g in the equation:
[tex] \sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \times 64.7196[/tex]
[tex]\sf \implies \eta = 2 \times 7.191[/tex]
[tex]\sf \implies \eta = 14.382 \: poise[/tex]
A 60.5-kg hiker starts at an elevation of 1280 m and climbs to the top of a peak 2570 m high.
(a) What is the hiker's change in potential energy?
(b) What is the minimum work required of the hiker?
(c) Can the actual work done be greater than this? Explain.
Answer:
A) Change in potential energy is approximately 766 KJ
B) The minimum work required by the hiker is 765 KJ
C). Yes, the actual work can be greater than this
Explanation:
A) The hiker's potential energy can be calculated at the two different elevations, and then subtracted.
P.E at 1280 m = m X g X h = 60.5kg X 9.81 m/s2 X 1280 = 759, 686 J
P.E at 2570 m = m X g X h = 60.5kg X 9.81 m/s2 X 2570 = 1, 525, 307.85 J
Change in P.E = 1, 525, 307.85 J -759, 686 J = 765, 621.85 J = 766 KJ
B) Work which was done = Force X distance moved.
The force can be got from the effect of gravity on the hiker's mass = 60.5 kg X 9.81 = 593.5 Newton.
The distance moved can be obtained by subtracting the two elevations = 2570 -1280 = 1290 m
Work done = 593 X 1290 = 765, 615 J
The minimum work required by the hiker is 765 KJ
C) Yes, the actual work can be greater than this. This is because we can now put into consideration the fact that the hiker is climbing up the elevation against the force of gravity. This will mean that the hiker is actually doing more work than if he covered the distance on flat terrain,
Name the Sl base units that are ilnportant in chem-istry. Give the Sl units for expressing the following:
(a) length.
(b) volume.
(c) mass,
(d) time,
(e) energy,
(f) temperature
Answer:
Explanation:
SI unit of
length=meter
volume =dm^3
mass =kilogram
time=second
energy= joule
temperature =kelvin
Exercise 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides along a floor (you wish to find c). You have a spring with spring constant k 5 N/m. You take the spring, you attach it to the mass and fix it to a wall. Then you pull on the spring and let the mass go. You find that the mass oscillates with frequency 1 Hz. What is the friction
Answer:
b = 0.6487 kg / s
Explanation:
In an oscillatory motion, friction is proportional to speed,
fr = - b v
where b is the coefficient of friction
when solving the equation the angular velocity has the form
w² = k / m - (b / 2m)²
In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction
let's call
w₀² = k / m
w² = w₀² - b² / 4m²
b² = (w₀² -w²) 4 m²
Let's find the angular velocities
w₀² = 5 / 0.1
w₀² = 50
w = 2π f
w = 2π 1
w = 6.2832 rad / s
we subtitute
b² = (50 - 6.2832²) 4 0.1²
b = √ 0.42086
b = 0.6487 kg / s
The coefficient friction of the mass during the measurement is 0.648 kg/s.
The given parameters;
mass, m = 0.1 kgspring constant, k = 5 N/mfrequency of the mass, F = 1 HzDuring oscillatory motion, friction is directly proportional to speed.
[tex]F_k = -vb[/tex]
where;
b is the coefficient of frictionThe angular velocity is given as;
[tex]\omega ^2 = \frac{k}{m} - \frac{b^2}{4m^2} \\\\\omega ^2 = \omega _0^2 - \frac{b^2}{4m^2}\ \ ---\ (1)[/tex]
From the equation above, we will have the following;
[tex]\omega_0^2 = \frac{k}{m} \\\\\omega_0^2 = \frac{5}{0.1} \\\\\omega_0^2 = 50[/tex]
Also, the instantaneous angular speed is calculated as;
[tex]\omega = 2\pi f\\\\\omega = 2\pi \times 1\\\\\omega = 2\pi\\\\\omega = 6.284 \ rad/s[/tex]
From equation (1), the coefficient of friction is calculated as follows;
[tex]\omega ^2 = \omega ^2_0 - \frac{b^2}{4m^2} \\\\ \frac{b^2}{4m^2} = \omega ^2_0 - \omega ^2 \\\\b^2 = 4m^2( \omega ^2_0 - \omega ^2)\\\\b= \sqrt{ 4m^2( \omega ^2_0 - \omega ^2)}\\\\b = \sqrt{ 4\times 0.1^2\times ( 50 - 6.284^2)}\\\\b = 0.648 \ \ kg/s[/tex]
Thus, the coefficient friction of the mass during the measurement is 0.648 kg/s.
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an electric field of magnitude 200 N/C in the positive x- direction. calculate the acceleration in (m/s^2) of a charged particle of mass 1g and charge 1mC that is released from rest in this field?
1/200 that should get your
answer
A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875 Hz. a) what is the max speed of that point in SI units? b) what is the max acceleration of the point in SI units?
Using
V = Amplitude x angular frequency(omega)
But omega= 2πf
= 2πx875
=5498.5rad/s
So v= 1.25mm x 5498.5
= 6.82m/s
B. .Acceleration is omega² x radius= 104ms²
Answer:
a
[tex]v _{max } = 6.82 \ m/s[/tex]
b
[tex]a_{max} = 37489.5 \ m/s^2[/tex]
Explanation:
From the question we are told that
The amplitude is [tex]A = 1.24 \ mm = 1.24 * 10^{-3} \ m[/tex]
The frequency is [tex]f = 875 \ Hz[/tex]
Generally the maximum speed is mathematically represented as
[tex]v _{max } = A * 2 * \pi * f[/tex]
=> [tex]v _{max } = 1.24*10^{-3} * 2 * 3.142 * 875[/tex]
=> [tex]v _{max } = 6.82 \ m/s[/tex]
Generally the maximum acceleration is mathematically represented as
[tex]a_{max} = A * (2 * \pi * f)[/tex]
=> [tex]a_{max} = 1.24*10^{-3} * (2 * 3.142 * 875 )^2[/tex]
=> [tex]a_{max} = 37489.5 \ m/s^2[/tex]
Which of the following statements is true regarding electromagnetic waves traveling through a vacuum?
a. All waves have the same wavelength.
b. All waves have the same frequency.
c. All waves have the same speed.
d. The speed of the waves depends on their wavelength.
e. The speed of the waves depends on their frequency.
Answer:
C. All waves have the same speed.
Explanation:
Wave equation is given as;
V = fλ
where;
V is the speed of the wave
f is the frequency of the wave
λ is the wavelength
The speed of the wave depends on both wavelength and frequency
The speed of the electromagnetic waves in a vacuum is 3 x 10⁸ m/s, this also the speed of light which is constant for all electromagnetic waves.
Therefore, the correct option is "C"
C. All waves have the same speed.
Which planet orbits in a different plane than all of the others?
Although they're all 'close', none of the planets orbits in the same plane as any other planet. They're all in slightly different planes.
The farthest out compared to all the others is Pluto, with an orbit inclined about 17 degrees compared to the ecliptic plane (Earth's orbit). But Pluto is officially not a planet, so I don't think it's a good answer.
The next greatest inclination compared to Earth's orbit is Mercury. That one is about 7 degrees.
The other six planets are all in different orbital planes inclined less than 7 degrees compared to Earth's orbit.
What is the wavelength (in 10-15 m) of a proton traveling at 13.2% of the speed of light?
Answer:
The wavelength is [tex]\lambda = 10.01 *10^{-15} \ m[/tex]
Explanation:
From the question we are told that
The speed is [tex]v = 0.132 c[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
Generally the wavelength is mathematically represented as
[tex]\lambda = \frac{h}{m* v }[/tex]
where m is the mass of the proton with the value [tex]m = 1.6726 ^{-27} \ kg[/tex]
h is the Planck's constant with value [tex]h = 6.626 *10^{-34} \ J\cdot s[/tex]
=> [tex]\lambda = \frac{6.626 *10^{-34}}{1.6726 *10^{-34}* 0.132*3.0*10^8 }[/tex]
=> [tex]\lambda = 10.01 *10^{-15} \ m[/tex]
The force of gravity will make it easier to stop your car if you are going uphill, and more difficult to stop your car if you are going downhill.
If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged?
Answer:
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
Explanation:
The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.
The first minimum occurs for m = 1, so the diffraction equation of a slit remains
a sin θ = λ
in general, the diffraction patterns occur at very small angles, so
sin θ = θ
θ = λ / a
in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.
θ = 1.22 λ /a
In this exercise we are told that the opening changes
a’ = 2 a
we substitute
θ ‘= 1.22 λ / 2a
θ' = (1.22 λ / a) 1/2
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
Monochromatic light with wavelength 588 nm is incident on a slit with width 0.0351 mm. The distance from the slit to a screen is 2.7 m. Consider a point on the screen 1.3 cm from the central maximum. Calculate (a) θ for that point, (b) α, and (c) the ratio of the intensity at that point to the intensity at the central maximum.
Answer:
0.276
0.9
0.756
Explanation:
Given that
Wavelength of the light, λ = 588 nm
Distance from the slit to the screen, L = 2.7 m
Width of the slit, a = 0.0351 mm
a point on the screen, y = 1.3 cm = 0.013 m
Sinθ = y/L
Sinθ = 0.013/2.7
sinθ = 0.0081
θ = sin^-1 0.00481
θ = 0.276°
α = (π.a.sinθ)/λ
α = (3.142 * 3.51*10^-5 * sin 0.276) / 588*10^-9
α = 5.3*10^-7 / 588*10^-9
α = 0.9 rad
I/i(m) = ((sinα)/α)²
I/I(m) = ((sin 0.9) / 0.9)²
I/I(m) = (0.783/0.9)²
I/I(m) = 0.87²
I/I(m) = 0.756
Note, our calculator has to be set in Rad instead of degree for part C, to get the answer
how many atoms of oxygen are there in one molecule of cardon dioxide , if the chemical formula is CO2
Answer:
2 oxygen 1 carbon
Explanation:
Answer:
2 oxygen, 1 carbon
Explanation:
A beak ball is thrown in an arc and in 2.0s its shadow on the ground travels 180 m in a straight line. What is the average speed of the shadow
Answer: 90 m/s
Explanation:
The formula for speed is distance/time. The distance is 180 m and the time it took for the ball to travel is 2 s.
180 m/2 s = 90 m/s
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object
Answer:
a). C = b/2 and C = b/4
b). [tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]
c). m = 63.4 kg (approx.)
Explanation:
Ex. 2.4.4
The total force acting on mass m is [tex]$ F = F_{spring }= -kx $[/tex] , where x is the displacement from the equilibrium position.
The equation of motion is [tex]$ m {\overset{..}x} + kx = 0 $[/tex]
or [tex]$ {\overset{..}x}+ \frac{k}{m}x=0 $[/tex] or [tex]$ {\overset{..}x} + w_0^2 x = 0 $[/tex] , where [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex]
The solution is [tex]$ x = A \cos (w_0t + \phi) $[/tex] , where A and Ф are constants.
A is amplitude of motion
[tex]$ w_0$[/tex] is the angular frequency of motion
Ф is the phase angle.
Now, [tex]$ w_0 = 2 \pi f_0 = \sqrt{k/m} $[/tex]
or [tex]$ m = \frac{k}{4\pi f_0^2} $[/tex]
Given [tex]$ f_0 = 0.8 Hz , k = 4 N/m $[/tex]
a). [tex]$ m = \frac{4}{4(3.14)^2(0.8)^2} = 0.158\ kg$[/tex]
b). [tex]$ w_0^2 = k/m $[/tex]
or [tex]$ m = k/ w_0^2 = k / (2\pi f_0)^2 = k / 4 \pi^2 f_0^2 $[/tex]
Ex. 2.4.5
a). Total force acting on the mass m is [tex]$F = F_{spring}+f $[/tex]
[tex]$ = -kx-bv $[/tex]
The equation of motion is [tex]$ m {\overset{..}x}= -kx-b{\overset{.}x} $[/tex]
or [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex] , angular frequency of the undamped oscillation.
γ = b/2m is called the damping coefficient (γ=C)
[tex]$ k = m w_0^2 = 4 \pi^2 m f_0^2 $[/tex]
for 1 kg weight (= 9.8 N), [tex]$ f_0$[/tex] = 1.1 Hz
k = 4 x (3.14)² x (9.8) x 1.1² = 4.6 x 10² N/m
For 2 kg weight (= 19.6 N), [tex]$ f_0$[/tex] = 0.8 Hz
k = 4 x 9.8596 x 2 x 9.8 x 0.8² = 5 x [tex]$ 10^7$[/tex] N/m
[tex]$ \gamma = \frac{b}{2m_1} = \frac{b}{2m_2} $[/tex]
or [tex]$ \gamma = \frac{b}{2 \times 1} = \frac{b}{2 \times 2} $[/tex]
γ = b/2 (for 1 kg) and γ = b/4 (for 2 kg)
C = b/2 and C = b/4
b). [tex]$ w_0^2 = \frac{k}{m} \Rightarrow \frac{k}{w_0^2} = \frac{k}{(2 \pi f_0)^2} = \frac{k}{4 \pi^2 f_0^2} $[/tex]
For two particle problem,
[tex]$ w'_0^2 = \sqrt{\frac{k(m_1+m_2)}{m_1 +m_2}} $[/tex]
[tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]
where, μ is the reduced mass.
This time period is same for both the particles.
c). [tex]$ m =\frac{k}{4 \pi^2 f_0^2}$[/tex]
[tex]$ = \frac{5 \times 10^2}{4 \times 9.14^2 \times 0.2} = 63.4\ kg $[/tex] ( approx.)
Solve for A if F=MA and F=100 and M=25?
Answer:
[tex] \boxed{ \boxed{ \bold{A = 4}}}[/tex]Explanation:
Given,
F = 100 , M = 25
Now, let's find the value of A
[tex] \sf{F = MA}[/tex]
plug the values
⇒[tex] \sf{100 = 25 A}[/tex]
Swap the sides of the equation
⇒[tex] \sf{25A = 100}[/tex]
Divide both sides of the equation by 25
⇒[tex] \sf{ \frac{25A}{25} = \frac{100}{25} }[/tex]
Calculate
⇒[tex] \sf{A = 4}[/tex]
Hope I helped!
Best regards!!
Given:-
Force,F = 100 NMass,m = 25 kgTo find out:-
Calculate the acceleration, a ?
Formula applied:-
F = m × a
Solution:-
We know,
[tex] \sf{F = m × a}[/tex]
Substituting the values of mass and acceleration,we get
[tex] \sf\implies \: 100 = 25 \times a[/tex]
[tex] \sf \implies a = \cancel \dfrac{100}{25} [/tex]
[tex] \sf \implies a = 4 \: ms {}^{ - 1} [/tex]
What amount of heat is required to increase the temperature of 75.0 grams of gold from 150°C to 250°C? The specific heat of gold is 0.13 J/g°C.A. 750 joulesB. 980 joulesC. 1300 joulesD. 1500 joulesE. 2500 joules
Answer:
B. 980 joulesExplanation:
Given the following data
initial temperature T1= 150 °C
final temperature T2= 250 °C
specific heat of gold c= 0.13 J/g°C
mass of gold m= 75.0 grams
we can use the expression stated below to solve for the quantity of heat
[tex]Q= mc(T2-T1)---------1[/tex]
Substituting our known data into the expression we can solve for the value of Q
[tex]Q= 75*0.13(250-150)---------1\\\\Q= 75*0.13(100)\\\\Q= 975 Joules[/tex]
The quantity of heat need to raise the temperature from 150°C to 250°C is 975 J
Answer:
B. 980 joules
Explanation:
A meter stick is supported by a pivot at its center of mass. Assume that the meter stick is uniform and that the center of mass is at the 50 cm mark.
a) If a mass m1 = 80 g is suspended at the 30 cm mark, at which cm mark would a mass m2 = 110 g need to be suspended for the system to be in equilibrium?
b) If a mass m1=80g is suspended at the 25cm mark,and a mass m2 =110g is suspended at the 60 cm mark, from what cm mark would a mass m3 = 45 g need to be suspended for the system to be in equilibrium?
Answer:
a) 800N × 20 cm = 1100N × x cm
16000= 1100x
x= 14.5
therefore it must be placed on the (50 + 14.5)cm mark
= 64.5 cm mark
b) 800N × 25 cm = (1100N × 10 cm)+(450N × x cm)
20000 = 11000 + 450x
450x = 9000
x = 20 cm
therefore it must be placed on the (50 + 20)cm mark
= 70 cm mark
a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.
b) The distance at which mass m₃(45 g) should be suspended is 70 cm.
What is meant by principle of moments?According to the Principle of Moments, when a body is balanced or is at equilibrium, the total clockwise and anticlockwise moments about a given point are equal.
a) m₁ = 80 g
m₂ = 110 g
r₁ = 30 cm
According to the Principle of Moments,
m₁r₁ = m₂r₂
Therefore, the distance,
r₂ = m₁r₁/m₂
r₂ = 80 x 20/110
r₂ = 14.54 cm
So, the distance at which mass m₂ should be suspended is,
r' = 50 + 14.54
r' = 64.54 cm
b) m₁ = 80 g
m₂ = 110 g
m₃ = 45 g
r₁ = 25 cm
r₂ = 60 cm
According to the Principle of Moments,
m₁r₁ = m₂r₂ + m₃r₃
80 x 25 = (110 x 10) + (45 x r₃)
45 x r₃ = 2000 - 1100
r₃ = 900/45
r₃ = 20 cm
So, the distance at which mass m₃ should be suspended is,
r' = 50 + 20
r' = 70 cm.
Hence,
a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.
b) The distance at which mass m₃(45 g) should be suspended is 70 cm.
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Suppose the width of your fist is 4.1 inches and the length of your arm is 35.4 inches. Based on these measurements, what will be the angular width (in degrees) of your fist held at arm’s length?
Answer:
7 degree
Explanation:
given data
width = 4.1 inches
length = 35.4 inches
solution
we consider as per fig
O is mid point of BC
so OB = 2.05 inches
and
AB = [tex]\sqrt{OB^2 + OA^2}[/tex]
AB = [tex]\sqrt{2.05^2 + 35.4^2}[/tex]
AB = 35.078 inches
so
[tex]sin \frac{\alpha }{2} = \frac{OB}{AB}[/tex]
[tex]sin \frac{\alpha }{2} = \frac{2.05}{35.078}[/tex]
[tex]\alpha = 7 degree[/tex]
an electromagnetic wave has an electric field with peak value 120. What is the averge energy delievered to a surface
Answer:
The average energy delivered to a surface is 19.116 W/m².
Explanation:
Given;
maximum electric field, E₀ = 120 v/m
The average energy delivered by the wave to a surface is given by
[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2}[/tex]
where;
c is the speed of light, = 3 x 10⁸ m/s
ε₀ is the permittivity of free space = 8.85 x 10⁻¹² c²/Nm²
[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2} \\\\I_{avg} = \frac{(3*10^8)(8.85*10^{-12})( 120)^2}{2}\\\\ I_{avg} =19.116 \ W/m^2[/tex]
Therefore, the average energy delivered to a surface is 19.116 W/m².
I don’t understand can someone break it down for me
Answer:
a = (v² – v₀²)/ 2(s – s₀)
Explanation:
v² = v₀² + 2a (s – s₀)
We can make 'a' the subject of the above expression as follow:
v² = v₀² + 2a (s – s₀)
Subtract v₀² from both side
v² – v₀² = v₀² + 2a (s – s₀) – v₀²
v² – v₀² = v₀² – v₀² + 2a (s – s₀)
v² – v₀² = 2a (s – s₀)
Divide both side by (s – s₀)
(v² – v₀²)/ (s – s₀) = 2a
Divide both side by 2
(v² – v₀²)/ (s – s₀) ÷ 2 = a
(v² – v₀²)/ (s – s₀) × 1/2 = a
(v² – v₀²)/ 2(s – s₀) = a
a = (v² – v₀²)/ 2(s – s₀)
A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i v, start subscript, i, end subscript. Later, its speed increases to 4v_i4v i 4, v, start subscript, i, end subscript. How does the magnitude of the car's centripetal acceleration change after the linear speed increases
Answer and Explanation: Centripetal Acceleration is the change in velocity caused by a circular motion. It is calculated as:
[tex]a_{c}=\frac{v^{2}}{r}[/tex]
v is linear speed
r is radius of the curve the object in traveling along
For its first lap:
[tex]a_{c}_{1}=\frac{v_{i}^{2}}{R}[/tex]
After a while:
[tex]a_{c}_{2}=\frac{(4v_{i})^{2}}{R}[/tex]
[tex]a_{c}_{2}=\frac{16v_{i}^{2}}{R}[/tex]
Comparing accelerations:
[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]
[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]
[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=16[/tex]
[tex]a_{c}_{2}=16a_{c}_{1}[/tex]
With linear speed 4 times faster, centripetal acceleration is 16 times greater.
A large, cylindrical water tank with diameter 3.00 m is on a platform 2.00 m above the ground. The vertical tank is open to the air and the depth of the water in the tank is 2.00 m. There is a hole with diameter 0.420 cm in the side of the tank just above the bottom of the tank. The hole is plugged with a cork. You remove the cork and collect in a bucket the water that flows out the hole.
A. When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank? Express your answer in millimeters.
B. How long does it take you to collect 1.00 gal of water in the bucket? Express your answer in seconds.
Answer:
1999.46 mm
45.59 s
Explanation:
given that
cylindrical water tank with diameter, D = 3 m
Height of the tank above the ground, h = 2 m
Depth of the water in the tank, d = 2 m
Diameter of hole, d = 0.420 cm
We start by calculating the volume of water in the tank, which is given as
Volume = πr²h
V = (πD²)/4 * h
V = (3.142 * 3²)/4 * 2
V = 28.278/4 * 2
V = 7.07 * 2
V = 14.14 m³
If 1.0 gal of water is equal to 0.0038m³, then
1 gal is 0.0038 = A * h
the area of the tank is 7.07 m²
therefore, 0.0038 = 7.07 * h
h₁ =0.00054 m = 0.54 mm is the height of water that flow out
the change in height of water in the tank = h - h₁ = 2 - 0.00054 = 1.99946 m
b)
Like we stated earlier, 1.0 gal of water is 0.0038m³
to solve this we use the formula
Q = Cd * A * √2gH
where Cd is a discharge coefficient, and is given by 0.9 for water
A is the area of the small hole
A = (πD²)/4
A = (π * 0.0042²)/4
A = 5.54*10^-5 / 4
A = 1.39*10^-5 m²
H= height of the hole from the tank water level = 2m - 0.0042 = 1.9958 m
g = 9.8 m/s²
Q = 0.9 * 1.39*10^-5 m² * √2 * 9.8 * 1.9958
Q = 1.251*10^-5 * 6.25
Q = 7.82*10^-5 m³/s
Q = V/t
t = V/Q = 0.0038m³ / 7.82*10^-5 m³/s
t = 45.59 s
If an electron in an atom moves from an energy level of 5 to an energy level of 10:____.
a. a photon of energy 5 is absorbed.
b. a photon of energy 15 is absorbed.
c. a photon of energy 5 is emitted.
d. a photon of energy 15 is emitted.
Answer
Answer:
a. a photon of energy 5 is absorbed
Explanation:
Because when an electron in a lower energy state absorbs energy, in form of photons it moves to higher energy stage in this case 5 photons because it moved from 5 to 10
A rectangular object was found to have a mass of 1.278 kg and density of 4.98 g/cm3. Suppose that you knew that the length was 47 mm and the width was 61 mm. Using this information, compute the height of the rectangle in cm.
Answer:
89.6 cm
Explanation:
From the question,
Volume of the rectangular object = Mass/Density.
V = m/D.................. Equation 1
Given: m = 1.278 kg, D = 4.98 g/cm³ = 4980 kg/m³
Substitute into equation 1
V = 1.278/4980
V = 2.57×10⁻⁴ m³.
But,
V = lwh............... Equation 2
Where l = length of the rectangular object, w = width of the rectangular object, h = height of the rectangular object.
make h the subject of the equation
h = V/lw........... Equation 3
Given: V = 2.57×10⁻⁴ m³, l = 0.047 m, w = 0.061 m.
Substitute into equation 3
h = 2.57×10⁻⁴/(0.047×0.061)
h = 0.896 m
h = 89.6 cm
(4)
The electric field inside an uncharged metal sphere is initially zero. If the sphere is
then charged positively, the field at the center of the sphere will be :
A) zero
B) finite and directed radially inward
C) nearly infinite
D) finite and directed radially outward
Answer:
Option (A) : Zero
Explanation:
Electric field Intensity inside a metallic body is always ZERO.