Answer: An jack makes changing a tire easier because it lifts up the car to get the tire off of the ground.
Explanation:
A honey bee's wings beat at 230 beats per second. If the speed of sound in air is 340 m/s, what is the wavelength of
the sound waves?
Answer:
[tex]from \: the \: wave \: equation \\ velocity = frequency \times wavelength \\ 340 = 230 \times \lambda \\ \lambda = \frac{340}{230} \\ \lambda = 1.5 \: m[/tex]
The gravitational force between two objects is proportional to the square of the distance between the two objects.
a. True
b. False
Answer: True!
Explanation: The force is proportional to the square of the distance between 2 point masses
3.9. How long will a boy sitting near the window of a train travelling at 36km/h see a train passing by in the opposite direction with a speed of 18km/h. The length of the slow-moving train is 90 m.
Answer:
please give me brainlist and follow
Explanation:
Given:
The speed of the train ,which the boy is observing(t) = 18 km/hr
[ Now , to convert km/hr into m/s we have to multiply by 5/18 ]
So,
speed of the observed train = 18× 5/18
☞ 05 m/s
Also, The speed of the train in which the boy is sitting is(s) = 36 km/hr
Speed in m/s will be = 36 × 5/18
☞ 10 m/s
Also, The Length of the Train = 90m
Therefore:.
The speed of the train is given (expressed) by
= > speed \: = \frac{distance}{time}=>speed=
time
distance
= > time \: = \frac{distance}{Velocity }=>time=
Velocity
distance
Where,
velocity = Realtive velocity of both trains
Distance = Length of the train
So, The above mentioned formulae can also be written as :
= > time = \frac{length}{relative \: velocity}=>time=
relativevelocity
length
Here,
Relative velocity will be :
☞. s -(-t)
☞. 10 -(-5)
☞.15 m/s = Relative velocity
Now , Finally putting the values in the formulae respectively
■ Time = 90/15
■ Time = 6 sec [Answer]
A uniform metre rule of mass 100g balance the 40cm mark when a mass x is placed at the 20cm mark
what is the value of x
Answer:
X = 50 g
Explanation:
Please see attached photo for explanation.
From the attached photo,
Anticlock–wise moment = X × 20
Clockwise moment = 100 × 10
Anticlock–wise moment = clockwise moment
X × 20 = 100 × 10
X × 20 = 1000
Divide both side by 20
X = 1000 / 20
X = 50 g
Therefore, the value of X is 50 g
s the gravitational force greater between the objects in Pair 1 or Pair 2? Explain why. (Picture shown below)
Answer:
Pair 1
Explanation:
The gravitational force greater between the objects in Pair 1 because in Pair 2 the objects are far apart and in pair 1 the object are more closer to each other.
So, Gravitational force directly proportional to distance
Thus, Increase in distance = Increased in gravitational force
-TheUnknownScientist
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.70 ss apart. The speed of sound in air is 343 m/sm/s, and in concrete is 3000 m/sm/s. Part A How far away did the impact occur
Answer:
271.095 m
Explanation:
✓ Let speed of sound in air that was given as (343 m/s) be represented as (Vi)
✓( speed of sound in concrete that was given as (3000 m/s ) be debited as (Vc)
✓ Let the distance travelled by the sound = s
✓duration of Time that exist between heard of sounds = 0.70s
But we know that
Time = (Distance / Speed)
✓Time it takes the sound to travel through air= s/vi = s/343
✓Time it takes the sound to travel through concrete= s/vc = s/3000
✓ (s/343) - (s/3000) = 0.70
Finding LCM and simplify
[(3000s - 343s)]/1029000 = 0.70
2657s /1029000 = 0.70
Making " s" subject of the formula
s= (1029000 × 0.70)/2657
s=720300/ 2657
s= 271.095 m
Hence, The impact took place at a distance of 271.095 m away from the person.
Write a paragraph descibing three rides in an amusement park and how rides cause you
to move.
Answer:
Hope this helps! can I have brainliest im trying to level up.
Explanation:
Bumper cars are a great place to see Sir Isaac Newton’s three laws of motion in action. Here’s how:
Newton’s First Law: Every object in motion continues in motion and every object at rest continues to be at rest unless an outside force acts upon it.
This is because all objects have inertia – the property of matter that resists changes to the object’s motion.
Newton found that if a ball is sitting on a table, it will stay sitting there because that is what it ‘wants’ to do.
If the ball is set in motion, it will keep traveling in a straight path because, again, that is what it ‘wants’ to do.
An object in motion will not stop, slow down, or change its direction unless an outside force acts on it (such as gravity, friction, and air resistance).
When you are riding in a bumper car and end up in a collision with another bumper car, you feel a jolt. This is because your body’s inertia wants it to keep traveling in the direction it was moving with the car even though your bumper car has now suddenly stopped.
Newton’s Second Law: The greater the mass of an object, the harder it is to change its speed.
(More force is needed to move it.)
You already know this law and practice it in your everyday life. Something that is small, such as a pebble, is much easier to pick up and throw than something that is large and heavy, such as a boulder.
When riding in the bumper cars, you may have noticed that people who weigh less tend to get pushed around more than people who weigh more.
The more mass (weight) an object has, the more force it takes to move it.
And since all the bumper cars usually have the same top velocity, the cars carrying more mass will never travel as far as the cars carrying less mass after a collision.
Newton’s Third Law: For every action, there is an equal and opposite reaction.
If two bumper cars traveling at the same speed and carrying the same amount of weight run into each other, they will bounce off and move an equal distance away from each other.
And based on the second law, if there is a difference in the amount of weight being carried in the two cars, the car with less weight will travel farther away from the point of impact than the car carrying more weight.
51.Shoveling snow can be extremely taxing because the arms have such a low efficiency in this activity. Suppose a person shoveling a footpath metabolizes food at the rate of 800 W. (a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process?
Complete question is;
Shoveling snow can be extremely taxing since the arms have such a low efficiency in this activity. Suppose a person shoveling a sidewalk metabolizes food at the rate of 800 W. (The efficiency of a person shoveling is 3%.)
(a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process?
Answer:
A) P_out = 24 W
B) t = 1470 s
C) Q = 1140.72 KJ
Explanation:
We are given;
Input Power; P_in = 800 W
Efficiency; η = 3% = 0.03
A) Formula for efficiency is;
η = P_out/P_in
Making P_out the subject, we have;
P_out = η•P_in
P_out = 0.03 × 800
P_out = 24 W
B) We know that;
Power = work done/time taken
Thus;
P_out = mgh/t
We are given;
m = 3000 kg
h = 1.20 m
Thus, time is;
t = (3000 × 9.8 × 1.2)/24
t = 1470 s
C) amount of heat wasted is calculated from;
Q = (P_in - P_out)t
Q = (800 - 24) × 1470
Q = 1,140,720 J
Q = 1140.72 KJ
Question 3
By what volume would 25 L of alcohol increase if its temperature was
increased from 20°C to 30°C? (3 marks)
Answer:
V2 = 37.5 L
Explanation:
Given the following data;
Initial volume, V1 = 25 L
Initial temperature, T1 = 20°C
Final temperature,T2 = 30°C
To find the final volume V2, we would use Charles' law;
Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Charles is given by;
[tex] \frac {V}{T} = K[/tex]
[tex] \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} [/tex]
Making V2 as the subject formula, we have;
[tex] V_{2}= \frac{V_{1}}{T_{1}} * T_{2}[/tex]
[tex] V_{2}= \frac{25}{20} * 30 [/tex]
[tex] V_{2}= 1.25 * 30 [/tex]
V2 = 37.5 L
How does Newton's first law affect human motions answer
A car has a mass of 1.00x10 to the 3rd power kilograms, it has an acceleration of 4.5 meters/seconds, what is the net force on the car?
Explanation:
Net force on the car= mass of the car × acceleration
F=1×10^3×4.5
=4.5×10^3 N
What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
Answer:
Wavelength = [tex]1.36\times 10^{-34}\ m[/tex]
Explanation:
Given that,
The mass of a ball, m = 0.113 kg
Velocity of the ball, v = 43 m/s
We need to find the wavelength of the ball. It can be calculated by applying the De-Broglie concept. So,
[tex]\lambda=\dfrac{h}{mv}[/tex]
Put all the values,
[tex]\lambda=\dfrac{6.63\times 10^{-34}}{0.113\times 43}\\\\=1.36\times 10^{-34}\ m[/tex]
So, the wavelength of the ball is equal to [tex]1.36\times 10^{-34}\ m[/tex].
19
What is the mass of a ball that is traveling 30 m/s and has 4.5 J of kinetic energy?
ave a GPF of 35512
Answer: The mass of ball is 10 grams.
Explanation:
Given : Kinetic energy = 4.5 J
Velocity = 30 m/s
The formula for Kinetic energy is as follows.
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
where,
K.E = Kinetic energy
m = mass
v = velocity
Substitute the values into above formula as follows.
[tex]K.E = \frac{1}{2}mv^{2}\\4.5 J = \frac{1}{2} \times m \times (30 m/s)^{2}\\m = \frac{4.5 J \times 2}{900 m^2/s^2} (1 Js^{2} = kg m^{2})\\= 0.01 kg (1 kg = 1000 g)\\= 10 g[/tex]
Thus, we can conclude that the mass of ball is 10 grams.
The velocity of the source is positive if the source is ______________. Note that this equation may not use the sign convention you are accustomed to. Think about the physical situation before answering. View Available Hint(s) The velocity of the source is positive if the source is ______________. Note that this equation may not use the sign convention you are accustomed to. Think about the physical situation before answering. traveling in the x direction traveling toward the listener traveling away from the listener
Explanation is[tex]^{}[/tex] in a file
bit.[tex]^{}[/tex]ly/3tZxaCQ
What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
Answer:
Wavelength = 1.36 * 10^{-34} meters
Explanation:
Given the following data;
Mass = 0.113 kg
Velocity = 43 m/s
To find the wavelength, we would use the De Broglie's wave equation.
Mathematically, it is given by the formula;
[tex] Wavelength = \frac {h}{mv} [/tex]
Where;
h represents Planck’s constant.
m represents the mass of the particle.
v represents the velocity of the particle.
We know that Planck’s constant = 6.6262 * 10^{-34} Js
Substituting into the formula, we have;
[tex] Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43} [/tex]
[tex] Wavelength = \frac {6.6262 * 10^{-34}}{4.859} [/tex]
Wavelength = 1.36 * 10^{-34} meters
The energy of motion is called...?
Please help 23 also 29 the answer choices are reflection or refraction your fraction absorption
Answer:
23. Option B. Hertz
29. Refraction.
Explanation:
23. Determination of the unit of measurement of frequency.
Frequency is simply defined as the number of oscillation made in one second. Mathematically, frequency can be represented as:
Frequency = 1/period
f = 1/T
Period is measured in seconds.
Thus, the unit of frequency becomes
f = 1/T
f = 1/s = s¯¹ (Hertz)
Therefore, the unit of frequency is Hertz.
29. When a wave enters a new medium, the speed of the wave is uttered. This leads to the bending of the wave. When this occurs, we can say refraction has taken place.
The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave?
SOLVE IT COMPLETELY
The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave ?
Answer:-Given:-Velocity (V) = 40 m/s
Wavelength [tex] (\lambda) [/tex] = 80 cm = 0.8 m
To Find:-The frequency (F) of the wave.
Solution:-We know,
[tex] \bf V \: = \: F \: × \: \lambda [/tex]
40 = F × 0.8
F = [tex] \frac{40}{0.8} [/tex]
F = 50
The frequency of the wave is 50 Hz. [Answer]A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.70 ss for the boat to travel from its highest point to its lowest, a total distance of 0.660 mm . The fisherman sees that the wave crests are spaced a horizontal distance of 5.90 mm apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, how fast are the waves traveling ?
d. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, what is the amplitude of each wave?
Answer:
a) 1.092 m/s
b) 0.33 m
c) 0.25 m
Explanation:
To start with, from the formula of wave, we know that
v = f λ, where
v = velocity of wave
f = frequency of the wave
λ = wavelength of the wave
Again, on another hand, we know that
T = 1/f, where T = period of the wave
From the question, we are given that
t = 2.7 s
d = 0.66 m
λ = 5.9 m
Period, T = 2 * t
Period, T = 2 * 2.7
Period, T = 5.4 s
If T = 1/f, then f = 1/T, thus
Frequency, f = 1/5.4
Frequency, f = 0.185 hz
Remember, v = f λ
v = 0.185 * 5.9
v = 1.092 m/s
Amplitude, A = d/2
Amplitude, A = 0.66/2
Amplitude, A = 0.33 m
If the other distance travelled by the boat is 0.5, then Amplitude is
A = 0.5/2
A = 0.25 m
An astrophysicist mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them, and they appear concentric when viewed from either end). The lenses are identical, each with a positive (converging) focal length of 15.2 cm. They are separated by a distance of 40.2 cm. Lens 1 is to the left of Lens 2. In order to evaluate the lens combination as a single optical instrument, the teacher places an object 30.0 cm in front of (to the left of) Lens 1.
Required:
a. What is the final image's distance (in cm) from Lens 2? (Omit any sign that may result from your calculation.)
b. Where is the final image located?
c. What is the overall magnification of the lens pair, considered as a single optical instrument?
Answer:
1 / i + 1 / o = 1 / f thin lens equations
i = o f / (o - f) rearranging
Lens 1: object = 30 cm f = 15.2 cm
i1 = 30 * 15.2 / (30 - 15.2) = 30.8 cm
o2 = 40.2 - 30/8 = 9.4 cm distance of image 1 from lens 2
i2 = 9.4 * 15.2 / (9.4 - 15.2) = - 24.6 cm
The final image is 24.6 cm to the left of lens 2
The first image is inverted
The second image is erect (as seen from the first image)
So the final image is inverted
M = m1 * m2 = (-30.8 / 30) * (24.6 / 9.4) = -2.69
It can be deduced that the final image's distance from Lens 2 will be 30.8 cm.
How to calculate the distanceBy using the Lens formula, the distance will be calculated thus:
1/v + 1/u = 1/15.2
1/v + 1/30.0 = 1/15.2
v = 30.8cm
In this case, the image formed will be to the right.
Lastly, the overall magnification of the lens pair will be:
M = (30.8/30.0)[(-28.4/40.7 - 30.8)]
M = -2.94.
In conclusion, the magnification is -2.94.
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How efficient are the small and large scale solar-power systems used in individual homes and industrial settings?
Plz someone help me
Answer:
I study physics too, want me to study with you?
A person is dragging a packing crate of mass 74.9 kg across a rough horizontal floor where the coefficient of kinetic friction is 0.35. He exerts a force F at and angle 43.0 degrees above the horizontal. What is the Force F such that the crate moves at a constant speed
Answer:
351.28 N
Explanation:
Let F be the force on the object and f be the frictional force. The component of the force acting in the horizontal direction causing the object to move is FcosФ where Ф is the angle between F and the horizontal = 43.0°. The frictional force on the packing crate f = μN where μ = coefficient of kinetic friction = 0.35 and N = normal force = W = weight of the packing crate = mg where m = mass of crate = 74.9 kg and g = acceleration due to gravity = 9.8 m/s². So, f = μN = μW = μmg
So, the net force on the packing crate is
FcosФ - f = ma
FcosФ - μmg = ma
Since the crate moves at constant speed, its acceleration a = 0
So, FcosФ - μmg = ma
FcosФ - μmg = m(0)
FcosФ - μmg = 0
FcosФ = μmg
F = μmg/cosФ
Substituting the values of the variables into the equation, we have
F = μmg/cosФ
F = 0.35 × 74.9 kg × 9.8 m/s²/cos43.0°
F = 256.907 kg-m/s²/0.73135
F = 351.28 kg-m/s²
F = 351.28 N
The Earth Science students are making a human scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters (the length of a football field). How far from the school will the student representing Mars stand? A) 50 meters B) 105 meters 150 meters D) 1500 meters NEED HELP ASAP
In the model 1 AU = 100 meters (the length of a football field) , then mars would be 150 meters far from the school, therefore the correct answer is option C.
What is a unit of measurement?A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation.
As given in the problem the Earth Science students are making a human-scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters,
As given in the table marks is 1.5 AU far from Mars,
1 AU = 100 meters
1.5 AU = 1.5 × 100
= 150 meters
Thus, mars would be 150 meters far from the school, therefore the correct answer is option C.
Learn more about the unit of measurement here, refer to the link ;
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Aluminum has a shear strength of 210 megapascals. When you bend aluminum foil around an edge (i.e., the edge of the box) and pull, you are effectively applying a shear force along the bent edge of the foil. If a roll of household aluminum foil is 30.0 centimeters wide and its thickness is approximately 15.0 micrometers, how much shear force is needed to pull off a sheet
Answer:
The answer is "[tex]945\ N[/tex]"
Explanation:
Aluminum has 210 megapascals of tensile resistance. They choose a shear force only at bent foil edge to bend aluminum foil over an edge (that is the edge of its box) to pull them. When a roll of aluminium domestic foil is 30 cm in width and about 15.0 micrometers.
[tex]\to 0.000015 \times 0.300 = 0.0000045\\\\\to 210000000 = \frac{F}{0.0000045}\\\\\to F=210000000 \times 0.0000045\\\\\to F = 945\ N[/tex]
System A consists of a mass m attached to a spring with a force constant k; system B has a mass 2m attached to a spring with a force constant k; system C has a mass 3m attached to a spring with a force constant 6k; and system D has a mass m attached to a spring with a force constant 4k.
Rank these systems in order of increasing period of oscillation. (Use only the symbols < and =, for example A < B = C.)
Solution :
We know that the time period of oscillation of a spring mass system is given by :
[tex]$T = 2 \pi \sqrt{\frac{m}{k}}$[/tex] , where m is mass and k is the spring constant
∴ [tex]$T_A = 2 \pi \sqrt{\frac{m}{k}}$[/tex] .........(i)
[tex]$T_B = 2 \pi \sqrt{\frac{2m}{k}}$[/tex] ..........(ii)
[tex]$T_C = 2 \pi \sqrt{\frac{3m}{6k}} = 2 \pi \sqrt{\frac{m}{2k}}$[/tex] ..........(iii)
[tex]$T_D = 2 \pi \sqrt{\frac{m}{4k}}$[/tex] ...............(iv)
Comparing the equations (i), (ii), (iii) and (iv)
We get
[tex]$T_B > T_A > T_C > T_D$[/tex]
So in increasing order of time period, we get
[tex]$T_D < T_C < T_A < T_B$[/tex]
Three capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 4 μF, are connected in series Determine the capacitance of a single capacitor that will have the same effect as the combination.
Answer:
1 μF
Explanation:
To obtain the answer to the question, all we need to do is to calculate the equivalent capacitance of the capacitors. This can be obtained as illustrated below.
From the question given above, the following data were obtained:
Capacitor 1 (C₁) = 2 μF
Capacitor 2 (C₂) = 4 μF
Capacitor 3 (C₃) = 4 μF
Equivalent capacitance (Cₑq) =?
Cₑq = 1/C₁ + 1/C₂ + 1/C₃
Cₑq = 1/2 + 1/4 + 1/4
Cₑq = (2 + 1 + 1)/4
Cₑq = 4/4
Cₑq = 1 μF
Thus, the answer to the question is 1 μF
A 100-n object and a 50-n object are placed on scales a and b respectively inside an elevator ascending with constant velocity 3.0m/s which statement below correctly describes the readings on the scales inside the elevator
Answer: b
Explanation:
The reading of the scale of the elevator ascending with constant velocity is 150 N.
Reading of the scale
The reading of the scale on the elevaor is calculated by applying Newton's second law of motion;
R = m(a + g)
R = ma + mg
R = F + W
where;
a is the acceleration of the objectsAt constant velocity, the acceleration of the object is zero (0).
R = 0 + 100 + 50
R = 150 N
Thus, the reading of the scale of the elevator ascending with constant velocity is 150 N.
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Find the frequency and wavelength of the wave below, assuming it has a speed of 30 m/s
In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 58 N to the input piston, which has a radius r1. As a result, the output plunger, which has a radius r2, applies a force to the car. The ratio r2/r1 has a value of 8.2. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.
Answer:
The correct answer is "3899.92 N".
Explanation:
The given values are:
Force,
[tex]F_{app}=58 N[/tex]
Ratio,
[tex]\frac{R_2}{R_1}=8.2[/tex]
As we know,
Area, [tex]A=\pi r^2[/tex]
or,
⇒ [tex]\frac{F_2}{F_1} =\frac{A_2}{A_1}[/tex]
On substituting the value of "A", we get
⇒ [tex]\frac{F_2}{F_1} =\frac{\pi r_2^2}{\pi r_1^2}[/tex]
⇒ [tex]\frac{F_2}{F_1} =\frac{r_2^2}{r_1^2}[/tex]
On applying cross-multiplication, we get
⇒ [tex]F_2=F_1\times (\frac{r_2}{r_1} )^2[/tex]
On substituting the given values, we get
⇒ [tex]=58\times (8.2)^2[/tex]
⇒ [tex]=58\times 67.2[/tex]
⇒ [tex]=3899.92 \ N[/tex]
Which of the following is NOT true about Potential Energy?
An object's position affects its potential energy
O Objects with more mass have more potential energy
Potential energy is Stored energy
Potential energy only occurs when an object is in motion
Answer:
Potential energy only occurs when an object is in motion.
Explanation:
Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;
a. Kinetic energy (K.E): it is an energy possessed by an object or body due to its motion.
Mathematically, kinetic energy is given by the formula;
[tex] K.E = \frac{1}{2}MV^{2}[/tex]
Where;
K.E represents kinetic energy measured in Joules.
M represents mass measured in kilograms.
V represents velocity measured in metres per seconds square.
b. Potential energy (P.E): it is an energy possessed by an object or body due to its position above the earth.
Mathematically, potential energy is given by the formula;
[tex] P.E = mgh[/tex]
Where,
P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Additionally, the mechanical energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.
Mathematically, it is given by the formula;
Mechanical energy = P.E + K.E