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
The energy of motion is called...?
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].
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]
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]
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.
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.
13. The percent of Earth's surface covered by high clouds
in January 1987 was closest to which of the following?
A. 13.09
B. 13.5%
C. 14.0%
D. 14.5%
16. Which of the following figures best represents the
monthly average cover of high. middle, and low clouds
in January 19922
E.
H.
cloud cover
okud cover
middle cloud
high clouds
high cloud
low clouds
low clouds
midle clouds
G.
J.
14. Based on Table I. a cosmic ray flux of 440.000 parti-
cles/m he would correspond to a cover of low clouds
that is closest to which of the following?
F. 28.75
G. 29.09
H. 29.35
J. 29.6%
ckud cover
cloud cover
high clouds
low clouds
middle cloud
high clouds
middle clouds
low clouds
How did the projected storm track change from image A TO B TO C TO D
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 certain gas is compressed adiabatically. The amount of work done on the gas is 800 J. What is the change in the internal (thermal) energy of the gas?
Answer:
800J
Explanation:
Using the formula for change in the internal energy of a system
∆U = Q - W
Q = heat added to the system
W =workdone by system.
We know the process is an adiabatic one then, there no addition/ removal of heat, then Q= 0
(∆U = -W )
Then substitute for W, we have
∆U = -[-800]
∆U= 800J
∆U = 800J
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
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.
un cubo de aluminio tiene un volumen de 45cm3 cuál es su masa en gramos
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)
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
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]
Question 4 of 10 A student measures the time it takes for two reactions to be completed. Reaction A is completed in 39 seconds, and reaction B is completed in 50 seconds. What can the student conclude about the rates of these reactions? A. The rate of reaction B is higher B. The rate of reaction A is higher. C. The rates of reactions A and B are equal.
Answer:
B. the rate of reaction is higher
Explanation:
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
Spring is a time when animals reproduce. What might you see a heron do? Build a nest A. Grow feathers A. Hibernate Migrate
Answer:
A. Build a nest.
Explanation:
It wont do any of the other things and they need a place to sleep and lay eggs plus it is a bird
The heron birds belong to Ardeidae family. Some of the species among them are referred to as egrets or bitterns rather than a heron. A heron bird build a nest during Spring. The correct option is A.
What is Heron bird?The heron is known as a long-legged, long-necked, fresh water and coastal birds. It a medium to large size bird with long necks and legs. They exhibit sexual dimorphism in size. The two male and female sex of the same species show distinct attributes beyond their difference in sexual organs.
Almost all the heron species are found in water and they are essentially non-swimming water birds which feed on the margins of lakes, ponds, sea, rivers and swamps. The small species of heron are generally referred to as the dwarf bittern.
The nesting season indicates the time of year during which birds build nests and lay eggs in them. In most cases bring up their young. It is in the spring season lot of food are available.
Thus the correct option is A.
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How do you make camping fun?
An ampere is measured in
Seconds
Hours
Answer:
Seconds
Explanation:
Hope it helps you in your learning process.
The rate of flow of charge (per second) through a conductor is called electric current.
S. I unit of electric current is Ampere (named after French Physicist Andre Marie Ampere)
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 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
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 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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What is the speed of an object moving around a 0.75 m radius circle that completes a revolution in 0.50 seconds?
Answer:
the speed of an object is 9.42 m/s
Explanation:
The computation of the speed of an object is given below:
v = 2πr ÷ t
where
v denotes the speed
r denotes the radius
r denotes the time
So,
= 2 × 3.14 × 0.75 ÷ 0.50
= 9.42 m/s
Hence, the speed of an object is 9.42 m/s
the same would be considered and relevant too
what is magnetic flux thru a closed surface
Answer: Zero
Explanation:
Gauss's law states that the total magnetic flux through a closed surface is equal to zero. Because the magnetic field lines are in continuous loops, so all closed surfaces have magnetic field lines coming out, hence the net magnetic flux is zero.
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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why do substance takes long to condense than freezing?