Explanation:
It is given that,
Wavelength of x-rays = 2 nm
Plane spacing, d = 0.281 nm
It is assumed to find the scattering angle for second order maxima.
For 2nd order, Bragg's law is given by :
[tex]2d\sin\theta=n\lambda[/tex]
For second order, n = 2
[tex]\sin\theta=\dfrac{n\lambda}{2d}\\\\\sin\theta=\dfrac{2\times 2\ nm}{2\times 0.28\ nm}\\\\\theta=\sin^{-1}(7.14)[/tex]
Here, θ is not defined. Also, the wavelength of x-rays is more than the plane spacing. It means that it cannot produce any diffraction maximum.
A person stands on a scale in an elevator. His apparent weight will be the greatest when the elevator(a) is standing still.(b) is moving upward at constant velocity.(c) is accelerating upward.(d) is moving downward at constant velocity.(e) is accelerating downward.
Answer:
(c) is accelerating upward
Explanation:
When a person stand in an elevator moving upwards, he feels heavier because the elevator's floor presses harder on his feet, and the scale will show a higher reading than when the elevator is at rest.
According to Newton's second law
R = mg + ma
where;
R is the reading of the scale = apparent weight of the person
mg is the normal weight of the person
ma is the upward force acting on the person
Therefore, his apparent weight will be the greatest when the elevator is accelerating upward
(c) is accelerating upward
g A person walks 70 m west, then turns and walks 30 m east. Find the average speed and velocity if the time for the total trip is 45 seconds.
Answer:
Average speed = 2.22m/s
Velocity = 0.89 m/s west
Explanation:
(i) Since speed is a scalar quantity, the direction of movement by the person is irrelevant. Therefore,
Average speed = total distance traveled / time taken
Where;
Total distance = 70m + 30m = 100m
Time taken = 45 seconds
Average speed = 100 / 45 = 2.22m/s
(ii) However, velocity is a vector quantity and so the direction of movement by the person is of utmost importance. Therefore,
Velocity = displacement / time taken
Where;
displacement = -70m + 30m [West direction is taken as negative and east is taken as positive]
displacement = -40m [The negative sign just means that the net displacement is in the direction of the West].
Thus, displacement can be written as 40m west.
Therefore,
Velocity = 40 / 45 = 0.89m/s west
A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. What is the change of the ball's momentum
Answer:
The change in momentum is Δp= 0.028 kg m/sExplanation:
An impulse describes a change in momentum. The change in momentum of an object is its mass times the change in its velocity.
The change in moment is given by the expression below
Δp=m⋅(Δv)=m⋅(vf−vi) .
Given data
mass m= 0.140-kg
initial velocity vi= 120 m/s
final velocity vf= 1 m/s
substituting we have
Δp=m⋅(Δv)=0.14⋅(1−1.2)
Δp=m⋅(Δv)=0.14⋅(-0.2)
Δp= 0.028 kg m/s
The change in momentum was found to be Δp= 0.028 kg m/s
At time t=0 , a cart is at x=10 m and has a velocity of 3 m/s in the −x -direction. The cart has a constant acceleration in the +x -direction with magnitude 3 m/s^2 < a < 6 m/s^2 . Which of the following gives the possible range of the position of the cart at t=1 s ?
Answer:
Explanation:
The minimum magnitude of acceleration = 3 m /s²
displacement at t = 1
s = ut + 1 /2 at²
= -3 x 1 + .5 x 3 x 1²
= - 3 + 1.5
= - 1.5 m
position at t = 1 s
= 10 - 1.5
= 8.5 m
The maximum magnitude of acceleration = 6 m /s²
displacement at t = 1
s = ut + 1 /2 at²
= -3 x 1 + .5 x 6 x 1²
= - 3 + 3
= 0
position at t = 1 s
= 10 +0
= 10 m
So range of position is 8.5 m to 10 m .
We want to find the range of possible positions for the cart at t = 1s.
The range is:
14.5m ≤ p ≤ 16m
Finding the cart's motion equations.
First, we know that the acceleration of the cart is a, a constant (but we don't know the exact value yet) so we write the acceleration as:
a(t) = a
To get the velocity equation we integrate over time, and we know that the velocity at t = 0s is 3m/s, so that will be the constant of integration.
v(t) = a*t + 3m/s
To get the position we integrate again, the initial position is x = 10m, so that will be the constant of integration.
p(t) = (a/2)*t^2 + (3m/s)*t + 10m
Now, to get the range of possible positions for t = 1s we need to use the minimum and maximum accelerations and evaluate the position equation in t = 1s.
The minimum acceleration is a = 3m/s^2, so we have the minimum position:
p(1s) = (3m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 14.5m
The maximum acceleration is 6m/s^2, so the maximum position is:
P(1s) = (6m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 16m
So the range of the position at t = 1s is:
14.5m ≤ p ≤ 16m
If you want to learn more about motion equations, you can read:
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Which is not an example of an external force acting on an object? (1 point)
• a meteor traveling unhindered through space
O a hockey puck striking the back of the net and coming to rest
O a billiard ball bouncing off the rail of a pool table
O a batter hitting a baseball
Answer:
A. a meteor traveling unhindered through space
Explanation:
If you are driving at a constant velocity of 5 m/s what is your
acceleration?
"Acceleration" means the rate at which velocity is changing.
If velocity is constant, then it isn't changing.
If velocity isn't changing, then acceleration is zero.
Pls help answer 7 and 8
Answer:
science is the knowledge gained through observations and experimentations
Characteristics
consistency
observability
predictability
testability
tentativeness
components of nature are soil,atmosphere and radiation
The Type K thermocouple has a sensitivity of about 41 micro-Volts/℃, i.e. for each degree difference in the junction temperature, the output changes by 41 micro-Volts. If you have a 32-bit ADC, what is the smallest temperature change you can detect if the ADC range is 10 V?
Answer:
5.68*10^-5 °C
Explanation:
ADC is an acronym that means analog to digital conversion.
The numbers of bit in any ADC is the level of Voltage it possess.
The question states 32 bits, this means that it's total voltage levels is 2^32
Also, the question gives us a range of 10 V for the ADC. Now we can evaluate the resolution of the voltage as:
Δv = 10 / (2^32)
Δv = 10 / 4.29^9
Δv = 2.33*10^-9
The thermocouple sensitivity is 41 micro volts/°C
Thermocouple sensitivity = Δv/ΔT, where ΔT is the temperature change
ΔT = Δv/thermocouple sensitivity
ΔT = 2.33*10^-9 / 41*10^-6
ΔT = 5.68*10^-5 °C
Thus, the smallest temperature change is 5.68*10^-5 °C
A resistor is connected in series with a power supply of 20.00 V. The current measure is 0.50 A. What is the resistance of the resistor
Answer:
40 Ω
Explanation:
The following data were obtained from the question:
Potential difference (V) = 20 V
Current (I) = 0.50 A
Resistance (R) =?
From ohm's law:
V = IR
Where:
V is the potential difference.
I is the current.
R is the resistance.
With the above formula, we can obtain the resistance of the resistor as follow:
Potential difference (V) = 20 V
Current (I) = 0.50 A
Resistance (R) =?
V = IR
20 = 0.5 × R
Divide both side by 0.5
R = 20/0.5
R = 40 Ω
Therefore, the resistance of the resistor is 40 Ω
Determine the weight in newtons of a woman whose weight in pounds is 157. Also, find her mass in slugs and in kilograms. Determine your own weight in newtons.
Answer:
Weight of the woman in Newton
[tex]x = 698.6 \ N[/tex]
Mass of the woman in slug
[tex]Mass = 4.86 \ slug[/tex]
Mass of the woman in kg
[tex]Mass = 16 \ kg[/tex]
My weight in Newton
[tex]W = 784 \ N[/tex]
Explanation:
From the question we are told that
The weight of the woman in pounds is [tex]W = 157 \ lb[/tex]
Converting to Newton
1 N = 0.22472 lb
x N = 157
=> [tex]x = \frac{157 * 1}{0.22472}[/tex]
=> [tex]x = 698.6 \ N[/tex]
Obtaining the mass in slug
[tex]Mass = \frac{W}{g}[/tex]
Here [tex]g = 32.2 ft/s^2[/tex]
So
[tex]Mass = \frac{157 }{32.2}[/tex]
[tex]Mass = 4.86 \ slug[/tex]
Obtaining the mass in kilogram
[tex]Mass = \frac{W}{g}[/tex]
Here [tex]g = 9.8 \ m/s[/tex]
So
[tex]Mass = \frac{157 }{9.8}[/tex]
[tex]Mass = 16 \ kg[/tex]
Generally weight is mathematically represented as
[tex]W = m * g[/tex]
Given that my mass is 80 kg then my weight is
[tex]W = 80 *9.8[/tex]
[tex]W = 784 \ N[/tex]
With the sun and the earth back in their regular positions,consider a space probe with mass mp = 125 kg launched from the earth towardthe sun. When the probe is exactly halfway between the earth andthe sun along the line connecting them, what is the direction ofthe net gravitational force acting on the probe? Ignore the effects of other massive objectsin the solar system, such as the moon and other planets.
A. The force is toward the sun.
B. The force is toward the earth.
C. There is no net force because neither the sunnor the earth attracts the probe gravitationally at themidpoint.
D. There is no net force because the gravitationalattractions on the probe due to the sun and the earth are equal insize but point in opposite directions, so they cancel each otherout.
Answer:
A
Explanation:
Gravitational force given by
F_ps =G(m_pM_s)/r^2
G= gravitational constant
m_p=mass of the probe =125 Kg
M_s= mass of sun
r= distance between probe and Sun.
Similarly, F_pe = G(m_pM_e)/r^2.
m_p = mass of probe
M_e = mass of earth
Clearly, M_s> M_e . Therefore, F_ps> F_pe.
Hence, A. The force is toward the sun.
GnuPG (GPG) uses __________ to create a random set of keys. keyrings hashing entropy steganography
Answer:
entropy
Explanation:
GPG is an acronym for GNU Privacy Guard, which is considered as free software, is a command-line tool for generating cryptographically secure keys.
These cryptographically secure keys can be created by using entropy, through reliable entropy source, like /dev/random, which in turn derives its entropy from physical devices connected to the machine, such as keyboards, mice, and disks.
Hence, the right answer is ENTROPY
A round coin has a diameter of 19.55 mm, a thickness of 1.55 mm, and a mass of 2.50 g. What is its density?
Answer:
Density, d = 5.37 g/cm³
Explanation:
Given that,
Diameter of coin is 19.55 mm
Thickness of the cone is 1.55 mm
Mass of the coin is 2.5 g
We need to find its density. Density is equal to mass upon volume. So,
[tex]d=\dfrac{m}{V}[/tex]
V is volume, V = A d, d is thickness
Radius = d/2 = (1.955/2) cm=0.9775 cm
Thickness, d = 1.55 mm = 0.155 cm
[tex]d=\dfrac{m}{\pi r^2 \times d}\\\\d=\dfrac{2.5}{\pi (0.9775\ cm)^2\times 0.155\ cm}\\\\d=5.37\ g/cm^3[/tex]
So, the density of the coin is 5.37 g/cm³.
what is an atomic nucleus
Answer:
The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.
Explanation:
In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube length is 25 cm. What is the magnitude of the overall magnification of the microscope?
Answer:
m = 312.5
Explanation:
Given that,
The focal length of the objecive lens, [tex]f_o=1\ cm[/tex]
The focal length of eye piece, [tex]f_e=2\ cm[/tex]
length of the tube, L = 25 cm
We need to find the magnitude of the overall magnification of the microscope. It is given by the formula as follows :
[tex]m=\dfrac{L}{f_o}\times \dfrac{D}{f_e}[/tex]
D = 25 cm
So,
[tex]m=\dfrac{25}{1}\times \dfrac{25}{2}\\\\m=312.5[/tex]
So, the overall magnification of the microscope is 312.5.
A virus has a diameter of 5.2×10−8m. What is the most appropriate prefix for reporting the diameter of the virus?
Answer:
5.2 centimicro meter or 5.2 cμm
Explanation:
Prefixes allow scientists report numbers logically depending on the order of the magnitude of the numbers. For example to represent 4 x 10⁻³m, it is more handy to say : 4 millimetre, where milli- is the prefix for 10⁻³.
Common prefixes are:
deci = 10⁻¹ deca = 10¹
centi = 10⁻² hecto = 10²
milli = 10⁻³ kilo = 10³
micro = 10⁻⁶ mega = 10⁶
nano = 10⁻⁹ giga = 10⁹
pico = 10⁻¹² terra = 10¹²
femto = 10⁻¹⁵ peta = 10¹⁵
Now, let's represent the given diameter: 5.2 x 10⁻⁸ m
Since the 10⁻⁸ prefix is not common (or maybe doesn't even exist), the way around this is to break it into bits such as:
10⁻⁸ = 10⁻² x 10⁻⁶ [10⁻² = centi, 10⁻⁶ = micro ]
10⁻⁸ = centimicro
Therefore, 5.2 x 10⁻⁸ m becomes 5.2 centimicro meter.
The symbol is 5.2 cμm
Where c = centi, μ = micro.
Which question cannot be answered through making measurements?
A. Should wolves be reintroduced into national parks?
B. What is the birthrate for wolves in their natural habitats?
O c. How would reintroducing wolves into a national park affect the
deer population there?
D. Which species in a national park do wolves hunt?
"Should wolves be reintroduced into national parks" is a question that cannot be answered through making measurements, 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. Any additional quantity of that type can be stated as a multiple of the measurement unit.
A length, for instance, is a physical quantity. A defined, predetermined length is represented by the length unit known as the meter.
A qualitative type of question cannot be answered while making a quantitative measurement.
Thus, the question of whether wolves should be reintroduced into national parks cannot be answered by taking measurements, so option C is the appropriate response.
Learn more about the unit of measurement from here
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Which statement is true? A Displacement can never be greater than distance. B Displacement can never be less than distance. C Displacement can never be equal to distance. D Displacement is always equal to distance.
Answer:
__________________
I think the answer is A.
__________________
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Answer:
The speed of q₂ is [tex]4\sqrt{10}\ m/s[/tex]
Explanation:
Given that,
Distance = 0.4 m apart
Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.
We need to calculate the speed of q₂
Using conservation of energy
[tex]E_{i}=E_{f}[/tex]
[tex]\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2[/tex]
[tex]\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})[/tex]
[tex]0.00075(400-v_{f}^2)=0.18 [/tex]
[tex]400-v_{f}^2=\dfrac{0.18}{0.00075}[/tex]
[tex]-v_{f}^2=240-400[/tex]
[tex]v_{f}^2=160[/tex]
[tex]v_{f}=4\sqrt{10}\ m/s[/tex]
Hence, The speed of q₂ is [tex]4\sqrt{10}\ m/s[/tex]
A parallel beam of light containing orange (610 nm) and violet (410 nm) wavelengths goes from fused quartz to water, striking the surface between them at a 60.0°60.0° incident angle. What is the angle between the two colors in water?
Answer:
0.98°
Explanation:
First we find the refracting angle of orange in water
Which is given as
စr= sin^-1. ( 1.456 sin60°/1.33)
= 71.2°
Then that of violet in water
စv=sin^-1. ( 1.468sin60°/1.342)
= 72.3°
So angle between boths colours is the difference
71.2- 72.3= 0.98°
A 0.450-kg ice puck, moving east with a speed of 5.34 m/s , has a head-on collision with a 0.990-kg puck initially at rest. Assume that the collision is perfectly elastic.1. What is the speed of the 0.450-kg puck after the collision?2. What is the direction of the velocity of the 0.450-kg puck after the collision?3. What ise the speed of the 0.990-kg puck after the collision?4. What is the direction of the velocity of the 0.990-kg puck after the collision?
(1) -2.0025m/s.
(2) It moves west.
(3) 3.3375m/s.
(4) It moves east.
Explanation:
Assuming the collision occurred in an isolated system. This means that total momentum of the system of pucks is conserved. Since no external forces are acting on these pucks, the momentum of the pucks before collision is equal to the momentum of the pucks after collision. i.e
(p₁)₀ + (p₂)₀ = (p₁)₁ + (p₂)₁ -------------(i)
Where;
(p₁)₀ = momentum of the 0.45kg puck before collision
(p₂)₀ = momentum of the 0.990kg puck before collision
(p₁)₁ = momentum of the 0.45kg puck after collision
(p₂)₁ = momentum of the 0.990kg puck after collision
But;
(p₁)₀ = m₁ u₁
[m₁ = mass of the 0.45kg, u₁ = speed of the 0.45kg before collision]
(p₂)₀ = m₂u₂
[m₂ = mass of the 0.990kg, u₂ = speed of the 0.990kg before collision]
(p₁)₁ = m₁v₁
[m₁ = mass of the 0.45kg, v₁ = speed of the 0.45kg after collision]
(p₂)₁ = m₂v₂
[m₂ = mass of the 0.990kg, v₂ = speed of the 0.990kg after collision]
Equation (i) then becomes;
m₁ u₁ + m₂u₂ = m₁v₁ + m₂v₂ ----------------(ii)
From the question:
m₁ = 0.450kg
u₁ = +5.34m/s [Taking east direction as positive]
m₂ = 0.990kg
u₂ = 0m/s [since the second puck is initially at rest]
Substitute these values into equation (ii)
(0.450 x 5.34) + (0.990 x 0) = 0.45 v₁ + 0.990 v₂
2.403 + 0 = 0.45 v₁ + 0.990 v₂
2.403 = 0.45 v₁ + 0.990 v₂ ------------------(iii)
Also, since the collision is perfectly elastic, it means that the kinetic energy is conserved. i.e the total kinetic energy before collision is equal to the total kinetic energy after collision.
=> [tex]\frac{1}{2}[/tex]m₁ u²₁ + [tex]\frac{1}{2}[/tex]m₂u²₂ = [tex]\frac{1}{2}[/tex]m₁v²₁ + [tex]\frac{1}{2}[/tex]m₂v²₂
Substitute the necessary values into the above equation:
[[tex]\frac{1}{2}[/tex] x 0.45 x 5.34²] + [0] = [[tex]\frac{1}{2}[/tex] x 0.45 x v²₁] + [[tex]\frac{1}{2}[/tex] x 0.990 x v²₂]
[6.41601] = [0.225 x v²₁] + [0.495 x v²₂] ------------------(iv)
Now let's solve equations (iii) and (iv) simultaneously
2.403 = 0.45 v₁ + 0.990 v₂
6.41601 = 0.225 x v²₁ + 0.495 x v²₂
let
v₁ = x
v₂ = y
2.403 = 0.45 x + 0.990 y ------------(5)
6.41601 = 0.225 x² + 0.495 y² -------------(6)
From equation (5), make x subject of the formula
2.403 = 0.45x + 0.990y
0.45x = 2.403 - 0.990y [divide through by 0.45]
x = 5.34 - 2.2y ----------------(m)
Substitute x into equation (6)
6.41601 = 0.225 (5.34 - 2.2y)² + 0.495 y² [expand bracket]
6.41601 = 0.225 [28.5156 - 23.496y + 4.84y²] + 0.495 y² [remove bracket]
6.41601 = 6.41601 - 5.2866y + 1.089y² + 0.495 y²
1.584y² - 5.2866y = 0
y(1.584y - 5.2866) = 0
y = 0 or 1.584y - 5.2866 = 0
y = 0 or 1.584y = 5.2866
y = 0 or y = 3.3375
Since y = v₂ cannot be zero because the puck will definitely move after collision, the second value of y = 3.3375 is considered.
Substitute this value into equation (m)
x = 5.34 - 2.2y
x = 5.34 - 2.2(3.3375)
x = 5.34 - 7.3425
x = -2.0025
Therefore,
x = v₁ = -2.0025m/s
y = v₂ = 3.3375m/s
(1) From the analyses above, the speed of the 0.450kg puck after collision is -2.0025m/s.
(2) Since the speed is negative, it shows that the 0.45kg puck moves opposite the direction at which it was moving before collision. It moves west.
(3) The speed of the 0.990kg puck after collision is 3.3375m/s.
(4) Since the speed is positive, it shows that the 0.990kg puck moves east. Remember that east has been taking as the positive direction.
Sapphire has an index of refraction of 1.77. The critical angle for sapphire in air with an index of refraction of 1.0 is _____ degrees.
Answer:
The critical angle is 34 degree.
Explanation:
Given the index refraction = 1.77
We have to find the critical angle of sapphire when the air has the index of refraction of 1.0.
Use the below formula to find the angle.
sin(θ)c = 1 / n
sin(θ) c = 1 / 1.77
(θ) c = Sin^-1 (0.565)
= 34.4 degree
= 34 degrees.
Thus, the critical angle is 34 degree.
Your friend asks you for a glass of water and you bring her 5 millilitersof water. Is this more or less than what she was probably expecting?Explain your reasoning
The period of a simple pendulum is 8 s. If the same pendulum experiment is repeated on a plant of g = 1/4 gE, the new period is
Answer:
The new period is 16 s
Explanation:
The period of a simple pendulum is given by
[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex]
Where [tex]T[/tex] is the period
[tex]L[/tex] is the length of the string
[tex]g[/tex] is the acceleration due to gravity
2π is constant
For the first experiment,
[tex]T =8 s[/tex]
[tex]g = gE[/tex]
Then,
[tex]8 = 2\pi \sqrt{\frac{L}{gE} } \\[/tex]
[tex]8\sqrt{gE} = 2\pi \sqrt{L}[/tex] ....... (1)
For the second experiment
[tex]T = T_{2}[/tex]
[tex]g = \frac{1}{4} gE[/tex]
Hence,
[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex] becomes
[tex]T_{2} =2\pi \sqrt{\frac{L}{\frac{1}{4}gE } }[/tex]
Then,
[tex]T_{2} \sqrt{\frac{1}{4}gE } = 2\pi\sqrt{L}[/tex] ....... (2)
Since 2π is constant and
The same pendulum is used for the second experiment, then [tex]L[/tex] is also constant.
∴ [tex]2\pi \sqrt{L}[/tex] is constant for both experiment.
Hence, we can equate equations (1) and (2) such that
[tex]8\sqrt{gE} = T_{2} \sqrt{\frac{1}{4}gE }[/tex]
Then,
[tex]\frac{8\sqrt{gE} }{\sqrt{\frac{1}{4}gE } } = T_{2} \\\frac{8\sqrt{gE} }{\frac{1}{2}\sqrt{gE} } } } = T_{2} \\ 16 = T_{2} \\T_{2} = 16 s[/tex]
Hence, the new period is 16 s
A simple pendulum swings for 8 seconds. The new period is determined by repeating the pendulum experiment on a plant with g = 1/4 gE.
Move to a position where the gravitational acceleration is greater.
- Reduce the pendulum's length.
Explanation: The frequency of a simple pendulum is given by where L is the length of the pendulum g is the gravitational acceleration
From the formula, we notice that - As the value of g increases, the frequency of the pendulum increases as well
- As the value of L increases, the frequency of the pendulum decreases
So, in order to increase the frequency of the pendulum, we can: - Move to a location where the acceleration due to gravity, g, is larger
To learm more about pendulum's visit: https://brainly.com/question/12495919
A ball starts from rest and undergoes uniform acceleration of 2.50m/s^2. What is the velocity of the ball 4s later?
Explanation:
Given:
v₀ = 0 m/s
a = 2.50 m/s²
t = 4 s
Find: v
v = at + v₀
v = (2.50 m/s²) (4 s) + 0 m/s
v = 10 m/s
The speed of the ball after 4s is 10 m/s.
There's not enough information to find its velocity. (We would need to know what direction it's moving.)
26500 in scientific notation
Answer:
in scientific notation 26500 is 2.65×10⁴
Answer:
2.56*10^4
26500
(We have to move the decimal 4 places)
2.6500*10^4
So the answer is 2.6500*10^4 or 2.56*10^4
Hope this helps! :)
Danielle exerts a 14.0 N force to compress a spring by a distance of 8.00 cm. What is the spring constant of this spring
Answer:
175 N/m
Explanation:
Given:
Force = F= 14.0 N
Distance = x = 8.00 cm = 0.08 m
To find:
spring constant
Solution:
spring constant is calculated by using Hooke's law:
k = F/x
Putting the values in above formula:
k = 14.0 / 0.08
k = 175 N/m
What is its inside radius, in centimeters, if it holds 360 g of coffee when filled to a depth of 8.5 cm
Answer:
1.14*10^-3m
Explanation:
We know that
Volume= mass/ density
But
Area x depth= mass/density
πr²*depth= mass/density
r²=m/πdp
So
r= √0.35/3.142*0.085*1000
r= 1.14*10^-3m
Two tiny, spherical water drops, with identical charges of -6.19 × 10-16 C, have a center-to-center separation of 1.22 cm. (a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?
Answer:
a
[tex]F = 2.32*10^{-17} \ N[/tex]
b
[tex]n =3869 \ electrons[/tex]
Explanation:
From the question we are told that
The charge on each water drop is [tex]q_1=q_2=q = - 6.19*10^{-16} \ C[/tex]
The distance of separation is [tex]d = 1.22\ cm = 0.0122 \ m[/tex]
Generally the electrostatic force between the water drops is mathematically represented as
[tex]F = \frac{k * q_1 * q_2 }{ d^ 2}[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^9 \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
So
[tex]F = \frac{9*10^9 * -6.19 *10^{-16} * (-6.19*10^{-16}) }{ 0.0122^ 2}[/tex]
[tex]F = 2.32*10^{-17} \ N[/tex]
Generally the quantity of charge is mathematically represented as
[tex]q = n * e[/tex]
Here n is the number of electron present
and e is the charge on one electron with value [tex]e = 1.60*10^{-19} \ C[/tex]
So
[tex]n = \frac{6.19 *10^{-16}}{1.60*10^{-19}}[/tex]
[tex]n =3869 \ electrons[/tex]
What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)
Answer:
E₁ / E₂ = M / m
Explanation:
Let the electric field be E₁ and E₂ for ions and electrons respectively .
Force on ions = E₁ e where e is charge on ions .
Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0
v² = u² - 2as
0 = u² - 2 x E₁ e d / M
u² = 2 x E₁ e d / M
Similarly for electrons
u² = 2 x E₂ e d / m
Hence
2 x E₁ e d / M = 2 x E₂ e d / m
E₁ / E₂ = M / m