Answer:
Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing.
Explanation:
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Answer:
Explanation:
a. the sound energy is converted into electrical signals using diaphragm present in the microphone and these signals then reach to the speakers and then convert- ed back to sound. The electrical current generated by a micro- phone is very small and referred to as MIC-level; this signal is typically measured in millivolts.
b. Thermodynamics
A 1510 kg car is experiencing a 2650 N friction force from the road. What force must be applied to the car in order for the car to move forward with a constant velocity of 50.0 km/hr?
Answer:
The force that must be applied to maintain a constant velocity is equal o the friction force which is 2,650 N
The total force required from rest is (2,650·t + 20,972.[tex]\overline 2[/tex])/t N
Explanation:
The given data of the car in motion are;
The mass of the car, m = 1,510 kg
The force of friction from the road, [tex]F_f[/tex] = 2,650 N
When the car is moving at 50.0 km/hr, we have;
v = 50.0 km/hr = 250/9 m/s ≈ 13.88889 m/s
According to Newton's first Law of motion, a body will continue in a state of rest or in uniform motion along a straight line unless acted on by a force
The force acting on the car in motion = 2,650 N
When the car is moving at a constant velocity, the forces acting on the car are in equilibrium and the net force is zero
Therefore;
The magnitude of the force that will keep the car moving at a constant velocity = The friction force of the road = 2,650 N
The direction of the force = In the direction of motion of the car
Where the time it takes the car to accelerate from rest to 50.0 km/hr = t seconds, we have;
The force required to move at a constant speed = 1,510 kg × (250/9 m/s - 0)/t s = 20,972.[tex]\overline 2[/tex]/t N
The total force required to move forward with a constant velocity of 50.0 km/hr becomes 20,972.[tex]\overline 2[/tex]/t N + 2,650 N = (2,650·t + 20,972.[tex]\overline 2[/tex])/t N
Where;
t = The time it takes the car to reach 50.0 km/hr from rest.
Why unit of pascal is called derived unit?
Answer:
This is because it is derived from newtons and metres as shown in the expression;
Explanation:
[tex]pressure = \frac{force}{area} \\ P= \frac{F}{A} \\ P= \frac{newtons}{ {m}^{2} } \\ = pascals[/tex]
Answer:
Unit of pascal is called derived unit because it involves three fundamental units i.e.meter,kg,and second.
Explanation:
formula of pascal=kg/m/s^2
state and explain any effect on sensitivity of liquid in a glass thermometer (i) reducing the diameter of capillary tube
Answer:
please give me brainlist and follow
Explanation:
The measuring sensitivity of liquid-in-glass thermometers increases with the amount of liquid in the thermometer. The more liquid there is, the more liquid will expand and rise in the glass tube. For this reason, liquid thermometers have a reservoir to increase the amount of liquid in the thermometer.
Force=k((q1q1)/r^2)
The force that one electrically charged object exerts on other charged objects can be calculated using Coulomb's law.
You can see this law here written in equation form. The q's represent the two charges, and the r the distance between
the charges. The constant kis 9109
Using this information, calculate the electric force exerted by a +2 coulomb charge on another +2 coulomb charge,
when the charges are 25 meters apart
A)1.44*10^11
B)5.76*10^11
C)8 Newton’s
D)64 Newton’s
Answer:
transferred from one object to another. Charged objects can exert forces on other charged ... Electric charge can be measured using the law for the forces between charges
The electric force exerted by the +2 C on another +2 C when at a distance of 25 m is 5.76×10⁷ N
Coulomb's law equationF = Kq₁q₂ / r²
Where
F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart How to determine the force Charge 1 (q₁) = 2 CCharge 2 (q₂) = 2 CElectric constant (K) = 9×10⁹ Nm²/C²Distance apart (r) = 25 mForce (F) =?F = Kq₁q₂ / r²
F = (9×10⁹ × 2 × 2) / (25)²
F = 36×10⁹ / 625
F = 5.76×10⁷ N
Learn more about Coulomb's law:
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pls somebody help me quick
Answer:
greater than
Explanation:
A bowling ball is bigger than a golf ball and greater than means bigger.
A uniform 1.5 T magnetic field points north. If an electron moves perpendicular (90 degrees) with a speed of 2.5x10 ^7 m/s, through this field, what force (magnitude and direction) will act on it?
Answer:
i think its 470
Explanation:
Ganymede, the largest of Jupiter’s moons,
is also the largest satellite in the solar system. Find the acceleration of
gravity on Ganymede if a simple pendulum with a length of 1.00 m has a period
of 10.5 s
Answer:hereeee for pointssss only
Explanation:
The acceleration of gravity on Ganymede will be [tex]g=0.35\ \frac{m}{s^2}[/tex]
What will be the acceleration of gravity on Ganymede?It is given that for a simple pendulum on Ganymede
Lenth= 1.00 m
the time period T= 10.5 sec
The formula for a time period of simple pendulum will be
[tex]T=2\pi\sqrt{\dfrac{l}{g} }[/tex]
By rearranging the formula
[tex]g=l\times (\dfrac{2\pi}{T} )^2[/tex]
[tex]g=1\times (\dfrac{2\pi}{10.5} )^2=0.3577\frac{m}{s^2}[/tex]
[tex]g=0.3577\ \frac{m}{s^2}[/tex]
Thus the acceleration of gravity on Ganymede will be [tex]g=0.35\ \frac{m}{s^2}[/tex]
To know more about simple pendulums follow
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A cart moving at 4.0 m/s has 44 J of kinetic energy. What is the cart's mass?
[tex]ke = \frac{1}{2} m {v}^{2} [/tex]
[tex]44 = \frac{m {4}^{2} }{2} \\ \frac{88}{16} = m \\ 5.5 = m[/tex]
Atmospheric pressure is measured by a device called
Answer:
barometer
Explanation:
Answer:
barometer
Explanation:
A student performed an experiment in the lab and discovered that the material used has a PH of 8.54. What is true about this material
a
it is a base
b
it is an acid
c
more information needed to know if the material is acid or base
Which of these statements describes protons and electrons?
The correct option is: "none of these statements is true".
To find out the correct option among all the options, we need to know more about the electron and proton.
What is electron?Electron is the fundamental particle of an atom.It is negatively charged.It is also a quantum particle. So, it can't stay at a place according to uncertainty principle i.e. it can move. What is proton?Proton is a fundamental particle of an atom present inside the nucleus.It's positively charged.It can move.Thus, we can conclude that the option (e) is correct.
Learn more about electron and proton here:
https://brainly.com/question/894705
#SPJ2
When do the heavier elements finally form? Give a few example.
Answer:
Some of the universe's heavier elements are created by neutron star collisions. ... Light elements like hydrogen and helium formed during the big bang, and those up to iron are made by fusion in the cores of stars. Some heavier elements like gallium and bromine need something more, such as a supernova.
Explanation:
MARK ME BRAINLIEST PZZZZZZZZZzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzA projectile is launched from the ground with an initial velocity of 12m/s at an angle of 30° above the horizontal. The projectile lands on a hill 7.5m away. The height at which the projectile lands is most nearly
Answer:
[tex]1.78\:\mathrm{m}[/tex]
Explanation:
Kinematics equations used:
[tex]v_f=v_i+at,\\\\\Delta y = v_it+\frac{1}{2}at^2,\\\\v_f^2=v_i^2+2a\Delta x[/tex]
Other equations used:
[tex]\Delta x =v_i\cdot t,\\\text{-basic trigonometry for right triangles},\\\text{-application of physics concepts}[/tex]
By finding the horizontal component of the projectile's initial launch, we can find how long it will take for the projectile to reach the hill.
Since the projectile is launched at 12 m/s 30 degrees above the horizontal, we can use basic trig for a right triangle to find the horizontal component of the launch:
[tex]\cos 30^{\circ}=\frac{x}{12},\\x=12\cos 30^{\circ}\approx 10.3923048454\:\text{m/s}[/tex]
Now we can use [tex]\Delta x =v_x\cdot t[/tex] to find the time it will take to reach the hill:
[tex]7.5=10.3923048454\cdot t,\\t=\frac{7.5}{10.3923048454}\approx 0.72168783648\:\text{s}[/tex]
To find the total flight time the projectile would take if it were to land back to the same height it started out with, we can use the following kinematics equation:
[tex]v_f=v_i+at[/tex]
We'll use basic trig for a right triangle again to find the vertical component of the velocity at launch:
[tex]\sin 30^{\circ}=\frac{y}{12},\\y=12\sin 30^{\circ}=6\:\text{m/s}[/tex]. (recall [tex]\sin 30^{\circ}=\frac{1}{2}[/tex])
Now plugging this value in:
*Note: the projectile will hit the ground with a vertical velocity equal in magnitude but opposite in direction than the projectile's initial vertical velocity at launch
[tex]-6=6+(-9.8\cdot t),\\t=\frac{-12}{-9.8},\\t=1.22448979592\:\text{s}[/tex]
^This is the total flight time the projectile would take if it were to land back to the same height it started with. Since acceleration is constant, we can divide this time by two to find the time it would take to reach maximum height.
[tex]1.22448979592\div 2=0.61224489795\:\text{s}[/tex]
Thus, it will take [tex]0.61224489795\:\text{s}[/tex] to reach maximum height, meaning that it will take [tex]0.72168783648 - 0.61224489795=0.10944293853\:\text{s}[/tex] to land on the hill after reaching maximum height. Now that we've found this, we can use the following kinematics equation to find the vertical displacement from the projectile's max height to the hill:
[tex]\Delta y =v_it+\frac{1}{2}at^2[/tex]
At maximum height, the projectile's vertical velocity will be zero. Therefore, [tex]v_i=0[/tex] and we have:
[tex]\Delta y =\frac{1}{2}at^2[/tex]
Plugging in values , we have:
[tex]\Delta y =\frac{1}{2}\cdot 9.8\cdot 0.10944293853^2=0.05869100829\:\text{m}[/tex]
This means that the hill is [tex]0.05869100829\:\text{m}[/tex] lower than the max. height the object will reach.
To find the height of the hill from the ground, we will use the following kinematics equation:
[tex]v_f^2=v_i^2+2a\Delta x,\\-6^2=0+2\cdot-9.8\cdot\Delta x,\\36=-19.6\Delta x,\\\Delta x =\frac{36}{-19.6},\\\Delta x =-1.83673469388\:\text{m}[/tex]
Therefore, the max. height that the object will reach is [tex]1.83673469388\:\text{meters}[/tex].
Thus, the hill's height is [tex]1.83673469388-0.05869100829=1.77804368559\:\text{m}\approx\boxed{1.78\:\mathrm{m}}[/tex].
An object is 15.0 cm from a convex lens. Its image is 12.0 cm from the lens on the other side. What is the focal length of the lens?
Question 22 options:
1.25 cm
8.6 cm
6.67 cm
27.0 cm
Answer:
6.67
Explanation:
A ray of light passes from air into cubic zirconia at an angle of 56.0° to the normal. The angle of
refraction is 22.00°.
What is the index of refraction of cubic zirconia?
Answer:
η = 2.2
Explanation:
The index of refraction is given by the following formula:
[tex]\eta = \frac{Sin\ \theta_i}{Sin\ \theta r}[/tex]
where,
η = index of refraction of cubic zirconia = ?
[tex]\theta_i[/tex] = angle of incidence = 56°
[tex]\theta_r[/tex] = angle of refraction = 22°
Therefore,
[tex]\eta = \frac{Sin\ 56^o}{Sin\ 22^o}\\\\\eta = \frac{0.829}{0.375}[/tex]
η = 2.2
A child on a sled has a combined mass of 97 kg. At the top of a 3.1 meter hill, the sled has a velocity of 8 m/s. Assuming there is no friction, what would the velocity of the child and sled be when they reach the top of a different 1.2 meter hill?
Answer:
Explanation:
This is an energy conservation problem in that the total energy available to a system s constant trhoughout the whole problem. That is, in equation form:
TE = PE + KE. This is a bit of a problem due to sig figs here, but I will keep with the rules for them all the way up til the end of the problem (I'll tell you when I veer away from the rule and why I did when I get there).
We first need to find the total energy available to the system and then use that value throughout the rest of the problem.
TE = PE + KE where
PE = mgh (mass times gravity times height of the object) and
KE = [tex]\frac{1}{2}mv^2[/tex] (one-half times the mass of the object times the square of the velocity of the object). We solve for PE first, rounding to the correct number of sig figs:
PE = 97(9.8)(3.1) and
PE = 2900 J (2 sig figs here since all the numbers given have 2 sig figs in them). Now for KE
[tex]KE=\frac{1}{2}(97)(8.0)^2[/tex] which, to 2 sig figs (notice I added a .0 to the 8), is
KE = 2200. This means that the total energy available to the sled throughout the whole trip is
TE = 2900 + 2200 so
TE = 5100 J
Now for the second part of the problem, the sled is at a different height so the PE is different, and we are asked to find the velocity at the top of this second hill. The total energy equation then is
5100 = PE + KE. Solving for PE first:
PE = (97)(9.8)(1.2) and, to 2 sig figs:
PE = 82. Now for the KE:
[tex]KE=\frac{1}{2}(97)v^2[/tex]. Plugging all of that into the total energy equation gives us:
[tex]5100=82+\frac{1}{2}(97)v^2[/tex] and isolating the v:
[tex]v=\sqrt{\frac{2(5100-82)}{97} }[/tex] and this is where I veer away from the sig figs. Doing the subtracting first, as we should, means that we would round to the hundreds place. Since the least significant place in the number 82 is the tens place and the least significant place in the 5100 is in the hundreds place, we round to the hundreds place. But I didn't keep to this rule because I'm not sure how adamant your physics teacher is about sig figs. I'll give you the answer that I would expect from my students (the right one) after I give you this one. I just simply subtracted 5100 - 82 and used the answer the calculator gave me which was 5018:
[tex]v=\sqrt{\frac{2(5018)}{97} }[/tex] gives us that the velocity at the top of the 1.2 meter hill is
10.2 m/s.
Doing this the right way, using the rules for sig figs properly:
[tex]v=\sqrt{\frac{2(1500)}{97} }[/tex] gives us, to 2 sig figs,
v = 1.0 × 10¹ m/s
Question 3.Two forces are acting on a 30-kg object, 200 N up and 300 N
down. What is the magnitude and direction of its acceleration?
A. It is 100 m/s2 upward,
B. It is 100 m/s2 downward.
C. It is 3 m/s2 upward.
D. It is 3 m/s2 downward.
Answer:
It is 3 m/s² downward.
Explanation:
The mass of an object, m = 30 kg
Upward force, F = 200 N
Downward force, F' = -300 N
We need to find the magnitude and direction of its acceleration.
Net force = F+F'
= 200+(-300)
= -100 N
We know that,
Force, F = ma, a is acceleration of the object
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{-100}{30}\\\\a=-3.33\ m/s^2\approx -3\ m/s^2[/tex]
Hence, the acceleration is -3 m/s² and it is acting in the downward direction.
A submarine is at an ocean depth of 3.00 km.
(a) Calculate the absolute pressure at this depth. Assume that the density of water is 1.025 ✕ 103 kg/m3 and that atmospheric pressure is 1.01 ✕ 105 Pa.
Pa
(b) Calculate the total force exerted at this depth on a circular submarine window with a diameter of 35.0 cm.
N
Answer:
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Explanation:
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Ms Jo Jo rubbed two balloons with a piece of wool. What will happen when the balloons are brought near each other?
A traveler at Newark airport pulls his 294N suitcase with constant velocity to the check-in counter applying a force of 159N at an angle of 37 degrees with the horizontal. Calculate the Normal force, frictional force and the coefficient of friction between the suitcase and the airport floor.
Answer:
N = 198.3 N, fr = 126.98 N and μ = 0.64
Explanation:
For this exercise we must use the translational equilibrium condition.
Let's set a reference system with the x-axis horizontally and the y-axis vertical. Let's use trigonometry to decompose the applied force
cos θ = Fₓ / F
sin θ = F_y / F
Fₓ = F cos θ
F_y = F sin θ
Fₓ = 159 cos 37 = 126.98 N
F_y = 159 sin 37 = 95.69 N
Y axis
N + F_y - W = 0
N = W - F_y
N = 294 - 95.69
N = 198.3 N
X axis
Fₓ -fr = 0
Fₓ = fr
fr = 126.98 N
the expression for the friction force is
fr = μ N
μ = fr / N
μ = 126.98 / 198.3
μ = 0.64
[tex] \underline{ \text{QUESTIONS}} : [/tex]
1. Write any three applications of electromagnetic waves. Complete the given ray diagram [ See the attached picture ]
2. The power of a bulb of car is 100 W. If 5A current flows through the filament of the bulb , calculate the voltage of the battery used in the car [ Ans : 20 V ] Please show your workings!
~Random/Irrelevant answers will be reported!
Answer:
Solution given:
any three applications of electromagnetic waves are:
Used for cooking purpose in microwave.Used in RADAR system for aircraft navigation.Used in communication.2.
Solution given;
power [P]=100W
Current [I]=5A.
Voltage [V]=?
we have
P=IV
100=5*V
V=[tex] \frac{100}{5} [/tex]=20V
the voltage of the battery used in the car is 20V.
Answer:
Explanation:
1. Electromagnetic waves are used to broadcast radio and TV signals. It is also used for wireless charging cell phones. Cell phone itself uses electromagnetic waves for its cell signal.
See attached drawing for ray diagram.
2. Given: the power of a bulb of car is 100 W and 5A current flows through the filament of the bulb.
Power of a bulb is given by voltage * current.
So 100W = voltage * 5A
Voltage = 100/5 = 20V
Which conclusions about alcohol and teen driving are supported by the data provided? Check all that apply. Rates of drinking and driving increased significantly between 2004 and 2014. Approximately 22 percent of teens admitted to driving after drinking in 2004. Fewer teens admitted to drinking and driving in 2013 than in 2004. In 2013, about one in five teens admitted to riding with a driver who had been drinking.
Answer: • Fewer teens admitted to drinking and driving in 2013 than in 2004.
• In 2013, about one in five teens admitted to riding with a driver who had been drinking.
Explanation:
The conclusions about alcohol and teen driving that are supported by the data provided will be:
• Fewer teens admitted to drinking and driving in 2013 than in 2004.
• In 2013, about one in five teens admitted to riding with a driver who had been drinking.
Other options are incorrect as the conclusion wasn't gotten from the data that was orovufed
Answer:
Fewer teens admitted to drinking and driving in 2013 than in 2004.
In 2013, about one in five teens admitted to riding with a driver who had been drinking.
Explanation:
A baseball is thrown at a 22.5° angle and an initial velocity of 65 m/s. What will be its horizontal
and vertical velocity?
Guys pls someone answer my question I have an exam and this question has 5 marks
Answer:
Horizontal component = 60.05 m/s
vertical component = 22.96 m/s
Explanation:
cosine(angle) = adjacent / hypotenuse or horizontal / angled velocity
angled velocity * cosine (angle) = horizontal velocity
65 m/s * cosine(22.5) = 60.05 m/s
sine(angle) = adjacent/ hypotenuse or vertical/ angled velocity
angled velocity * sine (angle) = vertical velocity
65m/s sine(22.5) = 22.96 m/s
can anyone help me please
Answer:
Gravity.
Rocket ships.
Ball.
Basketball.
Explanation:
Gravity has to do a lot with air. It puts the planets in there area.
Rocket Ship has to do a lot with air. If i'm right, they calculate the area, weather, about the air.
A ball gets throwed in the air, which gravity comes into place.
Basketball is also a similar example to a ball.
When light is incident on a boundary between two mediums, it often slows
down or speeds up. When it does this, the path the light travels can change.
This is called:
A. diffraction
B. reflection.
c. translocation.
D. refraction
Answer:
the answer is
D.refraction
How do we calculate efficiency?
Explanation:
Efficiency is often measured as the ratio of useful output to total input, which can be expressed with the mathematical formula r=P/C, where P is the amount of useful output ("product") produced per the amount C ("cost") of resources consumed.
i dont know what the answer really is but this is what i think
Calculate the critical angle for Zircon which has a refractive index of n = 2*
Answer:
Given that refractive index of the material is √2. i.e. n = √2. Hence, critical angle for the material is 45°
Mercury -197 is used for kidney scans and has a half-life of 3 days. If the amount of mercury-197 needed for a
study is 1.0 gram and the time allowed for shipment is 15 days, how much mercury-197 will need to be ordered?
Answer:
32 g
Explanation:
Applying,
R = (R')2ᵃ/ⁿ............... Equation 1
Where R = Amount of mercury-197 needed to be ordered, R' = amount of mercury-197 needed for the study, a = time allowed for shipment, n = half life.
From the question,
Given: R' = 1 g, a = 15 days, n = 3 days
Substitute these values into equation 1
R = 1(2¹⁵/³)
R = 1(2⁵)
R = 32 g
The different between capactance wnd capacitor
Answer:
The parallel plate capacitor is the simplest form of capacitor. ... The property of a capacitor to store charge on its plates in the form of an electrostatic field is called the Capacitance of the capacitor. Not only that, but capacitance is also the property of a capacitor which resists the change of voltage across it.
Explanation:
credits :- adultsscience
Explanation:
Capacitance exists wherever conductive material is separated by insulating material.
Capacitive structures have the ability to store energy as an electric field;
when a capacitive structure has been designed as an electrical component that has a specified amount of capacitance, it is called a capacitor.
Capacitance symbol is C
you can get a meter for Capacitance
Capacitance formula is C = ε(A/d) where ε represents the absolute permittivity of the dielectric material being used.
simple piece of wire has very low capacitance