Answer:
586 K
Explanation:
Let P is the initial pressure.
Initial temperature, T₁ = 20°C = 293 K
Final pressure, P₂ = 2P
We need to find the final temperature of the gas.
The relation between the pressure and the temperature is as follows
[tex]P\propto T\\\\or\\\\\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}[/tex]
Put all the values,
[tex]\dfrac{P}{2P}=\dfrac{293}{T_2}\\\\\dfrac{1}{2}=\dfrac{293}{T_2}\\\\T_2=2\times 293\\\\T_2=586\ K[/tex]
So, the final temperature of the gas is 586 K.
A metre rule is used to measure the length of a piece of string in a certain experiment. It is found to be 20 cm long to the nearest millimeter. How should thisresult be recorded in a table of results? a. 0.2000m b. 0.200m c. 0.20m d. 0.2m
Answer:
C
Explanation:
20 cm = 0.2m
since uncertainty is 0.1 cm (0.001 m), should be recorded to same number of decimal place as uncertainty
therefore it's 0.200m
A TV satellite dish is designed to receive radio waves of wavelength
0.0644 meters. What is the frequency of the waves it receives? _______GHz
Give your answer in gigahertz (GHz). 1 GHz = 10^9 Hz.
Give your answer to the nearest tenth of a GHz (one place after the decimal). Just enter the number; do NOT use scientific notation.
Answer:
4.7 GHz
Explanation:
Applying,
v = λf................. Equation 1
Where v = velocity of the radio wave, λ = wavelength, f = frequency
make f the subject of the equation
f = v/λ.............. Equation 2
Note: A radio wave is an electromagnetic wave, as such it moves with a velocity of 3.00 x 10⁸ m/s
From the question,
Given: λ = 0.0644 meters
Constant: v = 3.00 x 10⁸ m/s
Substitute these values into equation 2
f = (3.00 x 10⁸)/0.0644
f = 4.66×10⁹ Hz
f = 4.7 GHz
If the cornea is reshaped (this can be surgically done or with contact lenses) to correct myopia, should its curvature be made greater or smaller? Explain. Also, explain how hyperopia can be corrected?
Answer:
Myopia curvature of the cornea if it is negative the curvatures are positive,
hypermetry,
Explanation:
Myopia is the visual defect that does not allow to see distant objects, which is why it is corrected with a divergent lens so that the image is formed on the retina, therefore, by reforming the curvature of the cornea if it is negative
therefore the curvature must decrease
To correct hypermetry, the curvatures are positive, consequently the curvature of the lens must increase
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.
Stationary waves are
A) transverse waves
B) longitudinal waves
C) mechanical waves
Answer:
stationary waves are transverse waves
The image of an object placed 30cm from a diverging lens is formed 10cm in front of the lens.
Calculate the focal length of the lens.
Answer:
15cm
Explanation:
Since the lens is a diverging lens, the image distance is negative (virtual)
v = -30cm
u = 10cm
Required
focal length f
Using the lens formula;
1/u + 1/v = 1/u
1/10 - 1/30 = 1/f
(3-1)/30 = 1/f
2/30 = 1/f
f = 30/2
f = 15cm
Hence the focal length of the lens is 15cm
Electrical charges are of two types. True False
Answer:
Electrical charges r of 2 types its true.they are positive and negative.hope it helps.stay safe healthy and happy..Answer: Think its true
An object moving along a horizontal track collides with and compresses a light spring (which obeys Hooke's Law) located at the end of the track. The spring constant is 52.1 N/m, the mass of the object 0.250 kg and the speed of the object is 1.70 m/s immediately before the collision.
(a) Determine the spring's maximum compression if the track is frictionless.
?? m
(b) If the track is not frictionless and has a coefficient of kinetic friction of 0.120, determine the spring's maximum compression.
??m
(a) As it gets compressed by a distance x, the spring does
W = - 1/2 (52.1 N/m) x ²
of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object's displacement). By the work-energy theorem, this work is equal to the change in the object's kinetic energy. At maximum compression x, the object's kinetic energy is zero, so
W = ∆K
- 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.118 m
(b) Taking friction into account, the only difference is that more work is done on the object.
By Newton's second law, the net vertical force on the object is
∑ F = n - mg = 0
where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives
n = mg = 2.45 N
and from this we get the magnitude of kinetic friction,
f = µn = 0.120 (2.45 N) = 0.294 N
Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:
W (friction) = - (0.294 N) x
W (spring) = - 1/2 (52.1 N/m) x ²
==> W (total) = W (friction) + W (spring)
Solve for x :
- (0.294 N) x - 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.112 m
For the 0.250 kg object moving along a horizontal track and collides with and compresses a light spring, with a spring constant of 52.1 N/m, we have:
a) The spring's maximum compression when the track is frictionless is 0.118 m.
b) The spring's maximum compression when the track is not frictionless, with a coefficient of kinetic friction of 0.120 is 0.112 m.
a) We can calculate the spring's compression when the object collides with it by energy conservation because the track is frictionless:
[tex] E_{i} = E_{f} [/tex]
[tex] \frac{1}{2}m_{o}v_{o}^{2} = \frac{1}{2}kx^{2} [/tex] (1)
Where:
[tex]m_{o}[/tex]: is the mass of the object = 0.250 kg
[tex]v_{o}[/tex]: is the velocity of the object = 1.70 m/s
k: is the spring constant = 52.1 N/m
x: is the distance of compression
After solving equation (1) for x, we have:
[tex] x = \sqrt{\frac{m_{o}v_{o}^{2}}{k}} = \sqrt{\frac{0.250 kg*(1.70 m/s)^{2}}{52.1 N/m}} = 0.118 m [/tex]
Hence, the spring's maximum compression is 0.118 m.
b) When the track is not frictionless, we can calculate the spring's compression by work definition:
[tex] W = \Delta E = E_{f} - E_{i} [/tex]
[tex] W = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} [/tex] (2)
Work is also equal to:
[tex] W = F*d = F*x [/tex] (3)
Where:
F: is the force
d: is the displacement = x (distance of spring's compression)
The force acting on the object is given by the friction force:
[tex] F = -\mu N = -\mu m_{o}g [/tex] (4)
Where:
N: is the normal force = m₀g
μ: is the coefficient of kinetic friction = 0.120
g: is the acceleration due to gravity = 9.81 m/s²
The minus sign is because the friction force is in the opposite direction of motion.
After entering equations (3) and (4) into (2), we have:
[tex]-\mu m_{o}gx = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2}[/tex]
[tex]\frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} + \mu m_{o}gx = 0[/tex]
[tex] \frac{1}{2}52.1 N/m*x^{2} - \frac{1}{2}0.250 kg*(1.70)^{2} + 0.120*0.250 kg*9.81 m/s^{2}*x = 0 [/tex]
Solving the above quadratic equation for x
[tex] x = 0.112 m [/tex]
Therefore, the spring's compression is 0.112 m when the track is not frictionless.
Read more here:
https://brainly.com/question/14245799?referrer=searchResultshttps://brainly.com/question/16857618?referrer=searchResultsI hope it helps you!
g How much buoyancy force, in N, a person with a mass of 70 kg experiences by just standing in air
Answer:
686.7N
Explanation:
Given data
Mass= 70kg
We know that the buoyant force experienced by the person is equal to the weight of the person
Hence the weight is
Weight = mass* Acceleration
Weight= 70*9.81
Weight= 686.7N
Therefore the weight is 686.7N
Please show steps as to how to solve this problem
Thank you!
Answer:
Torques must balance
F1 * X1 = F2 * Y2
or M1 g X1 = M2 g X2
X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7
X2 = 22.4 cm
Torque = F1 * X2 =
62.3 gm* 980 cm/sec^2 * 22.4 cm = 137,000 gm cm^2 / sec^2
Normally x cross y will be out of the page
r X F for F1 will be into the page so the torque must be negative
A magnetohydrodynamic (MHD) drive works by applying a magnetic field to a fluid which is carrying an electric current.
a. True
b. False
Answer:
True
Explanation:
A magnetohydrodynamic drive or MHD accelerator is a method which is used for propelling the vehicles using only by applying the electric and magnetic fields. It has no moving parts. It accelerates an electrically conductive propellant (liquid or gas) with magnetohydrodynamics.
Its working principle is same as an electric motor except that in an MHD drive, the moving rotor is replaced by the fluid acting directly as the propellant.
An MHD accelerator is reversible.
So, the statement is true.
This diagram shows the magnetic field lines near the ends of two magnets. There is an error in the diagram.
Two bar magnet with the north pole of one near the south pole of the second. field lines are leaving the north pole and bent away from the south pole of the other. Field lines are leaving the south pole of one and bending away from the north pole of the other.
Which change will correct the error in the diagram?
a)changing the N to S
b)reversing the arrows on the left to point toward the N
c)changing the S to N
d)reversing the arrows on the right to point toward the S
Answer:
changing the N to S. that's how the error will be corrected
Answer:
C is the correct answer
Explanation:
i took the test
Assuming the atmospheric pressure is 1 atm at sea level, determine the atmospheric pressure at Badwater (in Death Valley, California) where the elevation is 86.0 m below sea level.
Answer:
Atmospheric pressure at Badwater is 1.01022 atm
Explanation:
Data given:
1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa
Elevation (h) = 86m
gravity (g) = 9.8 m/s2
Density of air P = 1.225 kg/m3
Therefore pressure at bad water Pb = Pi + Pgh
Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)
Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa
hence:
Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm
An electric heater is madde of a wire of resistance 100π and connected to a 240v mains supply. Determine the power rating of the heater
Answer:
Power = 576 Watts
Explanation:
The electrical power of an electric circuit can be defined as a measure of the rate at which energy is either produced or absorbed in the circuit.
Mathematically, electrical power is given by the formula;
[tex] Electrical \; power = current * voltage [/tex]
This ultimately implies that, the quantity (current times voltage ) is electrical power and it is measured (S.I units) in Watt (W).
Given the following data;
Resistance = 100 ohms
Voltage = 240 V
To find the power rating of the heater;
Power = V²/R
Where;
V is the voltage.
R is the resistance.
Substituting into the formula, we have;
Power = 240²/100
Power = 57600/100
Power = 576 Watts
what is Friction
short note on friction
Answer:
Explanation:
Friction can be defined as a force that resists the relative motion of two objects when there surface comes in contact. Thus, it prevents two surface from easily sliding over or slipping across one another. Also, friction usually reduces the efficiency and mechanical advantage of machines but can be reduced through lubrication.
Generally, there are four (4) main types of friction and these includes;
I. Static friction.
II. Rolling friction.
III. Sliding friction.
IV. Fluid friction.
crushing chalk into powder is and irreversible change. is this example a physical or chemical change?Why?
Answer:
It is a example of physical change
Question 4(Multiple Choice Worth 4 points)
(02.04 MC)
Which explanation justifies why the theory of evolution is a theory and not a law?
Predicts an organism's ability to adapt to its environment
It can be expressed as a simple mathematical statement
Explains the existence of diverse forms of life on Earth
O Additional evidence will change the theory into a law
Answer:
A(predicts an organisms ability to adapt to its enviroment, it is not a fact that each organization can adapt)
Explanation:
An ice chest at a beach party contains 12 cans of soda at 3.78 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 6.48-kg watermelon at 29.4 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.
Answer:
T = 13.25°C
Explanation:
From the law of conservation of energy:
Heat Lost by Watermelon = Heat Gained by Cans
[tex]m_wC_w\Delta T_w = m_cC_c\Delta T_c[/tex]
where,
[tex]m_w[/tex] = mass of watermelon = 6.48 kg
[tex]m_c[/tex] = mass of cans = (12)(0.35 kg) = 4.2 kg
[tex]C_w[/tex] = specific heat capacity of watermelon = 3800 J/kg.°C
[tex]C_c[/tex] = specific heat capacity of cans = 4200 J/kg.°C
[tex]\Delta T_w[/tex] = Change in Temprature of watermelon = 29.4°C - T
[tex]\Delta T_c[/tex] = Change in Temperature of cans = T - 3.78°C
T = final temperature = ?
Therefore,
[tex](4.2\ kg)(3800\ J/kg.^oC)(29.4^oC-T)=(6.48\ kg)(4200\ J/kg^oC)(T-3.78^oC)\\469224\ J-(15960\ J/^oC)T = (27216\ J/^oC)T-102876.48\ J\\469224\ J + 102876.48\ J = (27216\ J/^oC)T+(15960\ J/^oC)T\\\\T = \frac{572100.48\ J}{43176\ J/^oC}[/tex]
T = 13.25°C
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 generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the generator has an area of 0.016 m2. If the magnetic field used in the generator has a magnitude of 0.052 T, how many turns of wire are needed
Answer:
The number of turns of wire needed is 573.8 turns
Explanation:
Given;
maximum emf of the generator, = 190 V
angular speed of the generator, ω = 3800 rev/min =
area of the coil, A = 0.016 m²
magnetic field, B = 0.052 T
The number of turns of the generator is calculated as;
emf = NABω
where;
N is the number of turns
[tex]\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s[/tex]
[tex]N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns[/tex]
Therefore, the number of turns of wire needed is 573.8 turns
When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of static friction between the car and the road is 0.88. Assuming your car doesn't skid, what is the force exerted on it by static friction?
Answer:
7772.72N
Explanation:
When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.
Now which direction is the static friction, assume that it is pointing inward so
Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N
Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.
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.
A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is the spring constant of the bungee cord, assuming it has negligible mass compared to that of the jumper in N/m
Answer:
2.80N/m
Explanation:
Given data
mass m= 56kg
perios T= 11.2s
The expression for the period is given as
T=2π√m/k
Substitute
11.2= 2*3.142*√56/k
square both sides
11.2^2= 2*3.142*56/k
125.44= 351.904/k
k=351.904/125.44
k= 2.80N/m
Hence the spring constant is 2.80N/m
Newspapers often talk about an energy crisis-about running out of certain energy sources in the not-so-distant future. About which kind of energy sources are they talking
Answer:
Nonrenewable energy
Explanation:
Renewable energy is also known as clean energy and it can be defined as a type of energy that are generated through natural sources or technology-based processes that are replenished constantly. Some examples of these natural sources are water (hydropower), wind (wind energy), sun (solar power), geothermal, biomass, waves etc.
Basically, a renewable energy source is sustainable and as such can not be exhausted.
On the other hand, a non-renewable energy refers to an energy source such as fossil fuels that takes a very long time to be created or their creation happened long ago and isn't likely to happen again e.g uranium.
For example, fossil fuels such as coal, oil, and natural gas, come from deep inside the Earth where they formed over millions of years ago.
In this scenario, the kind of energy the newspaper sources are talking about is a nonrenewable energy source because they are capable of being exhausted in the not-so-distant future.
A rigid tank contains 10 lbm of air at 30 psia and 60 F. Find the volume of the tank in ft3. The tank is now heated until the pressure doubles. Find the heat transfer in Btu.
Answer:
Hence the amount of heat transfer is 918.75 Btu.
Explanation:
Now,
what is the average velocity if the initial velocity is at rest and the final velocity is 16 m/s
Answer:
8m/s
Explanation:
Vavg= 16-0/2=8m/s
You place an 8 kg ball on the top of your 2 cm^2 finger tip. Calculate the
PRESSURE. Show MATH, answer and unit.
Answer:
the pressure exerted by the object is 392,000 N/m²
Explanation:
Given;
mass of the object, m = 8 kg
area of your finger, A = 2 cm² = 2.0 x 10⁻⁴ m²
acceleration due to gravity, g = 9.8 m/s²
The pressure exerted by the object is calculated as;
[tex]Pressure = \frac{F}{A} = \frac{mg}{A} = \frac{8 \times 9.8}{2\times 10^{-4}} = 392,000 \ N/m^2[/tex]
Therefore, the pressure exerted by the object is 392,000 N/m²
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resistance. But driving slowly in dry sand is another story. If a 1500 kg car is driven in sand at 4.9 m/s , the coefficient of rolling friction is 0.060. In this case, nearly all of the energy that the car uses to move goes to overcoming rolling friction, so you can ignore air drag in this problem.
Required:
a. What propulsion force is needed to keep the car moving forward at a constant speed?
b. What power is required for propulsion at 5.0 m/s?
c. If the car gets 15 mpg when driving on sand, what is the car's efficiency? One gasoline contains 1.4×10 ^8 J of chemical energy.
Answer:
a) [tex]F_p=882N[/tex]
b) [tex]P=4410W[/tex]
c) [tex]V_p'=24135[/tex] ,[tex]n=15.2\%[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=1500kg[/tex]
Velocity [tex]v=4.9m/s[/tex]
Coefficient of Rolling Friction [tex]\mu=0.06[/tex]
a)
Generally the equation for The Propulsion Force is mathematically given by
[tex]F_p=\mu*mg[/tex]
[tex]F_p=0.06*1500*9.81[/tex]
[tex]F_p=882N[/tex]
b)
Therefore Power Required at
[tex]V_p=5.0m/s[/tex]
[tex]P=F_p*V_p[/tex]
[tex]P=882*5[/tex]
[tex]P=4410W[/tex]
c)
[tex]V_p' =15mpg[/tex]
[tex]V_p'=15*\frac{1609}[/tex]
[tex]V_p'=24135[/tex]
Generally the equation for Work-done is mathematically given by
[tex]W=F_p*V_p'[/tex]
[tex]W=882*15*1609[/tex]
[tex]W=2.13*10^7[/tex]
Therefore
Efficiency
[tex]n=\frac{W}{E}*100\%[/tex]
Since
Energy in one gallon of gas is
[tex]E=1.4*10^8J[/tex]
Therefore
[tex]n=\frac{2.1*10^7}{1.4*10^8}*100\%[/tex]
[tex]n=15.2\%[/tex]
Two projectiles A and B are fired simultaneously from a level, horizontal surface. The projectiles are initially 62.2 m apart. Projectile A is
fired with a speed of 19.5 m/s at a launch angle 30° of while projectile B is fired with a speed of 19.5 m/s at a launch angle of 60°. How long
it takes one projectile to be directly above the other?
Let the point where A is launched act as the origin, so that the horizontal positions at time t of the respective projectiles are
• A : x = (19.5 m/s) cos(30°) t
• B : x = 62.2 m + (19.5 m/s) cos(60°) t
These positions are the same at the moment one projectile is directly above the other, which happens for time t such that
(19.5 m/s) cos(30°) t = 62.2 m + (19.5 m/s) cos(60°) t
Solve for t :
(19.5 m/s) (cos(30°) - cos(60°)) t = 62.2 m
t = (62.2 m) / ((19.5 m/s) (cos(30°) - cos(60°))
t ≈ 8.71 s
12) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen, what things could you do to help yourself (before calling me over to assist you?)
Answer:
See the answer below
Explanation:
After seeing an object on a slide at the low-power objective of the microscope and it disappears on changing to high power, the following can be done to resolve the problem
1. Drop a few drops of immersion oil on the slide and view again under high the power objective.
2. If the object is still not visible after the action above, return the microscope to the low-power objective and make sure the object is refocused and centered. Then carefully change back to the high power objective and use the fine adjustment to bring it into focus.
