Answer:
1.07 × 10⁸ m/s
Explanation:
Using the relativistic Doppler shift formula which can be expressed as:
[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]
here;
[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v
[tex]\lambda_s =[/tex] observed wavelength from the source's frame.
Given that:
[tex]\lambda _o[/tex] = 656.3 nm
[tex]\lambda_s =[/tex] 953.3 nm
We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.
From the above equation, let's make (v/c) the subject of the formula: we have:
[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]
[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]
[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]
[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]
[tex]\dfrac{v}{c} =0.357[/tex]
v = 0.357 c
To m/s:
1c = 299792458 m/s
∴
0.357c = (299 792 458 × 0.357) m/s
= 107025907.5 m/s
= 1.07 × 10⁸ m/s
By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect the earth's rotational effects
Answer:
[tex]Weight\ loss=1.6321N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=85.9kg[/tex]
Altitude [tex]h= 6.33 km[/tex]
Let
Radius of Earth [tex]r=6380km[/tex]
Gravity [tex]g=9.8m/s^2[/tex]
Generally the equation for Gravity at altitude is mathematically given by
[tex]g_s=9.8(\frac{6380}{6380+6.33})^2[/tex]
[tex]g_s=9.781m/s^2[/tex]
Therefore
Weight at sea level
[tex]W_s=9.8*85.9[/tex]
[tex]W_s=841.82N[/tex]
Weight at 6.33 altitude
[tex]W_a=9.781*85.9[/tex]
[tex]W_a=840.2N[/tex]
Therefore
[tex]Weight loss=W_s-W_b[/tex]
[tex]Weight loss=841.82-840.2[/tex]
[tex]Weight loss=1.6321N[/tex]
You’re working with a patient who suddenly falls. You should?
A force of 350 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 20 centimeters to 50 centimeters
Answer:
52.5 J
Explanation:
Applying,
Hook's law,
F = ke............... Equation 1
Where F = Force, k = spring constant, e = extension.
make k the subject of the equation
k = F/e............ Equation 2
From the question,
Given: F = 350 Newtons, e = 30 cm = 0.3 m
Substitute these values into equation 2
k = 350/0.3 N/m
Also,
W = 1/2(ke²).................. Equation 3
Where W = work done in stretching the spring.
Also given: e = (50-20) cm = 30 cm = 0.3 m, k = 350/0.3 N/m
Substitute these values into equation 3
W = 1/2(350/0.3)(0.3²)
W = 350×0.3/2
W = 52.5 J
What is sieving? Give an example where this method is used. (2)
Answer:
sieving is when you separate particles of different sizes.
Explanation:
separating sand mixtures
separating chaffs from local garri
the boiling point of F2 much lower than the boiling point of NH3
Answer:yeah it A
Explanation:
Compared to its weight on Earth, a 5kg object on the moon will weigh
The same amount
Less
More
Answer:
Less
Explanation:
Weight is a force measurement. The object's mass is 5kg not its weight. To find its weight you have to take the mass of an object and multiply it by the acceleration of gravity. The acceleration of gravity is greater on earth than on the moon so therefore the object will weigh less on the moon.
A 40-kg crate is being lowered with a downward acceleration is 2.0 m/s2 by means of a rope. (a) What is the magnitude of the force exerted by the rope on the crate
Answer:
F = 312 N
Explanation:
Given that,
The mass of a crate, m = 40 kg
Acceleration of the crate, a = 2 m/s²
As the carte is falling downward, the net force exerted by the rope on the carte is given by :
F = m(g-a)
Put all the values,
F = 40(9.8-2)
F = 312 N
Hence, the required force exerted by the rope on the crate is equal to 312 N.
Dos cargas puntuales iguales y negativas, q1=q2=-24micro C se localizan en x=0 y y=38m y x=0 y y=-7m, respectivamente. Calcula la magnitud de la fuerza electrica total en N que ejercen estas dos cargas sobre una tercera, tambien puntual, Q=26micro C en y=0 y x=16m
Answer:
F_net = 9.87 10⁻⁴ N
Explanation:
Let's use that force is a vector magnitude
∑ F = F₁₃ + F₂₃
De bold arfe vectros. The force is the electric force, we use that charges of the same sign repel and when the charges are of a different sign they attract
the charges q1 and q2 are negative and the charge q3 is positive with the positions y1 = 38 m, y2 = -7m, y3 = 16 m
∑ F = F₁₃ - F₂₃
F_net = [tex]k \frac{q_1q_3}{r_{13}^2 } - k \frac{q_2q_3}{r_{23}^2 }[/tex]
in this case q₁ = q₂ = q
F_net = k q q₃ ( )
let's look for the distance
r₂₃ = y₂ - y₃
r₂₃ = -7 -16
r₂₃ = - 23 m
r₁₃ = 38 - 16
r₁₃ = 22 m
let's calculate
F_net = 9 10⁹ 24 26 10⁻¹² ( )
F_net = 5.616 ( 1.758 10⁻⁴ )
F_net = 9.87 10⁻⁴ N
Each of the four expansion models (recollapsing, critical, coasting, and accelerating) predict different ages for the universe, given the current expansion rate. Why is this
Answer:
This is because the age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.
Explanation:
The age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.
This is due to the fact that the expansion rate in the coasting model is constant and never changes. Because the cosmos is growing faster now than during the old days, recollapsing and critical models give shorter ages. According to the accelerating model, the universe is growing at a slower rate currently than in the past, implying an older age.
Iron is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron forms a crystal with an fcc unit cell and a lattice constant . Calculate the density of Iron.
The question is incomplete. The complete question is :
Iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms a crystal with an fcc unit cell and a lattice constant, a = 0.352 nm. Calculate the density of Iron β.
Solution :
The density is given by :
[tex]$\rho = \frac{ZM}{a^3N_0} \ \ g/cm^3$[/tex] ..................(i)
Here, Z = number of atoms in a unit cell
M = atomic mass
[tex]$N_0$[/tex] = Avogadro's number = [tex]$6.022 \times 10^{23}$[/tex]
a = edge length or the lattice constant
Now for FCC lattice, the number of atoms in a unit cell is 4.
So, Z = 4
Atomic mass of iron, M = 55.84 g/ mole
Given a = 0.352 nm = [tex]$3.52 \times 10^{-8}$[/tex] cm
From (i),
[tex]$\rho = \frac{ZM}{a^3N_0} $[/tex]
[tex]$\rho = \frac{4 \times 55.84}{(3.52 \times 10^{-8})^3 \times 6.022 \times 10^{23}} $[/tex]
[tex]$= 8.51 \ \ g \ cm ^{-3}$[/tex]
Therefore, the density of Iron β is [tex]$ 8.51 \ \ g \ cm ^{-3}$[/tex].
Is a measurement is precuse it must also be accurate
Which statements are true of noble gases?
Check all that apply.
A. They are metalloids.
B. Their valence shells are full of electrons.
C. They are not very reactive.
D. All of the noble gases have at least two electron shielding layers.
Consider a solid sphere and a solid disk with the same radius and the same mass. Explain why the solid disk has a greater moment of inertia than the solid sphere, even though it has the same overall mass and radius.
Answer:
Explanation:
In a Solid sphere; the moment of inertia around its geometrical axis can be expressed by using the formula:
[tex]\mathtt{I_s = \dfrac{2}{5} M_s R^2_s}[/tex]
For the solid disk; the moment of inertia around the central axis is:
[tex]\mathtt{I_D= \dfrac{1}{2}M_DR_D^2}[/tex]
Suppose [tex]M_D = M_S[/tex]; then we can say both to be equal to M
As well as [tex]R_D = R_S[/tex]; then that too can be equal to R
Now;
[tex]\mathtt{I_s = \dfrac{2}{5} M R^2} --- (1)[/tex]
[tex]\mathtt{I_D= \dfrac{1}{2}MR^2}---(2)[/tex]
Multiplying equation (1) by 2, followed by dividing it by 2; we have:
[tex]\mathtt{I_s= \dfrac{2}{5}MR^2} \times \dfrac{2}{2}[/tex]
[tex]I_s = \dfrac{4}{5} \times \dfrac{1}{2}MR^2 \\ \\ I_s = \dfrac{4}{5}\times I_D \\ \\ I_s > I_D[/tex]
Wave 1
WWW
m
Resulting
Wave
Wave 2
Wave 2 Phase Shift
The resulting wave has the largest possible amplitude when Wave-1 and Wave-2 are exactly in step ... their peaks both happen at the same time and their troughs both happen at the same time.
This means that Wave-1 and Wave-2 have the same frequency, and the phase shift from one wave to the other is zero.
When all of that happens, the amplitude of the resulting wave is the sum of the amplitudes of Wave-1 and Wave-2. If Wave-1 and Wave-2 have the same amplitude, then the resulting wave will have double that amplitude.
A 4.9 A current is set up in a circuit for 4.7 min by a rechargeable battery with a 12 V emf. By how much is the chemical energy of the battery reduced
Answer:
E = 16581.6 J
Explanation:
Given that,
Current, I = 4.9 A
Time for which the current is set up, I = 4.7 min = 282 s
The voltage of the battery, V = 12 V
We need to find how much chemical energy of the battery reduced. Let It is E. We know that,
E = P t
Where
P is power of battery, P = VI
So,
[tex]E=VIt[/tex]
Put all the values,
[tex]E=12\times 4.9\times 282\\E=16581.6\ J[/tex]
So, 16581.6 J of chemical energy of the battery is reduced.
A 3.50 kg basket of cookies sits on a 2.00 m high shelf. What is the gravitational potential energy of the basket?
pls help
Answer:
68.6 J
Explanation:
Applying,
P.E = mgh............... Equation 1
Where P.E = Potential Energy of the basket, m = mass of the basket, g = acceleration due to gravity of the basket, h = height of the basket
From the question,
Given: m = 3.5 kg, h = 2.00 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
P.E = 3.5×2×9.8
P.E = 68.6 J
Hence the potential energy of the basket is 68.6 J
PLEASE HELPPPPPP ME PLS PLS
If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
This question is incomplete, the complete question is;
In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration?
b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
Answer:
a) the angular acceleration is 80 rad/s²
b) the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
Explanation:
Given the data in the question;
from the graph below;
Angular Velocity at time 0s [tex]w_o[/tex] = 12 rad/s
Angular Velocity at time 0.15s [tex]w_f[/tex] = 24 rad/s
a) What is the angular acceleration;
Angular acceleration ∝ = ( [tex]w_f[/tex] - [tex]w_o[/tex] ) / dt
we substitute
Angular acceleration ∝ = ( 24 - 12 ) / 0.15
Angular acceleration ∝ = 12 / 0.15
Angular acceleration ∝ = 80 rad/s²
Therefore, the angular acceleration is 80 rad/s²
b)
If the ball is 0.60 m from her shoulder, i.e s = 0.6 m
the tangential acceleration of the ball will be;
a = ∝ × s
we substitute
a = 80 × 0.6
a = 48 m/s²
a = ( 48 / 9.8 )g
a = 4.9 g
Therefore, the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
C.
A palm fruit dropped to the ground from the top of
a tree 45m tall. How long does it take to reach the
ground? A. 9s B. 4.5s C. 6 D. 7.5s E. 35
(g = 10ms2).
Answer:
b 4.5
Explanation:
time=distance/speed
Keesha is looking at a beetle with a magnifying glass. She wants to see an upright, enlarged image at a distance of 25 cm. The focal length of the magnifying glass is +5.0 cm. Assume that Keesha's eye is close to the magnifying glass.
(a) What should be the distance between the magnifying glass and the beetle?
(b) What is the angular magnification?
Answer:
a) p = 4.167 cm, b) m = + 6
Explanation:
a) For this exercise we must use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
In this case the distance to the image q = 25 cm and the focal length is f = 5.0 cm
Since the object and its image are on the same side of the lens, the distance to the image by the sign convention must be negative.
[tex]\frac{1}{p } = \frac{1}{f} - \frac{1}{q}[/tex]
[tex]\frac{1}{p} = \frac{1}{5} - \frac{1}{-25}[/tex]
[tex]\frac{1}{ p}[/tex] = 024
p = 4.167 cm
b) angular magnification
m = h ’/ h = - q / p
m = - (-25) /4.167
m = + 6
the positive sign indicates that the image is straight and enlarged
If two dogs are pulling a bone with force-20Newtons in opposite direction, then the resultant force is
Answer:
Newtons third law of motion: Balanced forces
Every action has a corresponding and opposing response, according to Newton's third law of motion. As a result, forces always work in pairs. Once more, tug-of-war is a prime illustration.
What force in opposite direction follow newton law?The third law of motion by Newton states that equal, but diametrically opposed forces always act in pairs. There is an equal but opposite reaction to every action, to put it another way.
The forces are balanced if the pullers are exerting equal force but going in the opposite direction on either side of the rope. There is hence no motion.
Although equal and opposite in nature, action and reaction forces cannot be balanced since they act on separate things and do not cancel one another out.
Therefore, This means that when you push against a wall, the wall pushes back against you with an equal amount of force.
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One coulomb represents how many electrons?
a. 1 electron
b. 100 electrons
C. 6.25 quintillion electrons
d. 6.25 million-million electrons
e, none of the above
Answer:
6.24 x 1018 electrons.
Explanation:
So I think C
Isotopes of the same element always have the same
(2 points)
O atomic mass number
O A-number
O Z-number
O neutrinos
Answer:
Z-number
Explanation:
The Z number is the number of protons in an atom, and this does not change when an isotope is created. I got it right on the test.
The index of refraction of n-propyl alcohol is 1.39. Find the angle of refraction of light in that medium if light comes from air with an angle of incidence of 55 degrees.
Answer:
36.11 degrees
Explanation:
index of refraction n = sin i/sinr
i is the angle of incidence
r is the angle of refraction
Substitute into the expression
1.39 = sin55/sin(r)
1.39 = 0.8191/sin(r)
sin(r) = 0.8191/1.39
sin(r) = 0.5893
r = arcsin(0.5893)
r = 36.11
hence the angle of refraction of light is 36.11 degrees
A voltage of 75 V is placed across a 150 Ω resistor. What is the current through the resistor?
Answer:
0.5 A
Explanation:
Applying,
V = IR.................. Equation 1
Where V = Voltage, I = current, R = Resistance.
make I the subject of the equation
I = V/R............... Equation 2
From the question,
Given: V = 75 V, R = 150 Ω
Substitute these values into equation 2
I = 75/150
I = 0.5 A.
Hence the cuurent through the resistor is 0.5 A
How do a parachutes work??4-5 sentences plsss help rn
Answer:
Explanation:
A parachute works by forcing air into the front of it and creating a structured 'wing' under which the canopy pilot can fly. Parachutes are controlled by pulling down on steering lines that change the shape of the wing, cause it to turn. The main forces acting on a parachute are gravity and drag. When you first release the parachute, the force of gravity pulls it downward, and the parachute speeds toward the ground. The faster the parachute falls, though, the more drag it creates.
Many adventures like to go rafting on the Colorado River through Grand Canyon National Park. There are many locations where the river becomes more narrow, both the distance between the canyon walls as well as the depth changes due to debris like boulders on the bottom of the river; this leads to changes in the water speed. In the park, the Colorado River has an average width of 100m and an average depth of 8m, and an average speed of 3 m/s. At the Lava Falls Rapids, the river has an average width of about 25m and an average depth of about 15m. What is the approximate speed of the water in this location
Answer:
6.4 m/s
Explanation:
Given that :
The average width of the Colorado river = 100 m
Average depth of the river is = 8 m
Therefore, area = [tex]$A_1= 100 \ m \times 8 \ m$[/tex]
Speed of the river, [tex]$v_1 = 3 \ m/s$[/tex]
After the lava falls on the river,
Width of the river becomes = 25 m
Depth of the river became = 15 m
Therefore, area = [tex]$A_2= 25 \ m \times 15 \ m$[/tex]
Now, since the volume flow rate of the Colorado river is same, then from the Continuity equation,
[tex]$Q_1=Q_2$[/tex]
[tex]$A_1v_1=A_2v_2$[/tex]
∴ [tex]$100 \times 8 \times3 = 25 \times 15 \times v_2$[/tex]
[tex]$v_2=\frac{100 \times 8 \times 3}{25 \times 15}$[/tex]
= 6.4 m/s
Therefore, the speed of the river in this location is 6.4 m/s
A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it up at 43 cm from the end of the plank with force F2. If the plank has a mass of 13.7 kg and its center of gravity is at the middle of the plank, what is the force F1
Answer: [tex]115.52\ N[/tex]
Explanation:
Given
Length of plank is 1.6 m
Force [tex]F_1[/tex] is applied on the left side of plank
Force [tex]F_2[/tex] is applied 43 cm from the left end O.
Mass of the plank is [tex]m=13.7\ kg[/tex]
for equilibrium
Net torque must be zero. Taking torque about left side of the plank
[tex]\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N[/tex]
Net vertical force must be zero on the plank
[tex]\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N[/tex]
Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.How far must the faster car travel before it has a 15-min lead on the slower car
Answer:
The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Explanation:
Given;
speed of the faster car, v₁ = 60 mi/h
speed of the slower car, v₂ = 55 mi/h
Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles
[tex]\frac{x}{55} - \frac{x}{60} = \frac{15}{60}[/tex]
Note: divide 15 mins by 60 to convert to hours for consistency in the units.
[tex]\frac{x}{55} - \frac{x}{60} = \frac{15}{60}\\\\multiple \ through \ by \ 660\\\\12x - 11x = 165\\\\x = 165 \ miles[/tex]
Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Answer this
a) which ink is likely to be pure? Why?
b) What does the chromatography tell us about ink Y
c) Why are the three different spots separated out from ink Y found at different heights?
Answer:
a) Ink X is likely to be pure because it only contain 1 spot.
b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.
c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.
The different spots from Y are found at various heights because they represent different compounds.
What is chromatography?The term chromatography has to do with a method of separating the component of a substance. The term chromatography originally means color writing.
We can see that the pure ink is the ink marked X. We can see from the chromatogram that Y is a mixture of colors. The different spots from Y are found at various heights because they represent different compounds.
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