Answer:
The rate of deceleration is 15 km/h^2.
Explanation:
Since the car comes in 90km/h, it will need deceleration for 15 km/h for 6 seconds to finally stop the car. 15*6=90
A wave travels at a constant speed.How does the frequency change if the wavelength is reduced by a factor of 3 The frequency decreases by a factor of 3 the frequency increases by a factor of 3 the frequency increases by a factor of 9 the frequency does not change
Answer:
The frequency increases by a factor of 3.
Explanation:
The relation between speed, wavelength and frequency of a wave is given by :
[tex]v=f\lambda[/tex]
or
[tex]f\propto \dfrac{1}{\lambda}[/tex]
A wave travels at a constant speed. If the wavelength is reduced by a factor of 3, it would mean that the frequency increases by a factor of 3 because there is an inverse relationship between wavelength and frequency.
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
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.
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un cubo de aluminio tiene un volumen de 45cm3 cuál es su masa en gramos
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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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]
True or false? Charges flow from high voltage to low voltage
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
A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of mass 1.51 kg and length 1.79 meters spinning clockwise with an angular velocity of 5.12 rad/s is dropped on the spinning disk and stuck to it (the centers of the disk and the rod coincide). The combined system continues to spin with a common final angular velocity. Calculate the magnitude of the loss in rotational kinetic energy due to the collision
Answer:
The loss in rotational kinetic energy due to the collision is 36.585 J.
Explanation:
Given;
mass of the disk, m₁ = 1.64 kg
radius of the disk, r = 0. 61 m
angular velocity of the disk, ω₁ = 17.6 rad/s
mass of the rod, m₂ = 1.51 kg
length of the rod, L = 1.79 m
angular velocity of the rod, ω₂ = 5.12 rad/s (clock-wise)
let the counter-clockwise be the positive direction
let the clock-wise be the negative direction
The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;
m₁ω₁ + m₂ω₂ = ωf(m₁ + m₂)
where;
ωf is the common final angular velocity
1.64 x 17.6 + 1.51(-5.12) = ωf(1.64 + 1.51)
21.1328 = ωf(3.15)
ωf = 21.1328 / 3.15
ωf = 6.709 rad/s
The moment of inertia of the disk is calculated as follows;
[tex]I_{disk} = \frac{1}{2} mr^2\\\\I_{disk} = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk} = 0.305 \ kgm^2[/tex]
The moment of inertia of the rod about its center is calculated as follows;
[tex]I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2[/tex]
The initial rotational kinetic energy of the disk and rod;
[tex]K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i= \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \ \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J[/tex]
The final rotational kinetic energy of the disk-rod system is calculated as follows;
[tex]K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J[/tex]
The loss in rotational kinetic energy due to the collision is calculated as follows;
[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J[/tex]
Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.
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
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.
PLEASE HELP
A ball, 1.8 kg, is attached to the end of a rope and spun in a horizontal circle above a student's head. As the student rotates the ball in a horizontal clockwise circle their lab partner counts 53 rotations in one minute. What is the ball's angular velocity in radians per second?
Answer:
The angular speed of the ball in radians per second is 5.55 rad/s.
Explanation:
Given;
mass of the ball, m = 1.8 kg
number of the ball's rotation per minute, n = 53 RPM
The angular speed of the ball in radians per second is calculated as follows;
[tex]\omega = 53\frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s } \\\\\omega = 5.55 \ rad/s[/tex]
Therefore, the angular speed of the ball in radians per second is 5.55 rad/s.
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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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
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].
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]
the electrostatic force between two objects is 40N. if the charge of one object is cut in half, and the distance is doubled, what is the new force?
Answer:
F1 = K Q1 Q2 / R1^2
F2 = K Q1 / 2 * Q2 / (2 R1)^2
F2 / F1 = 1/2 / 4 = 1/8
The new force is 5N (1/2 due to charge and 1/4 due to distance)
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
The energy of motion is called...?
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
11) A tank of kerosene with density of 750 kg/m3 has a syphon used to remove the fluid that then exits into the local atmosphere, with pressure of 101 kPa. The pressure above the kerosene in the tank is 120 kPa absolute. The syphon tube has a diameter of 2 cm, exits the tank rising to 10 cm above the level of the kerosene and then drops down to 15 cm below the level of the kerosene where it exits into the atmospheric pressure. Calculate the exit velocity from the tube.
Answer:
[tex]7.32\ \text{m/s}[/tex]
Explanation:
[tex]v_1[/tex] = Velocity at initial point = 0
[tex]P_1[/tex] = Pressure in tank = 120 kPa
[tex]P_2[/tex] = Pressure at outlet = 101 kPa
[tex]\rho[/tex] = Density of kerosene = [tex]750\ \text{kg/m}^3[/tex]
[tex]Z_1[/tex] = Tank height = 15 cm
[tex]Z_2[/tex] = Height of pipe exit = 0
[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From Bernoulli's equation we have
[tex]\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+Z_2\\\Rightarrow \dfrac{P_1}{\rho g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}\\\Rightarrow v_2=\sqrt{2g(\dfrac{P_1}{\rho g}+Z_1-\dfrac{P_2}{\rho g})}\\\Rightarrow v_2=\sqrt{2\times 9.81(\dfrac{120\times 10^3}{750\times 9.81}+0.15-\dfrac{101\times 10^3}{750\times 9.81})}\\\Rightarrow v_2=7.32\ \text{m/s}[/tex]
The exit velocity from the tube is [tex]7.32\ \text{m/s}[/tex].
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
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
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 difference between 1 celcius and 1 kelvin
Answer:
One degree unit on the Celcius scale is equivalent to one degree unit on the kelvin scale. The only difference between these two scales is the zero point. The zero point on the Celcius scale was defined as the freezing point of water, which means that there are higher and lower temperatures around it.
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]
To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as:
Answer:
dorsiflexion
Explanation:
To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as: dorsiflexion
Answer:
dorsiflexion
Explanation:
Hope this helps
45
А ______ is a solid, liquid or gas that a wave travels through.
Answer:
medium
Explanation:
I am not really sure
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.
A string with a length of 7.25 m is fixed at both ends. If the string is plucked so that there are five nodes along the string in addition to those at either end, what is the wavelength of the interfering waves for this mode?
Answer:
the wavelength of the interfering waves for this mode is 2.4168
Explanation:
The computation of the wavelength of the interfering waves for this mode is shown below:
Given that
The length is 7.25m
And, the number of nodes is 6
So, the wavelength should be
= 2 × 7.25m ÷ 6
= 2.4168 m
Hence, the wavelength of the interfering waves for this mode is 2.4168
The same would be considered