Answer:
Density → Mass and Length.
Acceleration → Length and Time.
Force → Mass, Length and Time.
Pressure → Mass, Length and Time.
Work → Mass, Length and Time.
Mains electricity is an ac supply. Explain the difference between direct and alternating potential difference.
Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction.
#LETS STUDYThe decrease in the intensity of light over distance, such as with increasing depth in water, is known as ______.
Answer:
Attenuation
Step By step Explanation:
The decrease in the intensity of light over distance, such as with increasing depth in water, is known as ______.
It is called attenuation.
The decraese in the amplitude of the signals is called attenuation.
(d)
The signals for the monitor unit are transmitted as electromagnetic waves with a
wavelength of 0.125 m.
Wave speed of electromagnetic waves = 3 * 108 m/s
Calculate the frequency of the signal.
Answer:
Frequency = 24 × 10⁸ Hz
Explanation:
Given the following data;
Speed = 3 × 10⁸ m/s
Wavelength = 0.125 meters
To find the frequency of the electromagnetic wave;
Mathematically, the speed of a wave is given by the formula;
Speed = Wavelength × frequency
Substituting into the formula, we have;
3 × 10⁸ = 0.125 × frequency
Frequency = (3 × 10⁸)/0.125
Frequency = 24 × 10⁸ Hz
A cylindrical container closed of both end has a radius of 7cm and height of 6cm A.)find the total surface area of the container B.) find the volume of the container
A marble is rolling across a smooth 1.2 m tall table at a velocity of 3 m/s. How far from the edge of the table does it land? (SHOW ALL OF YOUR WORK)
Answer:
S = 1/2 g t^2 where t is the time to fall 1.2 m
t = (2 S / g)^1/2 = (2 * 1.2 / 9.8) = .495 s
Sy = Vy T = 3 m/sec * .495 sec = 1.48 m distance from edge of table
(Rotational speed has no effect since table is smooth)
three condensers are connected in series across a 150 volt supply. The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10^-8(a) calculate the capacitance of each condenser (b)calculate the effective capacitance of the combination
Explanation:
Given that,
The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.
(a) Capacitance of capacitor 1,
[tex]C_1=\dfrac{Q}{V_1}\\\\C_1=\dfrac{6\times 10^{-8}}{40}\\\\C_1=1.5\times 10^{-9}\ F\\\\C_1=1.5\ nF[/tex]
Capacitance of capacitor 2,
[tex]C_2=\dfrac{Q}{V_2}\\\\C_2=\dfrac{6\times 10^{-8}}{50}\\\\C_2=1.2\times 10^{-9}\ F\\\\C_2=1.2\ nF[/tex]
Capacitance of capacitor 3,
[tex]C_3=\dfrac{Q}{V_3}\\\\C_3=\dfrac{6\times 10^{-8}}{60}\\\\C_3=1\times 10^{-9}\ F\\\\C_3=1\ nF[/tex]
(b) The equivalent capacitance in series combination is :
[tex]\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\\\\dfrac{1}{C}=\dfrac{1}{1.5}+\dfrac{1}{1.2}+\dfrac{1}{1}\\\\C=0.4\ nF[/tex]
Hence, this is the required solution.
A projectile is fired horizontally from a gun that is 58.0 m above flat ground, emerging from the gun with a speed of 170 m/s. (a) How long does the projectile remain in the air
Answer:
t = 3.44 s
Explanation:
We are given;
Fired from rest, and so; u = 0 m/s
Final speed; v = 170 m/s
Height above flat ground; y_o = 58 m
Height at starting point; y = 0 m
Thus, from Newton's equation of motion, we have;
y - y_o = ut - ½gt²
(since it's motion is against gravity)
Plugging in the relevant values, we have;
0 - 58 = 0 - (½ × 9.8 × t²)
-58 = -4.9t²
t² = 58/4.9
t = √(58/4.9)
t = 3.44 s
A plane is flying a circular path at a speed of 55.0 m/ s, with a radius of 18.3 m. The centripetal force needed to maintain this motion is 3000 N. What is the plane's mass?
The plane has a centripetal acceleration a of
a = v ²/r
where v is the plane's tangential speed and r is the radius of the circle. By Newton's second law,
F = mv ²/r
Solve for the mass m :
m = Fr/v ² = (3000 N) (18.3 m) / (55.0 m/s)² ≈ 18.1 kg
Which items did Mendeleev write down on note cards while he was observing
different elements?
A. Buoyancy
B. Density
C. Color
D. Atomic mass
Mendeleev wrote down atomic mass
Answer:
A. Buoyancy.
hope it helps
stay safe healthy and happy..Fluorometers are designed so that the path of the excitation light is at a right angle to the path of the emitted light. What is the purpose of this design
Answer: prevent excitation light from reaching the detector
Explanation:
A fluorometer refers to the device that's used in the measurement of parameters that are of visible spectrum fluorescence. They also prevent excitation light from reaching the detector.
These parameters are used in the identification of the amount and presence of molecules in a medium.
A satellite of mass 5460 kg orbits the Earth and has a period of 6520 s
A)Determine the radius of its circular orbit.
B)Determine the magnitude of the Earth's gravitational force on the satellite.
C)Determine the altitude of the satellite.
Answer:
what if I do and b then someone else c I don't have enough time pls
What is the acceleration of a motorcycle that starts from rest and reaches a velocity of 24 m/s in 8.5 seconds?
Answer:
2.82 m/s²
Explanation:
[tex]v = u + at \\ 24 = 0 + a(8.5) \\ a = 2.82 \: ms {}^{ - 2} [/tex]
The acceleration of a motorcycle is 2.82 m/s^2.
What is acceleration?Acceleration is the rate at which speed and direction of velocity vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates.
Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration for all other motions.
Given that:
Initial velocity of the motor cycle: u = 0 m/s.
Final velocity of the motorcycle = 24 meter/second.
Time taken to reach this velocity = 8.5 second.
Hence, acceleration of the motor cycle = change in velocity/time interval
= ( final velocity - initial velocity)/time interval
= ( 24 m/s - 0 m/s)/8.5 s
= 2.82 m/s^2.
Its acceleration is 2.82 m/s^2.
Learn more about acceleration here:
brainly.com/question/12550364
#SPJ2
Example Problem
The potential energy of an object is given by U(x) = 8x2 - x4, where U is in joules and x is in
(a) Determine the force acting on this object.
(b) At what positions is this object in equilibrium?
(c) Which of these equilibrium positions are stable and which are unstable?
metres.
111 Unit 2 Concepts and Definitions Prof Mark Lester
Exam Part B Example
A neutron of mass m moving with velocity v collides head-on and elastically with a stationary nucleus of mass M.
(a) Show that the velocity of the nucleus after the collision, U, is given by
U= 2m v (m+M)
(b) Hence show that the neutron loses a fraction f of its energy where
f= 4mM (m+M)
10marks 5 marks
(c) A fast neutron enters a target of carbon nuclei which may be assumed to have masses 12 times that of the neutron. How many head-on collisions will it take
before the neutron loses 95% of its energy?
4 marks
(d) Suggest one reason why in a real reactor a neutron is likely to make more
collisions with the moderator nuclei before losing this much energy
2
1 mark
Answer:
Part A
a) F = -16x + 4, b) x = 0.25 m, c) STABLE
Explanation:
Part A
a) Potential energy and force are related
F = [tex]- \frac{dU}{dx}[/tex]- dU / dx
F = - (8 2x -4)
F = -16x + 4
b) The object is in equilibrium when the forces are zero
0 = -16x + 4
x = 4/16
x = 0.25 m
c) An equilibrium position is called stable if with a small change in position, the forces make it return to the initial position, in case the forces make it move away it is called unstable.
In this case there is only one equilibrium point
by changing the position a bit
x ’= x + Δx
we substitute
F ’= - 16 x’ + 4
F ’= - 16 (x + Δx) + 4
F ’= (-16x +4) - 16 Δx
at equilibrium position F = 0
F ’= 0 - 16 Δx
we can see that the body returns to the equilibrium position, therefore it is STABLE
PART B
This is an exercise in body collisions, let's define the system formed by the two bodies in such a way that the forces during the collisions are internal and the moment is conserved
initial instant. Before the shock
p₀ = m v
final instant. After the crash
p_f = (m + M) v_f
We have two possibilities: an elastic collision in which the bodies separate, each one maintaining its plus, and an INELASTIC collision where the neutron is absorbed by the nucleus and the final mass is M '= m + M, in this case they indicate that the collision is elastic
p₀ = pf
mv = mv ’+ M v_f
in the case of the elastic collision, the kinetic energy is conserved
K₀ = K_f
½ m v² = ½ m v’² + ½ M v_f²
we write the system of equations
mv = mv ’+ M v_f (1)
m (v² -v'²) = M v_f ²
m (v - v ’) = M v_f
m (v-v ’) (v + v’) = M v_f
v + v ’= v_f
we substitute in equation 1 and solve
v ’=[tex]\frac{m -M }{m+M } \ vo[/tex]
v_f = [tex]\frac{2m}{m+M} \ v_o[/tex]
the mechanical energy of the neutron is
initial
Em₀ = K = ½ m v²
final moment
Em_f = K + U = ½ m v_f ² + U
U is the energy lost in the collision
total energy is conserved
Em₀ = Em_f
½ m v² = ½ m v_f ² + U
U = ½ m (v² -v_f ²)
U = ½ m [v² - ( [tex]\frac{m-M}{m+M}[/tex] v)² ]
U = ½ m v² [1- ( [tex]\frac{m-M}{m+M}[/tex] )² ]
U = ½ m v2 [ [tex]\frac{2M}{m+M}[/tex]]
U = [tex]\frac{2 mM}{m +M } \ v^2[/tex]
Let's do the same calculations for the nucleus
initial Em₀ = 0
final Em_f = K + U = ½ M v_f ² + U
Em₀ = Em_f
0 = K + U
U = -K
U = - ½ M v_f ²
U = - ½ M [ [tex]\frac{2m}{m+M} \ v[/tex] ]²
U = [tex]\frac{2 m M }{m+M} \ v^2[/tex]
We can see that we obtain the same result, that is, the potential energy lost by the neutron is equal to the potential energy gained by the nucleus.
b) the fraction of energy lost
f = U / Em₀
f = 4 m M / m + M
c) let's calculate the fraction of energy lost in a collision
m = 1.67 10⁻²⁷ kg
M = 12 1.67 10⁻²⁷= 20 10⁻²⁷ kg
f = 4 1.6 20 / (1.6+ 20) 10⁻²⁷
f = 5.92 10⁻²⁷ J
the energy of a fast neutron is greater than 1 eV
Eo = 1 eV (1.67 10⁻¹⁹ J / 1eV) = 1.67 10⁻¹⁹ J
Let's use a direct portion rule if in a collision f loses in how many collisions it loses 0.95Eo
#_collisions = 0.95 Eo / f
#_collisions = 0.95 1.67 10⁻¹⁹ / 5.92 10⁻²⁷
#_collisions = 2.7 10⁷ collisions
A car is stationary. It accelerates at 0.8 ms^2
for 10 s and then at 0.4 ms^2
for a further 10 s. Use
the equations of motion to deduce the car’s final displacement. You will have to split the journey
into two parts, since the acceleration changes after 10 s.
Answer:
the car’s final displacement is 60 m
Explanation:
Given;
initail velocity of the car, u = 0
acceleration of the car, a = 0.8 m/s²
time of motion, t = 10 s
The first displacement of the car:
[tex]x_1 = ut + \frac{1}{2} at^2\\\\x_1 = 0 + \frac{1}{2} (0.8)(10)^2\\\\x_1 = 40 \ m[/tex]
The second displacement of the car;
acceleration, a = 0.4 m/s², time of motion, t = 10 s
[tex]x_2 = ut + \frac{1}{2} at^2\\\\x_2 = 0 + \frac{1}{2} (0.4)(10)^2\\\\x_2 = 20 \ m[/tex]
The final displacement of the car;
x = x₁ + x₂
x = 40 m + 20 m
x = 60 m
Therefore, the car’s final displacement is 60 m
what do you think will happen to the people in the airplane if their location is unknown?
Question:
What do you think will happen to the people in the airplane if their location is unknown?
My opinion:
I mean what are the circumstances? Is it just a rogue pilot that takes a random plane with all the passengers Hostage or is It bad whether an they lose connection to the airport? I Mean either way, If I was on a plane being held hostage, or the plane crashing, I Would be a in a panic type state of mind. I Would be worried about what will happen when the plane crashes, I would worry about my friends or family that is aboard the plane with me, I Would be scared for my life because I Don't know if i would make it out alive.
Also:
How would they know their location is unknown wouldn't the Pilot be the only one to know exactly where there with their navigating system, But then again there was this time I headed to Florida an They had mini tv's on the seats of each chair, where you could: Watch movies, listen to music, an Also see exactly where you are in the world. It was really amazing actually.
A boy im50kg at rest on a skateboard is pushed by another boy who exerts a force of 200 N on him. If the first boy's
final velocity is 8 m/s, what was the contact time?
seconds
Answer:
Time, t = 2 seconds
Explanation:
Given the following data;
Mass, m = 50 kg
Initial velocity, u = 0 m/s (since it's starting from rest).
Final velocity, v = 8 m/s
Force, F = 200 N
To find the time, we would use the following formula;
[tex] F = \frac {m(v - u)}{t} [/tex]
Making time, t the subject of formula, we have;
[tex] t = \frac {m(v - u)}{F} [/tex]
Substituting into the formula, we have;
[tex] t = \frac {50(8 - 0)}{200} [/tex]
[tex] t = \frac {50*(8)}{200} [/tex]
[tex] t = \frac {400}{200} [/tex]
Time, t = 2 seconds
A car's bumper is designed to withstand a 5.04-km/h (1.4-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.255 m while bringing a 890-kg car to rest from an initial speed of 1.4 m/s.
Answer:
3420.39 N
Explanation:
Applying,
Fd = 1/2(mv²-mu²)................. Equation 1
Where F = force on the bumber, d = distance, m = mass of the car, v = final velocity, u = initial velocity.
make F the subject of the equation
F = (mv²-mu²)/2d............... Equation 2
From the question,
Given: m = 890 kg, v = 0 m/s (to rest), u = 1.4 m/s, d = 0.255 m
Substitute these values into equation 2
F = [(890×0²)-(890×1.4²)]/(2×0.255)
F = -1744.4/0.51
F = -3420.39 N
The negative sign denotes that the force in opposite direction to the motion of the car.
In 'coin on card' experiment a smooth card is used.
Answer:
In coin card experiment smooth card is used so that the card can slide easily from glass
A ball tied on a string rotates in a circular path as shown above. The only forces acting on the ball at any point are the weight and of the string. What is the equation for the net centripetal force at point C?
Answer:
the third one T-W
Explanation:
the direction of the Tension and weight are opposite
A 25.0kg girl pushes a 50.0kg boy with a force of 100.0N. What is the acceleration of the girl?
Answer:
im pretty sure it should be 50.0
An ammeter with a resistance of 5.0 ohm is connected in series with a 3.0V cell and a lamp rated at 300 mA, 3V. Calculate the current that the ammeter will measure.
plz solve this, I'll mark you as brainliest
Answer:
I = 0.2 A
Explanation:
Lamp is rated at 300 mA
I_lamp = 300 mA = 0.3 A
Voltage is; V = 3V
Thus; Resistance is given by;
R = V/I
R = 3/0.3
R = 10 ohms
Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;
R_eq = 10 + 5
R_eq = 15 ohms
Ammeter current will be;
I = V/R_eq
I = 3/15
I = 0.2 A
An object 2cm high is placed 3cm in front of a concave lens of focal length 2cm, find the magnification?
Answer:
0.4
Explanation:
A concave lens is a diverging lens, so it will always have a negative focal length. Image distance is always negative for a concave lens because it forms virtual images.
From the lens formula;
1/f = 1/u+ 1/v
- 1/2 = 1/3 - 1/v
1/v = 1/3 + 1/2
v= 6/5
v= 1.2 cm
Magnification = image distance/object distance
Magnification = 1.2cm/3cm
Magnification = 0.4
please help.. i got it wrong on my last attempt
Answer:
The answer is C.
Explanation:
A drone traveling horizontally at 100 m/s over flat ground at an elevation of 3000 meters must drop an emergency package on a target on the ground. The trajectory of the package is given by x=100t , y=−4.9t2+3000 ,t≥0 where the origin is the point on the ground directly beneath the drone at the moment of release. How many horizontal meters before the target should the package be released in order to hit the target? Round to the nearest meter.
Answer:
The package should be dropped 244.7 meters before the target
Explanation:
We need to find how long the package will take to hit the ground then solve for position.
y=0
-4.9t^2=-3000
t= sqrt (3000/4.9)
t= 24.74 seconds.
x= 247.4 meters.
While traveling north on an expressway, a car traveling 60 mph (miles per hour) slows down to 30 mph in 12 minutes due to traffic conditions
Answer:
acceleration = - 150 m/s^2
distance = 9 miles.
Explanation:
initial speed, u = 60 mph
time, t = 12minutes = 0.2 hour
final speed, v = 30 mph
Let the acceleration is a and the distance is s.
By the first equation of motion
v = u + at
30 = 60 + a x 0.2
a = - 150 m/s^2
Let the distance is s.
Use third equation of motion is
[tex]v^2 = u^2 + 2 a s \\\\30^2 = 60^2 + 2 \times 150\times s\\\\s = 9 miles[/tex]
A water wave passes by a floating leaf that is made to oscillate up and down two complete cycles each second, which means that the wave's frequency is
Answer:
2 Hz.
Explanation:
Frequency is simply defined as the number of appearances of a periodic event occurring per time. It is usually measured in cycles/second.
Now, in this question, we are told that there are 2 cycles for each second.
Thus, we can say that the frequency is 2 cycles/1 s = 2 Hz.
Define standard 1 killogram
" standard 1 meter
" standard 1 second
Answer:
-standard 1 kg : Kilogram (kg), basic unit of mass in the metric system.
-standard 1 meter: The standard metre is the length of the path travelled by light in vaccum during a time interval of 1/299792458 of a second.
- standard 1 second : The second (abbreviation, s or sec) is the Standard International ( SI ) unit of time.
Explanation:
HOPE IT HELPS YOU !!
A squirrel jumps into the air with a velocity of 4 m/s at an angel of 50 degrees. What is the maximum height reached by the squirrel?
Answer:
Explanation:
Assuming the squirrel is jumping off the ground, here's what we know but don't really know...
v₀ = 4.0 at 50.0°
So that's not really the velocity we are looking for. We are dealing with a max height problem, which is a y-dimension thing. Therefore, we need the squirrel's upward velocity, which is NOT 4.0 m/s. We find it in the following way:
[tex]v_{0y}=4.0sin(50.0)[/tex] which gives us that the upward velocity is
v₀ = 3.1 m/s
Moving on here's what we also know:
a = -9.8 m/s/s and
v = 0
Remember that at the very top of the parabolic path, the final velocity is 0. In order to find the max height of the squirrel, we need to know how long it took him to get there. We are using 2 of our 3 one-dimensional equations in this problem. To find time:
v = v₀ + at and filling in:
0 = 3.1 - 9.8t and
-3.1 = -9.8t so
t = .32 seconds.
Now that we know how long it took him to get to the max height, we use that in our next one-dimensional equation:
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:
Δx = [tex]3.1(.32)+\frac{1}{2}(-9.8)(.32)^2[/tex] and using the rules for adding and subtracting sig fig's correctly, we can begin to simplify this:
Δx = .99 - .50 so
Δx = .49 meters
A stream leaving a mountain range deposits a large part of its load in a __
Answer:
(n) alluvial fan sandbar
Explanation:
Wavelengths of incoming solar radiation are __________________ the wavelengths of reradiated heat. Which term best completes the sentence
Explanation:
Hydraulic Pressure-Control, On-Off Deluge Valve
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