A spring of spring constant 25 N/m is hung vertically and a 0.300 kg mass is attached to one end, causing a displacement of the end of the spring of _____ m.
a. 0.01
b. 0.12
c. 0.77
d. all of the above
Answer:
The displacement of the spring is 0.12 m.
Explanation:
Given;
spring constant, k = 25 N/m
mass attached to the spring, m = 0.300 kg
displacement of the spring, x = ?
Apply Hook's law, weight of the mass attached (applied force) is directly proportional to the extension of the spring.
F = kx
mg = kx
0.3 x 9.8 = 25x
2.94 = 25x
x = 2.94 / 25
x = 0.12 m
Therefore, the displacement of the spring is 0.12 m.
When an element ejects an alpha particle, the mass number of that element
A.reduce by 4.
B.remains unchanged.
C.increase by 1.
D. reduce by 1.
Answer:
A.reduce by 4.
Explanation:
In Chemistry, an alpha particle which is typically a helium nucleus has an atomic mass (number of nucleons) of four (4). Therefore, the ejection of an alpha particle by any chemical element simply means that, the mass number of that particular chemical element would be reduced by four (4).
Basically, this process is referred to as an alpha decay and can be defined as a radioactive disintegration of a chemical element that causes it to emit an alpha particle and as a result reducing its mass number by four (4).
Hence, when an element ejects an alpha particle, the mass number of that element reduces by 4.
What would be the magnitude of a star with a temperature of 25,000 located on
the main sequence?
Answer:
hvkhvkhvkvlhvgjvgv
Explanation:
A police officer standing at the side of the road uses a radar emitting frequency of 24.15 GHz. A car is going away from the officer at a speed of 50mph. What will be the difference in frequency of the beam reflected by the car received back by the radar?
a. 4.0 kHz (lower frequency)
b. 1.8 kHz (higher frequency)
c. 4.0 kHz (higher frequency)
d. 1.8 kHz (lower frequency)
Answer:
The correct answer is C
Explanation:
From the question we are told that
The frequency of the radar is [tex]f = 24.15 \ GHz = 24.15 *10^{9} \ Hz[/tex]
The speed of the car is [tex]v = 50 mph = \frac{50}{2.237} = 22.35 \ m/s[/tex]
Generally the difference frequency reflected by the car and the frequency which the radar receives back is mathematically represented as
[tex]\Delta f = \frac{f * 2 * v }{ c }[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
=> [tex]\Delta f= \frac{2 * 24.15 *10^{9} * 22.35}{ 3.0*10^{8}}[/tex]
=> [tex]\Delta f = 4000 \ Hz[/tex]
=> [tex]\Delta f = 4.0 \ kHz[/tex]
Given that the value is positive then it a higher frequency
I
2800000 kg
Fig. 2.1
The total mass of the rocket and its full load of fuel is 2.8 106 kg. The constant force provided by
the rocket's motors is 3.2 x 107N.
32000000 N.
(a) Calculate
(i) the total weight of the rocket and the fuel.
weight =
Answer:
W = 27468000 [N] or 27.47 [kN]
Explanation:
The weight of a body is defined as the product of mass by gravitational acceleration.
[tex]W = m*g[/tex]
where:
W = weight of the rocket with fuel [kg]
m = mass = 2800000 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now we can solve:
[tex]W = 2800000*9.81\\W = 27468000 [N][/tex]
What is the condition required of the phase difference (in radians) between two waves with the same wavelength if these waves interfere constructively?
a. (2m +1)π where m= 0, +1, +2, etc.
b. mπ where m = 0, +1, +2, etc.
c. 2mπ where m = 0, +1, +2, etc.
d. (m+1)π where m = 0, +1, +2, etc.
Answer:
c.
Explanation:
In order to two waves with the same wavelength can interfere constructively, their crests and valleys must coincide in space, so the phase difference must be equal to an integer number of wavelengths, i.e. m *(2 π rad), where m= 0, +1, +2, etc.This is equal to the stated by the answer c) , so c) it's the right answer.1+1=?
FIRST ONE TO ANSWER GETS BRAINLIEST!
Answer:
Thx for points merry Christmas answer 2
Explanation:
Answer:2
Explanation:
A tsunami, an ocean wave generated by an earthquake, propagates along the open ocean at 700 km/hr and has a wavelength of 750 km. What is the frequency of the waves in such a tsunami?A. 6.8 HzB. 0.93 HzC. 0.00026 HzD. 1.1 HzE. 0.15 Hz
Answer:
C) 0.00026 Hz
Explanation:
In any wave, there exists a fixed relationship between the speed v, the frequency f and the wavelength λ, as follows:[tex]v = \lambda * f (1)[/tex]
Replacing by the givens in (1), and solving for f, we get:[tex]f = \frac{v}{\lambda} = \frac{700km/hr}{750 km} = 0.93 1/hr (2)[/tex]
Converting the units to Hz (1/sec), we get:[tex]f = 0.93 \frac{1}{hr} *\frac{1 hr}{3600sec} = 2.6e-4 = 0.00026 Hz (2)[/tex]
The answer C. is the right one.Find the current if 55 C of charge pass a particular point in a circuit in 5 seconds.
Answer:
The current is 11 Amperes
Explanation:
Electric Current
The electric current is defined as a stream of charged particles that move through a conductive path.
The current intensity can be calculated as:
[tex]\displaystyle I=\frac{Q}{t}[/tex]
Where:
Q = Electric charge
t = Time taken by the charge to move through the conductor
The current intensity is often measured in Amperes.
The charge passing through a point in a circuit is Q= 55 c during t=5 seconds, thus the current intensity is:
[tex]\displaystyle I=\frac{55}{5}[/tex]
I = 11 Amp
The current is 11 Amperes
what is mean by combination reaction ?
[tex] \underline{\purple{\large \sf Combination \: reaction :-}} [/tex]
Those reaction in which two or more substances combine to form a one new substance are called Combination reaction
In this reaction, We can add :
Two or more elements can combine to form a compound.Two or more compounds can combine to from a one new compound.An element and a compound can combine to form a new compound.[tex] \underline{\green{\large \sf For\: example :}} [/tex]
[tex] \sf 2H_{2} + O_{2} \: \underrightarrow{Combination} \: 2H_{2}o[/tex]
In this, Hydrogen is an element and Oxygen is another element. Both are combined to form compound 'Hydrogen oxide'. Hydrogen oxide is commonly known as water.
true or false The most extreme tides occur during a new
moon and full moon.
Answer:
true
Explanation:
The speed of sound in air is 10 times faster than the speed of a wave on a certain string. The density of the string is 0.002kg/m. The tension in the string is __________.
Answer:
The tension on the string is 2.353 N.
Explanation:
Given;
the speed of sound in air, v₀ = 343 m/s
then, the speed of sound on the string, v = 343 / 10 = 34.3 m/s
mass per unit length, m/l = μ = 0.002 kg/m
The speed of sound on the string is given as;
[tex]v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu[/tex]
where;
T is the tension on the string
T = (34.3)²(0.002)
T = 2.353 N
Therefore, the tension on the string is 2.353 N.
Through what angle in degrees does a 33 rpm record turn in 0.32 s?
63°
35°
46°
74°
Answer:
1 rev = 2(pi) rad pi(rad) = 180 degrees
so 33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s = 63.36 degrees
Explanation:
63.36 estimated to 63 so 63
The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.
Calculation of the angle:Since we know that1 rev = 2(pi) rad
So here pi(rad) = 180 degrees
Now for 33 rpm it should be like
= 33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s
= 63.36 degrees
= 63 degrees
hence, The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.
Learn more about an angle here: https://brainly.com/question/3394547
Suppose a uniform solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The linear velocity of the sphere at the bottom of the incline depends on?
Answer:
None of the mass or the radius of the sphere
Explanation:
When a uniform solid sphere of any given mass, say M and any given radius, say R, rolls without slipping downwards an inclined plane that starts from rest. The linear velocity of the sphere at about the bottom of the inclined happens not to depend on either of its mass or that of the radius of its sphere.
PHYSICS! HELP ME PLEASE ASAP!
1.) If an ambulance passes you while you're walking your dog, a change in pitch is ____?
A.) not perceived by either the pedestrian or you
B.) perceived by the you
C.) perceived by the driver
2.) In the star wars movies, the soundtrack is very important. musicians and sound technicians spend weeks perfecting the noise of the light sabers, lasers, tie fighters and explosions. in real life, there would be no sound at all in space. Why?
A.) mechanical waves require a medium to travel
B.) sound is an electromagnetic wave
C.) mechanical waves do not require a medium to travel
D.) sound is a product of vibration
3.) what is a compression?
A.) region of zero pressure in a medium caused by a wave passing
B.)region of a high altitude in a medium caused by a passing wave
C.) region of a high pressure in a medium cause by a passing wave
D.) region of low pressure in a medium caused by a passing wave
Answer:
1B, 2B and 3C hope this helps
Explanation:
A gas cylinder holds 0.10 mol of O2 at 150 C and a pressure of 3.0 atm. The gas expands adiabatically until the pressure is halved
Part A
What is the final volume?
Part B
What is the final temperature?
Answer:
V2 = 1.899*10^-3 m^3
T2 = 347.125 K
Explanation:
Using gas law, we know that
PV = nRT,
Where
V1 = 0.00115743 m^3.
gamma = 1.4
Now, when we solve for final volume, V2 we get
V2 = V1/((P2/P1)^(1/gamma))
V2 = 1.899*10^-3 m^3
Using the same law and method, when we try to solve for the temperature, we find that the final temperature, T2 is
T2 = T1*((V1/V2)^(gamma-1))
T2 = 347.125 K
The final volume is 1.899*10^-3 m^3
And, the final temperature is 347.125 K
Gas law:here we used gas law,
we know that
PV = nRT,
Here
V1 = 0.00115743 m^3.
gamma = 1.4
Now final volume is
V2 = V1/((P2/P1)^(1/gamma))
V2 = 1.899*10^-3 m^3
Now the final temperature is
T2 = T1*((V1/V2)^(gamma-1))
T2 = 347.125 K
Learn more about volume here: https://brainly.com/question/13011592
20 pts.
Which of the following statements is true?
O Electromagnets use electrlcity and magnets.
O Magnetic fields are strongest around the poles of a magnet.
O The south pole of a magnet will repel the south pole of another magnet.
O all of the above
Answer:
all are true so d is right
Explanation:
Electromagnets use electrlcity and magnets is true.
Magnetic fields are strongest around the poles of a magnet is true.
The south pole of a magnet will repel the south pole of another magnet is true
and since all of them is true the answer is d all of the above
An astronaut stands on the surface of an asteroid. The astronaut then jumps such that the astronaut is no longer in contact with the surface. The astronaut falls back down to the surface after a short time interval. Which of the following forces CANNOT be neglected when analyzing the motion of the astronaut?
Asteroids are known through the help of artificial gravity, to have small gravity. The forces that cannot be neglected when analyzing the motion of the astronaut is that the gravitational force between the astronaut and the asteroid.
The gravitational force between two objects is said to be inversely proportional to the distance between them when squared. Therefore, when an individual halve the distance then the force increases by four times.
Unbalanced forces are simply known to be brought about due to a change in motion, speed, and/or direction. If two forces act in the same direction on an object, the net force is said to be equal to the sum of the two forces.
Learn more from
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What electromagnetic wave is a is a TV remtoe
Answer:
infared radiation
Explanation:
g A communication system operating at a power of 510 W produces electromagnetic radiation with a wavelength of 10.5 cm . Calculate the number of photons produced by the system per minute.
Answer:
The value is [tex]N = 2.69 *10^{26} \ photons[/tex]
Explanation:
From the question we are told that
The power is [tex]P = 510 \ W[/tex]
The wavelength is [tex]\lambda = 10.5 \ cm = 0.105 \ m[/tex]
Gnerally the number of photon produced is mathematically represented as
[tex]N = \frac{P * \lambda }{ h * c }[/tex]
Here [tex]h[/tex] is the Planck's constant with a value [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
c is the speed of light with value [tex]c = 3.0*10^{8 } \ m/s[/tex]
So
[tex]N = \frac{ 510 * 0.105 }{ 6.62607015 * 10^{-34} * 3.0 *10^{8} }[/tex]
=> [tex]N = 2.69 *10^{26} \ photons[/tex]
Help me please,
A ball is thrown straight up in the air. What is the velocity and acceleration at the top of the path?
A) v 0m/s, = 0m/s/s
B) v = 0m/s, a 10m/s/s
C) v = 10m/s, a 10m/s/s
D) v = 10m/s, a = 0m/s/s
E) None of the above
Option B
Explanation:
no distance was given only the acceleration due to the fact that it went up (10m/s/s)
s0 it is
0 m/s and 10m/s/s (option B)
Define conductor and insulator, including how the resistance is different in the two, and give at least one example of each.
Answer:
Those substances which can conduct electricity are called conductors, while those substances which don't conduct electricity are called insulators.
Resistance is the obstruction provided by the material through which the current passes,so since conductors conduct electricity and insulators don't,so the obstruction i.e resistance provided by the conductor must be less,while insulators being unable to conduct electricity,has very high resistance.
Example of conductor is copper
Example of insulator is plastic
A carousel at the local carnival rotates once every 45 seconds.
(a) What is the linear speed of an outer horse on the carousel, which is 2.75 m from the axis of rotation?
(b) What is the linear speed of an inner horse that is 1.75 m from the axis of rotation?
Answer:
We know that the carousel does a complete rotation in 45 seconds.
Then the frequency of this carousel will be f = 1/45 seconds.
And the angular frequency will be 2*pi times the frequency, then we have:
angular frequency = w = 2*3.14*(1/45s) = 0.1396 s^-1
Now, the linear speed of an object that rotates with a radius R, and an angular frequency W is:
S = R*W
then:
a) in this case the radius is 2.75m, then the linear speed is:
S = 2.75m*0.1396 s^-1 = 0.3839 m/s
b) in this case the radius is 1.75m, then the linear speed here is:
S = 1.75m*0.1396 s^-1 = 0.2443 m/s
(a) The linear speed of an outer horse on the carousel is 0.384 m/s.
(b) The linear speed of an inner horse on the carousel is 0.244 m/s.
Given data:
The time interval for the rotation of carousel is, t = 45 s.
The distance of the outer horse from the axis of rotation is, r = 2.75 m.
The distance of an inner horse from the axis of rotation is, r' = 1.75 m.
(a)
The linear speed in this problem can be obtained from the concept of rotational mechanic, in which the ratio of the circumference and the time gives required linear speed. So,
v = 2 π r/t
Solving as,
v = 2 π (2.75) / 45
v = 0.384 m/s
Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.384 m/s.
(b)
Now similarly the linear speed of an inner horse is calculated as,
v' = 2 π r' / t
Solving as,
v' = 2 π (1.75) / 45
v' = 0.244 m/s
Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.244 m/s.
Learn more about the rotational mechanics here:
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Which structures are found in the middle ear? Check all that apply.
A. hammer
B. anvil
C. ear canal
D. cochlea
E. ear drum
F. stirrup
Answer:
E, A, B, F
Explanation:
The Middle Ear contains:
eardrum.
cavity (also called the tympanic cavity)
ossicles (3 tiny bones that are attached) malleus (or hammer) - long handle attached to the eardrum. incus (or anvil) - the bridge bone between the malleus and the stapes. stapes (or stirrup) - the footplate; the smallest bone in the body.
Answer: rawr
Explanation:
hammer
anvil
ear drum
stirrup
two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of 20.g and sphere B has a mass of 10.g.
If both spheres leave the edge of the table at the same instant, sphere A will land
a. at some time after sphere B.
b. at the same time as sphere B.
c. at some time before sphere B.
d. There is not enough information to decide.
A would land before since its heavier
A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 6.0 m/s in the opposite direction. If the object is in contact with the wall for 2.0 ms, what is the magnitude of the average force on the object by the wall?
a. 9.8 kN.
b. 8.4 kN.
c. 7.7 kN.
d. 9.1 kN.
e. 1.2 kN.
Given that,
Mass of the object, m = 1.2 kg
Initial speed of the object, u = 8 m/s
Final speed of the object, v = -6 m/s (in opposite direction)
Time, t = 2 ms
To find,
The average force on the object by the wall.
Solution,
Let F be the force. Using Newton's second law of motion,
F = ma, a is acceleration
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{1.2\times ((-6)-8)}{2\times 10^{-3}}\\\\=8400\ N[/tex]
or
F = 8.4 N
So, the magnitude of average force in the object by the wall is 8.4 N.
*
If a rock falls for 3 seconds off of a bridge, how far will the rock fall?
-30 m
-45m
-60m
-75m
A disk rotates at a constant angular velocity of 30 degrees per second. Consider a point on the edge of the disk. Through how many degrees has it rotated after 3 seconds?
Answer:
The disk covers a rotation of 90º after 3 seconds.
Explanation:
Since the disk rotates at constant angular speed, we can determine the change in angular position ([tex]\Delta \theta[/tex]), measured in sexagesimal degrees, by the following kinematic formula:
[tex]\Delta \theta = \omega\cdot \Delta t[/tex] (1)
Where:
[tex]\omega[/tex] - Angular velocity, measured in sexagesimal degrees per second.
[tex]\Delta t[/tex] - Time, measured in seconds.
If we know that [tex]\omega= 30\,\frac{\circ}{s}[/tex] and [tex]\Delta t = 3\,s[/tex], then the change in angular position is:
[tex]\Delta \theta = \left(30\,\frac{\circ}{s} \right)\cdot (3\,s)[/tex]
[tex]\Delta \theta = 90^{\circ}[/tex]
The disk covers a rotation of 90º after 3 seconds.
A 10.0kg object is moving at 1 m/s when a force is applied in the direction of the objects motion, causing it to speed up to 4 m/s. If the force was applied for 5s what is the magnitude of the force
Answer:
F = 6[N].
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.
[tex]P=m*v\\or\\P=F*t[/tex]
where:
P = impulse or lineal momentum [kg*m/s]
m = mass = 10 [kg]
v = velocity [m/s]
F = force [N]
t = time = 5 [s]
Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.
[tex](m_{1}*v_{1})+F*t=(m_{1}*v_{2})[/tex]
where:
m₁ = mass of the object = 10 [kg]
v₁ = velocity of the object before the impulse = 1 [m/s]
v₂ = velocity of the object after the impulse = 4 [m/s]
[tex](10*1)+F*5=10*4\\10+5*F=40\\5*F=40-10\\5*F=30\\F=6[N][/tex]
A popular ride at an amusment park lifts
customers up to a height of 50 m and then
drops them threw a displacement of 50 m
before slowing them to a stop. How fast
are the customers going at the 50 m
mark?
Answer:
[tex]31.32\ m/s[/tex]
Explanation:
[tex]We\ are\ given\ that:\\Height\ to\ which\ there're\ lifted=50m\\Displacement\ during\ the\ descent=50m\\Now,\\In\ order\ to\ find\ the\ velocity\ of\ the\ customers\ at\ 50\ m,\\We\ can\ use\ the\ Third\ Equation\ Of\ Motion,\ which is:\\2as=v^2-u^2\\As\ we\ know\ that,\\Acceleration\ due\ to\ gravity=9.81\ m/s^2\ or\ roughly\ 10\ m/s^2\\Displacement=50\ m\\Initial\ velocity=0\ m/s^2\\ [As\ they\ stop\ when\ they\ reach\ the\ maximum\ height\ of\ 50\ m\\ and\ begin\ their\ descent][/tex]
[tex]By\ reconstructing\ the\ Third\ Equation\ Of\ Motion,\ we\ have:\\2gs=v^2\\Hence,\\v^2=2*9.81*50 \\v^2=981\ m^2/s^2 \\v=\sqrt{981\ m^2/s^2} \\v=31.32\ m/s[/tex]