How long does it take a P-wave to travel 7,000 km? ______ minutes b) How long does it take an S-wave to travel 7,000 km? ______ minutes c) How long does it take a Love wave to travel 7,000 km? ______ minutes d) How long does it take a Rayleigh wave to travel 7,000 km? ______minutes
Answer:
A. 8.64 secs.
B. 14.58 secs.
C. 26.002 secs.
D. 33.46secs.
Explanation:
A. P wave would travel 7000km
p-wave travels on a speed of 13.5km/s
= 7000km/13.5km/s
= 8.64 secs.
B. S-wave time to travel 7000km
s-wave travels on a speed of 8km/s
= 7000km/8km/s
= 14.58 secs.
C Love wave travels at a speed of 10,000m/s ( 2.7778 m/s ).
= 7000km to miles
= 4349.598m/2.788m/s
= 26.002 secs.
D. Rayleigh wave to travel 7,000 km
10,000m/s ( 2.1667 m/s ).
= 7000km to miles
= 4349.598m/2.1667m/s
= 33.46secs.
In each of the given pairs, choose which element will have the bigger atom. Give reasons for your choice. (a) Mg (atomic number 12) or Cl (atomic number 17) (b) Na (atomic number 11) or K (atomic number 19)
Answer:
Mg (atomic number 12)
K (atomic number 19)
Explanation:
The size of an atom is estimated in terms of its atomic radius.
The atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance (d) between two nuclei in the solid - state of metals.
Across a period, atomic radii decrease progressively from left to right. This is due to the progressive increase in the nuclear charge without increase in the number of electronic shells. Down a group, atomic radii increase progressively due to the successive shells of electrons being added which have been compensated for by the increase in nuclear charge.Cl is further right of Mg in the third period
K is below Na in the first group
How long is a day in Neptune
Answer: the long day in neptune would be .18383562 years!
Explanation:also for every day is 16 hours
how do you Convert 50 g to kg in an equation for physics
Answer:
Divide by 1000
Explanation:
A truck with a mass of 1330 kg and moving with a speed of 15.0 m/s rear-ends a 805 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.
Answer:
The Speed of the vehicles is 9.34m/s
Explanation:
For an elastic collision the two bodies move with similar velocities after collision
Given
M1=1330kg
V1=15m/s
M2=805kg
V2=0(the car is parked on neutral)
The formula is
M1V1+M2V2=(M1+M2)V
1330*15+805*0=(1330+805)V
19950+0=2135V
2135V=19950
divide both sides by 2135
V=19950/2135
V=9.34m/s
Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus.
What is the strength of the nucleus' electric field at the orbital radius of the electrons?
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?
Answer:
1. [tex] E = 1.14 \cdot 10^{13} N/C [/tex]
2. [tex]E_{k} = 9.1 \cdot 10^{-17} J[/tex]
Explanation:
1. The strength of the nucleus' electric field (E):
[tex]E = \frac{kq}{r^{2}}[/tex]
Where:
k: is the Coulomb constant = 9x10⁹ Nm²/C²
q: is the proton charge = 1.6x10⁻¹⁹ C
r: is the radius = 10⁻¹⁰ m
[tex]E = \frac{kq}{r^{2}} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*79*1.6 \cdot 10^{-19} C}{(10^{-10} m)^{2}} = 1.14 \cdot 10^{13} N/C[/tex]
2. The kinetic energy (Ek) of an electron is the following:
[tex] E_{k} = \frac{1}{2}mv^{2} [/tex]
Where:
m is the electron's mass = 9.1x10⁻³¹ kg
v: is the speed of the electron
We can find the speed of the electron by equaling the centripetal force (Fc) and the electrostatic force (Fe):
[tex] F_{c} = F_{e} [/tex]
[tex] \frac{mv^{2}}{r} = \frac{kq^{2}}{r^{2}} = qE [/tex]
[tex] v^{2} = \frac{qEr}{m} = \frac{1.6 \cdot 10^{-19} C*1.14 \cdot 10^{13} N/C*10^{-10} m}{9.1 \cdot 10^{-31} kg} = 2.00 \cdot 10^{14} m^{2}/s^{2} [/tex]
Now, we can find the kinetic energy:
[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*2.00 \cdot 10^{14} m^{2}/s^{2} = 9.1 \cdot 10^{-17} J [/tex]
I hope it helps you!
Suppose a signal from a vibrating motor is to be sampled discretely with a digital data acquisition system to determine the RPM of the motor. It is known that the maximum possible RPM is 1800. What is the minimum sampling frequency (in Hz) with which the signal can be sampled in order to measure the RPM of the motor
Answer:
f > 60 Hz
Explanation:
According to Nyquist's Sampling Theorem, to be fully reconstructed without any aliasing, the signal must be sampled at least more than twice during the period of the maximum frequency of the signal.In this case, the signal to be sampled has only one frequency.However, as we have the information in RPM, we need to convert this to cycles/sec (Hz) first, as follows:[tex]f = \frac{1800 rev}{min} * \frac{1 min}{60 sec} = 30 Hz[/tex]
Per Nyquist, fs > 2*30 Hz⇒ fs > 60 Hz
the equation of motion is given for a particle when s is in meters and t is in seconds. Find the acceleration after 4.5 seconds s
Answer:
Explanation:
The question is incomplete.
The equation of motion is given for a particle, where s is in meters and t is in seconds. Find the acceleration after 4.5 seconds.
s= sin2(pi)t
Acceleration = d²S/dt²
dS/dt = 2πcos2πt
d²S/dt² = -4π²sin2πt
A(t) = -4π²sin2πt
Next is to find acceleration after 4.5 seconds
A(4.5) = -4π²sin2π(4.5)
A(4.5) = -4π²sin9π
A(4.5) = -4π²sin1620
A(4.5) = -4π²(0)
A(4.5) = 0m/s²
The car alarm on a stationary car emits sound waves with a frequency of 450 Hz. If you are moving away from the stationary car at 20 m/s, what alarm frequency do you perceive
Answer:
Explanation:
The frequency of wave is directly proportional to velocity
f = kV
k = f/V
f1/V1 = f2/V2
Given
f1 = 450Hz
V1 = 343m/s
f2 = ?
V2 = 20m/s
Substitute into the formula
450/343 = f2/20
Cross multiply
343f2 = 450×20
343f2 = 9000
f2 = 9000/343
f2 = 26.24Hz
A person drops a marble from the top of a skyscraper. After falling four floors the marble has gained a certain speed. How many more floors will the marble have to fall to triple this speed?
a. 8
b. 12
c. 32
d. 48
Answer:
B. 12
Explanation:
4 x 3 = 12
What must be true if a wave's wavelength is short?
A
Its frequency is low.
B
It does not carry energy.
с
Its frequency is high.
D
The waves are visible.
Answer:
im sure its A
Explanation:
What is the force of a 12 kg object that is accelerating 6 m/s
We are given:
Mass of object (m) = 12 kg
acceleration (a) = 6 m/s²
Solving for the Force:
From newton's second law of motion:
F = ma
replacing the variables
F = 12*6
F = 72N
an air filled parallel plate capacitor has a capacitor of 1.3 pico farad. the separation of the plates is doubled and wax is inserted between them, the new capacitance is 2.6 pico farad. find the dielectric constant of the wax
You measure the radius of a sphere as (6.45 ± 0.30) cm, and you measure its mass as (1.79 ± 0.08) kg. What is the density and uncertainty in the density of the sphere, in kilograms per cubic meter?
Answer:
[tex](1630.13\pm 300.10)\ kg/m^3[/tex]
Explanation:
Given that,
The radius of a sphere is (6.45 ± 0.30) cm
Mass of the sphere is (1.79 ± 0.08) kg
Density = mass/volume
For sphere,
[tex]d=\dfrac{m}{V}\\\\d=\dfrac{m}{\dfrac{4}{3}\pi r^3}\\\\d=\dfrac{1.79\ kg}{\dfrac{4}{3}\pi (6.4\times 10^{-2}\ m)^3}\\\\d=1630.13\ kg/m^3[/tex]
We can find the uncertainty in volume as follows :
[tex]\dfrac{\delta V}{V}=3\dfrac{\delta r}{r}\\\\=3\times \dfrac{0.3\times 10^{-2}}{6.45\times 10^{-2}}\\\\=0.1395[/tex]
Uncertainty in mass,
[tex]\dfrac{\delta m}{m}=\dfrac{0.08}{1.79}\\\\=0.0446[/tex]
Now, the uncertainty in density of sphere is given by :
[tex]\dfrac{\delta d}{d}=\dfrac{\delta m}{m}+\dfrac{\delta V}V}\\\\=0.0446+0.1395\\\\\dfrac{\delta d}{d}=0.1841\\\\\delta d=0.1841\times d\\\\\delta d=0.1841\times 1630.13\\\\\delta d = 300.10\ kg/m^3[/tex]
Hence, the density pf the sphere is [tex](1630.13\pm 300.10)\ kg/m^3[/tex]
Light of wavelength 580 nm falls on a slit that is 3.70×10−3mm wide. Part A Estimate how far the first brightest diffraction fringe is from the strong central maximum if the screen is 10.0 m away.
Answer:
Explanation:
wavelength λ = 580 x 10⁻⁹ m
slit width d = 3.7 x 10⁻⁶ m
distance of screen D = 10 m
distance of first bright fringe = 1.5 x λ D / d
= 1.5 x 580 x 10⁻⁹ x 10 / 3.7 x 10⁻⁶
= 2351.34 x 10⁻³ m
= 2351.34 mm .
the neuron is considered a (a. Cell. (B.artery. (C. Vein
Answer:
A Cell
Explanation:
A 849-kg car starts from rest on a horizontal roadway and accelerates eastward for 5.00 s when it reaches a speed of 35.0 m/s. What is the average force exerted on the car during this time
Answer:
The average force exerted on the car during this time is 5,943 N
Explanation:
Given;
mass of the car, m = 849 kg
initial velocity of the car, u = 0
time of motion of the car, t = 5.00 s
final velocity of the car, v = 35 m/s
The average force exerted on the car during this time is given by;
[tex]F = ma \\\\F = \frac{m(v-u)}{t}\\\\F = \frac{849(35-0)}{5}\\\\F = \frac{849(35)}{5}\\\\ F = 849*7\\\\F = 5,943 \ N[/tex]
Therefore, the average force exerted on the car during this time is 5,943 N
Answer:
5943N
Let's say (+x) = eastward
Average horizontal acceleration
ax = vx -v0x/5.00s
= 35.0m/s-0/5.00s
= +7.09m/s
From here we apply the second law of newton
During this period average horizontal force acting on car
Summation x = max = (849kg)(+7.09m/s²)
= 5943N
+5.943x10³N
= 5.94kN east ward.
What is Triangle ill send a picture
The feeling of weightlessness occurs because _____________________.
there is no supporting force under your mass.
there is no gravity present.
there is only a small amount of gravity present.
Answer:
there is only a small amount of gravity present.
Explanation:
this is because the only force acting upon your body during free fall is the force of gravity which is a non contact force.
a man weighing 490 n on earth weighs only 81.7 n on the moon. His mass on the moon is__kg. (Use g=9.8 m/s2
Answer:
m = 50 [kg]
Explanation:
In order to solve this problem we must be clear about the difference between weight and mass. Weight is the product of mass by the acceleration of the planet or the star. While the mass is always preserved it never changes regardless of where it is located.
So for the earth we have:
g = gravity acceleration = 9.8 [m/s^2]
m = mass [kg]
W = weigth = 490 [N]
therefore the mass will be:
m = W/g
m = 490/9.8
m = 50 [kg]
Now it is important to remember that the mass will be the same on the moon or on the earth, but the weight will be different, because the gravity acceleration of the moon is different from the gravity acceleration on earth
So the gravity on the moon is equal to:
81.7 = 50 * gm
gm = 1.634 [m/s^2]
Which of the following is true about the following lever?
Answer:
It will rotate counter-clockwise.
Explanation:
The reason is that there is more Nm on the left side which will lift the lever towards the left side.
The lever would rotate counterclockwise as the torque on the left is higher, so, option B is correct.
What is torque?The force which causes the object to rotate about any axis is called perpendicular distance.
Torque is a twisting or turning force that frequently results in rotation around an axis, which may be a fixed point or the center of mass. The ability of something rotating, such as a gear or a shaft, to overcome turning resistance is another way to think of torque.
Given:
The force on the left side, f = 40 N,
The force on the right side, F = 80 N,
The distance of the pivot from the left side, d = 9 m,
The distance of the pivot from the right side, d = 3 m,
Calculate the torque on both side as shown below,
Torque on the left side = 40 × 9
Torque on the left side = 360 Nm
Torque on the right side = 80 × 3
Torque on the right side = 240 Nm,
Here, the torque on the left side is more,
Thus, the lever will rotate counterclockwise.
To know more about torque:
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A race car accelerates from rest to a velocity of +90 m/s over a distance of 423m. Determine the acceleration of the race car.
Answer:
9.57m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Final velocity = 90m/s
Distance = 423m
Unknown:
Acceleration of the race car = ?
Solution:
To solve this problem, we should apply one of the appropriate motion equations;
V² = U² + 2as
Where V is the final velocity
U is the initial velocity
a is the acceleration
s is the distance
90² = 0² + 2 x a x 423
8100 = 846a
a = 9.57m/s²
a box with a Constance velocity has a 5 N of force applied to it from all sides and direction. what will happen to the motion of the box as result?
A-the object will come to rest
b-the velocity of the object will remain the same
c- the velocity of the object will decrease
d- the velocity of the object will increase
Answer:
b-the velocity of the object will remain the same
Explanation:
Forces from opposite sides cancel each other, so there is no net force on the box that would affect its motion. The velocity of the box will remain unchanged.
__
(The box may be crushed, but it will continue in the same direction at the same speed.)
Answer:
b the velocity of the object will remain the same
Explanation:
use your brain:)
Define reflection of sound?
A tennis ball is dropped from a roof 16 meters from the ground. How long does it take
for the ball to reach the ground?
3.3 seconds would be the right answer
The time taken for the tennis ball dropped from the roof-top to reach the ground level is 1.8 seconds.
What is Motion?Motion is simply the change in position of an object or particle over time.
From the Second Equation of Motion;
s = ut + (1/2)gt²
Where s is the distance from ground level, u is initial velocity, t is time elapsed and g is acceleration due to gravity ( g = 9.8m/s² ).
Given the data in the question;
Since the ball was initially at rest before it was dropped.
Initial velocity u = 0Height or distance from gound level s = 16mTime taken to reach the gound t = ?We substitute our values into the expression above.
s = ut + (1/2)gt²
16m = ( 0 × t ) + ( (1/2) × 9.8m/s² × t² )
16m = 0.5 × 9.8m/s² × t²
16m = 4.9m/s² × t²
t² = 16m / 4.9m/s²
t² = 3.2653s²
t = √(3.2653s²)
t = 1.8s
Therefore the time taken for the tennis ball dropped from the roof-top to reach the ground level is 1.8 seconds.
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A 20m length wire 1.5mm in diameter has a resistance of 2.5 ohm what is the resistance of a 35m length of wire 3mm in diameter made of the same material?
From the calculations, the resistance of the material is 1.1 ohm.
What is the resistance of the wire?Given that;
R α l/A
R = ρl/A
R = resistance
l = length
A = Area
ρ = resistivity
Now;
A = πr^2
A = 3.142 * (1.5 * 10^-3/2)^2
A = 1.77 * 10^-6 m^2
ρ = RA/l
ρ = 2.5 * 1.77 * 10^-6 /20
ρ = 2.2 * 10^-7 ohm/m
Now;
R =ρl/A
A = 3.142 * (3 * 10^-3/2)^2
A = 7.1 * 10^-6 m^2
Thus
R = 2.2 * 10^-7 * 35/ 7.1 * 10^-6
R = 1.1 ohm
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The acceleration of the spacecraft in which the Apollo astronauts took off from the moon was 3.4 m/s2 m / s 2 . On the moon, g g = 1.6 m/s2 m / s 2 . what's the apparent weight
Complete Question
The acceleration of the spacecraft in which the Apollo astronauts took off from the moon was 3.4 m/s2. On the moon, g = 1.6 m/s2. What was the apparent weight of a 75 kg astronaut during takeoff?
Answer:
The value is [tex]N = 375 \ N[/tex]
Explanation:
From the question we are told that
The acceleration is [tex]a = 3.4 \ m/s^2[/tex]
The acceleration due to gravity in the moon is [tex]g = 1.6 m/s^2[/tex]
The mass of the astronaut is [tex]m = 75 \ kg[/tex]
Generally the apparent weight is mathematically represented as
[tex]W = ma + mg[/tex]
=> [tex]W = 3.4 * 75 + 1.6 * 75[/tex]
=> [tex]W = 375 \ N[/tex]
I need help with this please
There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?
Answer:
We are given:
V1 = 5.5L T1 = -38 C or 235 k
V2 = 1.3L T2 = T
From the gas equation:
PV = nRT
Since the pressure (P) , number of moles (n) and the universal gas constant (R) are constants, we can write the same equation as:
V / T = k (where k is a constant)
so a bit more insight, since the values noted above are constant, when multiplied by each other, they will provide us with a constant number irrespective of the value of the variables
Changing the variables for the first case:
V1 / T1 = k (where k is the same constant) ----------------(1)
Similarly,
V2 / T2 = k (again, k has the same value)----------------(2)
From (1) and (2):
k is the common value
V1 / T1 = V2 / T2
Replacing the variables
5.5 / 235 = 1.3 / T
T = 1.3 * 235 / 5.5
T = 55.54 k
Therefore, at 55.54 K the gas will have a volume of 1.3L
Thanks to me, you can see straight through the wall. What am I?
Answer:
A window
Explanation:
Answer:
a window.. duh
Explanation: