In Mathematics and Physics, scalar is a quantity or a single number that shows the measurement of a medium in magnitude only (It does not include direction as vectors do); examples of scalars are voltage, mass, temperature, speed, volume, power, energy, and time.
Two examples of scalars I have used recently are Degrees Celsius to measure the temperature of my living room and Cubic Feet to measure the volume of my mug.
A scalar is a quantity represented only by a magnitude and devoid of direction. Examples of scalars include temperature and energy, which are significant in physics and everyday measurements. Unlike vectors, scalars do not change with coordinate system rotations.
Explanation:Definition of Scalar
A scalar is a physical quantity that is represented only by a magnitude (or numerical value) but does not involve any direction. In contrast to vectors, which have both magnitude and direction, scalars are not affected by coordinate system rotations or translations. Scalars come in handy when representing physical quantities where direction is non-applicable.
Two examples of scalars that I have used recently are:
Temperature: I noted the room temperature this morning, which was 20°C. The temperature is a scalar because it has no direction.Energy: I consumed a snack bar which had an energy content of 250 kilocalories, or 250 Calories. Energy is a scalar quantity as it does not possess a specific direction.When dealing with physics problems or everyday measurements, it's crucial to understand whether you're working with a scalar or a vector to correctly interpret the quantity in question.
The uncertainty in a proton's position is 0.015 nm. a) What is the minimum uncertainty Δp in its momentum? b) What is the kinetic energy of a proton whose momentum is equal to this uncertainty (Δp=p)? (Unit: meV)
Answer:
[tex]\Delta P_{x}=3.51 *10^{-24}[/tex] kg m/s
[tex]E = 3.68 * 10^{-21}[/tex] J
Explanation:
given data
uncertainty in proton position [tex]\Delta x[/tex] = 0.015 nm
according to Heisenberg's principle of uncertainty
[tex]\Delta x \Delta P_{x}= \frac{\frac{h}{2\pi }}{2}[/tex]
Where h is plank constant = [tex]6.6260 * 10^{-34}[/tex] j-s
[tex]\Delta P_{x}= \frac{\frac{h}{2\pi }}{2\Delta x}[/tex]
[tex]\Delta P_{x}= \frac{\frac{6.6260 * 10^{-34}}{2\pi }}{2*0.015*10^{-9}}[/tex]
[tex]\Delta P_{x}=3.51 *10^{-24}[/tex] kg m/s
b) kinetic energy of proton whose momentum
[tex]P =\Delta p[/tex]
[tex]E =\frac{\Delta p^{2}}{2m}[/tex]
where m is mass is proton
[tex]E =\frac{(3.51*10^{-24})^{2}}{2*1.67*10^{-27}}[/tex]
[tex]E = 3.68 * 10^{-21}[/tex] J
A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun is 300.0 m/s, how far vertically will the bullet have dropped by the time it hits the target? Ignore the effects of air resistance.
Answer:
The bullet will have been dropped vertically h= 0.54 meters by the time it hits the target.
Explanation:
d= 100m
V= 300 m/s
g= 9.8 m/s²
d= V*t
t= d/V
t= 0.33 s
h= g*t²/2
h=0.54 m
A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine the object's acceleration & charge magnitude. Assume the acceleration is due to the E-field (i.e., ignore all other forces).
Explanation:
It is given that,
Mass of the object, m = 0.8 g = 0.0008 kg
Electric field, E = 534 N/C
Distance, s = 12 m
Time, t = 1.2 s
We need to find the acceleration of the object. It can be solved as :
m a = q E.......(1)
m = mass of electron
a = acceleration
q = charge on electron
"a" can be calculated using second equation of motion as :
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
[tex]s=0+\dfrac{1}{2}at^2[/tex]
[tex]a=\dfrac{2s}{t^2}[/tex]
[tex]a=\dfrac{2\times 12\ m}{(1.2\ s)^2}[/tex]
a = 16.67 m/s²
Now put the value of a in equation (1) as :
[tex]q=\dfrac{ma}{E}[/tex]
[tex]q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}[/tex]
q = 0.0000249 C
or
[tex]q=2.49\times 10^{-5}\ C[/tex]
Hence, this is the required solution.
An FM radio consists of a series RLC circuit with a 6.57 pF (peco – 10-12) capacitor. If a person is dialed into a station broadcasting at 103.9 MHz, what is the inductance needed?
Answer:
3.6 x 10⁻⁷ H
Explanation:
C = capacitance of the capacitor = 6.57 x 10⁻¹² F
L = inductance of the inductor = ?
f = frequency of broadcasting system = 103.9 MHz = 103.9 x 10⁶ Hz
frequency is given as
[tex]f = \frac{1}{2\pi \sqrt{LC}}[/tex]
[tex]103.6\times 10^{6} = \frac{1}{2(3.14)\sqrt{(6.57\times 10^{-12})L}}[/tex]
Inserting the values
L = 3.6 x 10⁻⁷ H
Electric power is to be generated by installing a hydraulic turbine generator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 2400 kg/s steadily. Determine the power generation potential.
Answer:
2.82 × 10^6 watt
Explanation:
H = 120 m,
Mass per second = 2400 kg/s
Power = work / time
Power = m g H / t
Power = 2400 × 9.8 × 120 / 1
Power = 2.82 × 10^6 Watt
The brakes of a 125 kg sled are applied while it is moving at 8.1 m/s, which exerts a force of 261 N to slow the sled down. How far does the sled travel while coming to a stop in meters? Round to one decimal place.
Answer:
15.7 m
Explanation:
m = mass of the sled = 125 kg
v₀ = initial speed of the sled = 8.1 m/s
v = final speed of sled = 0 m/s
F = force applied by the brakes in opposite direction of motion = 261
d = stopping distance for the sled
Using work-change in kinetic energy theorem
- F d = (0.5) m (v² - v₀²)
- (261) d = (0.5) (125) (0² - 8.1²)
d = 15.7 m
An earthquake 45 km from a city produces P and S waves that travel outward at 5000 m/s and 3000 m/s, respectively. Once city residents feel the shaking of the P wave, how much time do they have before the S wave arrives in seconds?
Answer:
The S wave arrives 6 sec after the P wave.
Explanation:
Given that,
Distance of P = 45 km
Speed of p = 5000 m/s
Speed of S = 3000 m/s
We need to calculate the time by the P wave
Using formula of time
[tex]t = \dfrac{D}{v}[/tex]
Where, D = distance
v = speed
t = time
Put the value in to the formula
[tex]t_{p} =\dfrac{45\times1000}{5000}[/tex]
[tex]t_{p} = 9\ sec[/tex]
Now, time for s wave
[tex]t_{s}=\dfrac{45000}{3000}[/tex]
[tex]t =15\ sec[/tex]
The required time is
[tex]\Delta t=t_{s}-t_{p}[/tex]
[tex]\Delta t=15-9[/tex]
[tex]\Delta t =6\ sec[/tex]
Hence, The S wave arrives 6 sec after the P wave.
A block of mass 0.245 kg is placed on top of a light, vertical spring of force constant 4 975 N/m and pushed downward so that the spring is compressed by 0.103 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.)
Answer:
10.99 m
Explanation:
m = mass of the block = 0.245 kg
k = spring constant of the vertical spring = 4975 N/m
x = compression of the spring = 0.103 m
h = height to which the block rise
Using conservation of energy
Potential energy gained by the block = Spring potential energy
mgh = (0.5) k x²
(0.245) (9.8) h = (0.5) (4975) (0.103)²
h = 10.99 m
A water wave of wavelength 859 cm travels a distance of 64 m in a time of 14 s. What is the period of the wave, in units of seconds?
Answer:
1.87 s
Explanation:
d = distance traveled by the water wave = 64 m
t = time taken to travel the distance = 14 s
[tex]v[/tex] = speed of water wave
Speed of water wave is given as
[tex]v=\frac{d}{t}[/tex]
[tex]v=\frac{64}{14}[/tex]
[tex]v[/tex] = 4.6 m/s
[tex]\lambda[/tex] = wavelength of the wave = 859 cm = 8.59 m
T = period of the wave
period of the wave is given as
[tex]T = \frac{\lambda }{v}[/tex]
[tex]T = \frac{8.59 }{4.6}[/tex]
T = 1.87 s
There are three identical train locomotives on separate tracks. Each applies equal force to begin moving. The first train consists of a single empty freight car, the second train consists of 50 empty freight cars, and the third train consists of 50 equally loaded freight cars. Rank the accelerations of the trains. All of the freight cars, when empty, are identical.
Answer:
From lowest to highest acceleration:
3rd train
2nd train
1st train
Explanation:
The acceleration of an object can be found by using Newton's second law:
[tex]a=\frac{F}{m}[/tex]
where
a is the acceleration
F is the net force on the object
m is the mass of the object
We notice that for equal values of the forces F, the acceleration a is inversely proportional to the mass, m. Therefore, greater mass means lower acceleration, and viceversa.
So, the train with lowest acceleration is the one with largest mass, i.e. the 3rd train consisting of 50 equally loaded freight cars. Then, the 2nd train has larger acceleration, since it consists of 50 empty freight cars (so its mass is smaller). Finally, the 1st train (a single empty car) is the one with largest acceleration, since it is the train with smallest mass.
The acceleration of the three trains can be ranked as follows: the single empty freight car train, the 50 empty freight cars train, and the 50 equally loaded freight cars train.
Explanation:The accelerations of the three trains can be ranked as follows:
The single empty freight car trainThe 50 empty freight cars trainThe 50 equally loaded freight cars trainWhen an equal force is applied to each train, the train with fewer cars will experience a greater acceleration. This is because the force is distributed over fewer cars, resulting in a larger net force acting on each car.
For example, consider two cars:
Single empty freight car train: The force is applied to only one car, resulting in a larger net force per car50 empty freight cars train: The force is distributed over 50 cars, resulting in a smaller net force per carTherefore, the single empty freight car train will have the highest acceleration, followed by the 50 empty freight cars train, and finally the 50 equally loaded freight cars train.
A force of 500 N is applied to a hydraulic jack piston that is 0.2 m in diameter. If the piston that supports the load has a diameter of 2.0 m, approximately how much mass can be lifted by the jack? Ignore any difference in height between the pistons.
Answer:
50,000 N
Explanation:
d = 0.2 m, r = 0.1 m, f = 500 N,
D = 2 m, R = 1 m
Let F be the weight lifted by the jack.
By use of Pascal's law
f/a = F / A
F = f x A / a
F = 500 x 3.14 x 1 x 1 / (3.14 x 0.1 x 0.1)
F = 50,000 N
A current runs through a wire in the +y direction along the y-axis. What is the direction of the magnetic field at a point (x,y,z) = (0,0,1.045 m)
Answer:
Magnetic field at given point is along + x direction
Explanation:
As we know that position of the wire is along Y axis
current is flowing along +y direction
so here we will have
[tex]\vec B = \frac{\mu_0 i (\vec {dl} \times \hat r)}{4\pi r^2}[/tex]
now the direction of length vector is along +Y direction
also the position vector is given as
[tex]\hat r = \hat k[/tex]
now for the direction of magnetic field we can say
[tex]\vec B = (\hat j \times \hat k)[/tex]
[tex]\vec B = \hat i[/tex]
so magnetic field is along +X direction
Using the right-hand rule for a wire carrying current in the +y direction along the y-axis, the magnetic field at the point (0,0,1.045 m) on the x-axis would be directed in the +x-direction.
When a current runs through a wire in the +y direction along the y-axis, we can use the right-hand rule to determine the direction of the magnetic field at a point near the wire. For a point on the x-axis such as (0,0,1.045 m), you would point your thumb in the direction of the current (upward along the y-axis) and curl your fingers.
Your fingers will curl in the direction of the magnetic field lines, which at the given point on the x-axis would circle around the wire. This means the magnetic field at that point would be directed in the +x-direction.
Since force is dp/dt, the force due to radiation pressure reflected off of a solar sail can be calculated as 2 times the radiative momentum striking the sail per second. In the vicinity of Earth's orbit around the Sun, the energy intensity of sunlight is about 1300 W/m2. What is the approximate magnitude of the pressure on the sail? (For comparison, atmospheric pressure is about 105 N/m2.)
The approximate magnitude of the pressure on the sail due to radiation pressure is [tex]\( 2600 \, \text{N/m}^2 \).[/tex]
To calculate the magnitude of the pressure on the sail due to radiation pressure, we can use the formula for pressure:
[tex]\[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} \][/tex]
The force due to radiation pressure is given by:
[tex]\[ \text{Force} = 2 \times \text{radiative momentum} \][/tex]
The radiative momentum can be calculated using the energy intensity of sunlight:
[tex]\[ \text{Radiative momentum} = \text{Energy intensity} \times \text{Time} \][/tex]
Given:
- Energy intensity of sunlight [tex](\( I \))[/tex] is approximately [tex]\( 1300 \, \text{W/m}^2 \),[/tex]
- Time [tex](\( t \)) is \( 1 \, \text{s} \)[/tex] (per second),
- Atmospheric pressure is about [tex]\( 105 \, \text{N/m}^2 \).[/tex]
Let's calculate the radiative momentum:
[tex]\[ \text{Radiative momentum} = 1300 \times 1 \, \text{N/s} = 1300 \, \text{N/s} \][/tex]
Now, let's calculate the force:
[tex]\[ \text{Force} = 2 \times 1300 \, \text{N/s} = 2600 \, \text{N/s} \][/tex]
Finally, let's calculate the pressure on the sail:
[tex]\[ \text{Pressure} = \frac{2600 \, \text{N/s}}{1 \, \text{m}^2} = 2600 \, \text{N/m}^2 \][/tex]
So, the approximate magnitude of the pressure on the sail due to radiation pressure is [tex]\( 2600 \, \text{N/m}^2 \).[/tex]
Wings on race cars push them into the track. The increased normal force makes large friction forces possible. At one Formula One racetrack, cars turn around a half-circle with diameter 190 m at 68 m/s. For a 610 kg vehicle, the approximate minimum static friction force to complete this turn is________.
Answer:
The minium static friction force to complete this turn is F= 29,690.94 N.
Explanation:
R= 95m
m= 610 kg
V= 68 m/s
an= V²/R
an= 48.67 m/s²
F= m * an
F= 29690.94 N
The approximate minimum static friction force to complete the turn is approximately 18806 N.
Explanation:When a race car turns a corner, the wings on the car push it into the track, increasing the normal force. This increased normal force allows for larger friction forces, which help the car maintain traction and complete the turn.
For the given Formula One racetrack with a half-circle turn of diameter 190 m and a speed of 68 m/s, we can find the approximate minimum static friction force using the centripetal force equation. The centripetal force in this situation is equal to the product of the mass (610 kg), the speed squared (68 m/s) squared, and the reciprocal of the radius (half of the diameter = 95 m).
The approximate minimum static friction force to complete this turn is approximately 18806 N.
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A 89.1 g piece of copper (specific heat 0.38 J/g･°C) is heated and then placed into 400.0 g of water initially at 20.7°C. The water increases in temperature to 22.2°C. What is the initial temperature of the copper? (The specific heat of water is 4.18 J/g･°C).
Answer:
The initial temperature of copper is 96.3 °C
Explanation:
Specific heat capacity is the energy needed to raise the temperature of one gram of material by one degree celsius. The energy absorbed or released by a material during temperature change can be calculated by the below formula:
[tex]Q=mc(T_{2}-T_{1})[/tex] , where:
Q = energy (J)
m = mass (g)
c = specific heat capacity (J/g·°C)
[tex]T_{1}[/tex] = initial temperature (°C)
[tex]T_{2}[/tex] = final temperature (°C)
In an ideal situation, it can be assumed that all the energy lost by the piece of copper is gained by the water, resulting in its temperature rise and that there is no change of mass/state for either material. Thus, the equation can be written as below:
[tex]+Q_{c}=-Q_{w}[/tex]
[tex]+m_{c}c_{c}(T_{c2}-T_{c1})=-m_{w}c_{w}(T_{w2}-T_{w1})[/tex]
It can also be assumed that the final temperature of both the copper and water are the same. Thus substituting below values in above equation will give:
[tex]m_{w}[/tex] = 400 g
[tex]c_{w}[/tex] = 4.18 J/g·°C
[tex]T_{w1}[/tex] = 20.7 °C
[tex]T_{w2}[/tex] = 22.2 °C
[tex]m_{c}[/tex] = 89.1 g
[tex]c_{c}[/tex] = 0.38 J/g·°C
[tex]T_{c1}[/tex] = ? °C
[tex]T_{c2}[/tex] = 22.2 °C
[tex]+89.1*0.38*(22.2-T_{c1})=-400*4.18(22.2-20.7)[/tex]
Solving for [tex]T_{c1}[/tex] gives:
[tex]T_{c1}[/tex] = 96.3 °C
Here, we are required to determine the initial temperature of the copper.
The initial temperature of the copper is;
T(c1) = 96.27 °C
The energy required to raise the temperature of one gram of a material by one degree celsius is termed the Specific Heat Capacity of that material.
Mathematically, we have;
Q=mc{T(2) -T(1)}
Q=mc{T(2) -T(1)} where:
Q=mc{T(2) -T(1)} where:Q = energy (J)Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:
Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)
Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)
m(c) × c(c) × (T2(c) - T1(c)) = -{m(w) × c(w) × (T2(w) - T1(w))}
However, the final temperature of the copper piece and water can be assumed to be equal
i.e. T2(c) = T1(w)
m(w) = 400gc(w) = 4.18 J/g·°CT (w1) = 20.7 °CT(w2) = 22.2 °Cm(c) = 89.1 gc(c) = 0.38 J/g·°CT(c1) = ? °CT(c2) = 22.2 °C+89.1*0.38*(22.2-T(c1)) = -400*4.18(22.2-20.7)
(22.2−T(c1) ) = −2508/33.858
(22.2−T(c1) ) = -74.07
T(c1) = 74.07 + 22.2
T(c1) = 96.27 °C.
Therefore, the initial temperature of the copper is; T(c1) = 96.27 °C
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When 23Na is bombarded with protons, the products are 20Ne and A. a neutron B. an alpha particle C. a deuteron D. a gamma ray particle E. two beta particles
Answer:
The correct option is B. an alpha particle.
Explanation:
When ²³Na is bombarded with protons [tex]_{1}^{1}\textrm{H}[/tex], ²⁰Ne and one alpha particle [tex]_{2}^{4}\textrm{He}[/tex] is released.
The reaction is as follows:
[tex]_{11}^{23}\textrm{Na} + _{1}^{1}\textrm{H} \rightarrow _{10}^{20}\textrm{Ne} + _{2}^{4}\textrm{He}[/tex]
Therefore, an alpha particle and ²⁰Ne are released when ²³Na is bombarded with protons.
A point source of light is 80.0 cm below the surface of
a body of water. Find the diameter of the circle at the surface
through which light emerges from the water.
Answer:
The diameter of the circle at the surface is 1.814 m.
Explanation:
Given that,
Depth = 80 cm
Let the critical angle be c.
From Snell's law
[tex]sin c=\dfrac{1}{n}[/tex]
[tex]c=sin^{-1}\dfrac{1}{n}[/tex]
The value of n for water,
[tex]n = \dfrac{4}{3}[/tex]
Put the value of n in the equation (I)
[tex]c =sin^{-1}\dfrac{3}{4}[/tex]
[tex]c=48.6^{\circ}[/tex]
We know that,
[tex]tan c=\dfrac{r}{h}[/tex]
We calculate the radius of the circle
[tex]r =h tan c[/tex]
[tex]r=80\times10^{-2}\times\tan48.6^{\circ}[/tex]
[tex]r =0.907\ m[/tex]
The diameter of the circle at the surface
[tex]d =2\times r[/tex]
[tex]d =2\times 0.907[/tex]
[tex]d =1.814\ m[/tex]
Hence, The diameter of the circle at the surface is 1.814 m.
Two long, parallel wires are separated by a distance of 2.2 cm. The force per unit length that each wire exerts on the other is 3.6 × 10^-5 N/m, and the wires repel each other. The current in one wire is 0.52 A. What is the magnitude of the current in the second wire? (Give your answer in decimal using "A" (Ampere) as unit)
Answer:
[tex]i_2 = 7.6 A[/tex]
Explanation:
As we know that the force per unit length of two parallel current carrying wires is given as
[tex]F = \frac{\mu_o i_1 i_2}{2\pi d}[/tex]
here we know that
[tex]F = 3.6 \times 10^{-5} N/m[/tex]
[tex]i_1 = 0.52 A[/tex]
d = 2.2 cm
now from above equation we have
[tex]3.6 \times 10^{-5} = \frac{4\pi \times 10^{-7} (0.52)(i_2)}{2\pi (0.022)}[/tex]
[tex]3.6 \times 10^{-5} = 4.73 \times 10^{-6} i_2[/tex]
[tex]i_2 = 7.6 A[/tex]
Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. What is the translational kinetic energy per mole of an ideal gas at (c) 27.8°C and (d) 143°C?
Answer:
a) [tex]k_{avg}=6.22\times 10^{-21}[/tex]
b) [tex]k_{avg}=8.61\times 10^{-21}[/tex]
c) [tex]k_{mol}=3.74\times 10^{3}J/mol[/tex]
d) [tex]k_{mol}=5.1\times 10^{3}J/mol[/tex]
Explanation:
Average translation kinetic energy ([tex]k_{avg} [/tex]) is given as
[tex]k_{avg}=\frac{3}{2}\times kT[/tex] ....................(1)
where,
k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K
T = Temperature in kelvin
a) at T = 27.8° C
or
T = 27.8 + 273 = 300.8 K
substituting the value of temperature in the equation (1)
we have
[tex]k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8[/tex]
[tex]k_{avg}=6.22\times 10^{-21}J[/tex]
b) at T = 143° C
or
T = 143 + 273 = 416 K
substituting the value of temperature in the equation (1)
we have
[tex]k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416[/tex]
[tex]k_{avg}=8.61\times 10^{-21}J[/tex]
c ) The translational kinetic energy per mole of an ideal gas is given as:
[tex]k_{mol}=A_{v}\times k_{avg}[/tex]
here [tex]A_{v}[/tex] = Avagadro's number; ( 6.02×10²³ )
now at T = 27.8° C
[tex]k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}[/tex]
[tex]k_{mol}=3.74\times 10^{3}J/mol[/tex]
d) now at T = 143° C
[tex]k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}[/tex]
[tex]k_{mol}=5.1\times 10^{3}J/mol[/tex]
Water from Old Faithful Geyser shoots upward to a height of about 50 m once every 90 minutes. What must be the pressure in the hot springs below the ground in order for the water to go that high?
Solution:
Given:
height of geyser, h = 50 m
speed of water at ground can be given by third eqn of motion with v = 0 m/s:
[tex]u^{2} = v^{2} + 2gh[/tex]
putting v = 0 m/s in the above eqn, we get:
u = [tex]\sqrt{2gh}[/tex] (1)
Now, pressure is given by:
[tex]\Delta p = \frac{1}{2}u^{2}\rho \\[/tex] (2)
where,
[tex]\Delta p[/tex] = density of water = 1000 kg/[tex]m^{3}[/tex]
g = 9.8 m/[tex]s^{2}[/tex]
Using eqn (1) and (2):
[tex]\Delta p = \frac{1}{2}(\sqrt{2gh})^{2}\rho[/tex]
⇒ [tex]\Delta p = \frac{1}{2}\rho gh[/tex]
[tex]\Delta p = \frac{1}{2}\times2\times 1000\times 9.8\times 50[/tex] = [tex]\Delta p[/tex] = 490000 Pa
[tex]p_{atm}[/tex] = 101325 Pa
For absolute pressure, p:
p = [tex]\Delta p[/tex] + [tex]p_{atm}[/tex]
p = 490000 - 101325
p = 388675 Pa
Therefore, pressure in hot springs for the water to attain the height of 50m is 388675 Pa
An object of mass 8.0 kg is attached to an ideal massless spring and allowed to hang in the Earth's gravitational field. The spring stretches 3.6 cm before it reaches its equilibrium position. If this system is allowed to oscillate, what will be its frequency?
Answer:
2.63 Hz
Explanation:
m = mass of the object = 8.0 kg
x = stretch in the spring = 3.6 cm = 0.036 m
k = spring constant of the spring
using equilibrium of force
Spring force = weight of object
k x = m g
k (0.036) = (8) (9.8)
k = 2177.78 N/m
frequency of oscillation is given as
[tex]f = \frac{1}{2\pi }\sqrt{\frac{k}{m}}[/tex]
[tex]f = \frac{1}{2\pi }\sqrt{\frac{2177.78}{8}}[/tex]
[tex]f [/tex] = 2.63 Hz
The frequency of the spring is about 2.6 Hz
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Further explanationSimple Harmonic Motion is a motion where the magnitude of acceleration is directly proportional to the magnitude of the displacement but in the opposite direction.
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The pulled and then released spring is one of the examples of Simple Harmonic Motion. We can use the following formula to find the period of this spring.
[tex]T = 2 \pi\sqrt{\frac{m}{k}}[/tex]
T = Periode of Spring ( second )
m = Load Mass ( kg )
k = Spring Constant ( N / m )
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The pendulum which moves back and forth is also an example of Simple Harmonic Motion. We can use the following formula to find the period of this pendulum.
[tex]T = 2 \pi\sqrt{\frac{L}{g}}[/tex]
T = Periode of Pendulum ( second )
L = Length of Pendulum ( kg )
g = Gravitational Acceleration ( m/s² )
Let us now tackle the problem !
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Given:
mass of the object = m = 8.0 kg
extension of the spring = x = 3.6 cm = 3.6 × 10⁻² m
Unknown:
frequency of the spring = f = ?
Solution:
Firstly, we will calculate the spring constant as follows:
[tex]F = kx[/tex]
[tex]mg = kx[/tex]
[tex]k = mg \div x[/tex]
[tex]k = \frac{mg}{x}[/tex]
[tex]\texttt{ }[/tex]
Next, we could calculate the frequency of the spring as follows:
[tex]f = \frac{1}{2 \pi}\sqrt{\frac{k}{m}}[/tex]
[tex]f = \frac{1}{2 \pi}\sqrt{\frac{mg / x}{m}}[/tex]
[tex]f = \frac{1}{2 \pi}\sqrt{\frac{g}{x}}[/tex]
[tex]f = \frac{1}{2 \pi}\sqrt{\frac{9.8}{3.6 \times 10^{-2}}}[/tex]
[tex]f = \frac{35\sqrt{2}}{6\pi} \texttt{ Hz}[/tex]
[tex]f \approx 2.6 \texttt{ Hz}[/tex]
[tex]\texttt{ }[/tex]
Learn moreModel for Simple Harmonic Motion : https://brainly.com/question/9221526Force of Simple Harmonic Motion : https://brainly.com/question/3323600Example of Simple Harmonic Motion : https://brainly.com/question/11892568[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Simple Harmonic Motion
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Keywords: Simple , Harmonic , Motion , Pendulum , Spring , Period , Frequency
At a certain temperature, the ????p for the decomposition of H2S is 0.841. H2S(g)↽−−⇀H2(g)+S(g) Initially, only H2S is present at a pressure of 0.259 atm in a closed container. What is the total pressure in the container at equilibrium?
Answer : The total pressure in the container at equilibrium is, 0.4667 atm
Solution : Given,
Initial pressure of [tex]H_2S[/tex] = 0.259 atm
Equilibrium constant, [tex]K_p[/tex] = 0.841
The given equilibrium reaction is,
[tex]H_2S(g)\rightleftharpoons H_2(g)+S(g)[/tex]
Initially 0.259 0 0
At equilibrium (0.259 - x) x x
Let the partial pressure of [tex]H_2[/tex] and [tex]S[/tex] will be, 'x'
The expression of [tex]K_p[/tex] will be,
[tex]K_p=\frac{(p_{H_2})(p_{S})}{p_{H_2S}}[/tex]
Now put all the values of partial pressure, we get
[tex]0.841=\frac{(x)\times (x)}{(0.259-x)}[/tex]
By solving the term x, we get
[tex]x=0.2077atm[/tex]
The partial pressure of [tex]H_2[/tex] and [tex]S[/tex] = x = 0.2077 atm
Total pressure in the container at equilibrium = [tex]0.259-x+x+x=0.259+x=0.259+0.2077=0.4667atm[/tex]
Therefore, the total pressure in the container at equilibrium is, 0.4667 atm
A moving particle encounters an external electric field that decreases its kinetic energy from 9650 eV to 8900 eV as the particle moves from position A to position B The electric potential at A is 56.0 V and that at B is 19.0 V Determine the charge of the particle Include the algebraic sign or with your answer.
Answer:
Charge, [tex]q=3.24\times 10^{-18}\ C[/tex]
Explanation:
A moving particle encounters an external electric field that decreases its kinetic energy from 9650 eV to 8900 eV as the particle moves from position A to position B The electric potential at A is 56.0 V and that at B is 19.0 V. We need to find the charge of the particle.
It can be calculated using conservation of energy as :
[tex]\Delta KE=-q(V_B-V_A)[/tex]
[tex]q=\dfrac{\Delta KE}{(V_B-V_A)}[/tex]
[tex]q=\dfrac{ 9650\ eV-8900\ eV}{(19\ V-56\ V)}[/tex]
q = -20.27 e
[tex]q =-20.27e\times \dfrac{1.6\times 10^{-19}\ C}{e}[/tex]
[tex]q=-3.24\times 10^{-18}\ C[/tex]
Hence, this is the required solution.
An electron and a proton are separated by a distance of 1.0 m. What happens to the force between them if the electron moves 0.5 m away from the proton?
Explanation:
It is given that, an electron and a proton are separated by a distance of 1.0 m i.e d = 1 m . At this position, F is the force between them
[tex]F=k\dfrac{q_1q_2}{1^2}[/tex]...............(1)
We need to find effect on force between them if the electron moves 0.5 m away from the proton. Let the force is F'.
[tex]F'=k\dfrac{q_1q_2}{0.5^2}[/tex]...............(2)
On dividing equation (1) and (2) we get :
[tex]\dfrac{F}{F'}=\dfrac{k\dfrac{q_1q_2}{1^2}}{k\dfrac{q_1q_2}{0.5^2}}[/tex]
[tex]\dfrac{F}{F'}=0.25[/tex]
F' = 4 F
So, the distance between the electron and the proton is 0.5 m, the new force will be 4 times of previous force.
Final answer:
The electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m can be calculated using Coulomb's Law. The force is inversely proportional to the square of the distance, so if the distance is increased, the force decreases. In this case, if the electron moves 0.5 m away from the proton, the new force can be calculated using Coulomb's Law.
Explanation:
The electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m can be calculated using Coulomb's Law. Coulomb's Law states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
So, if the distance between the charges is increased to 1 m, the force between them will decrease. The force is inversely proportional to the square of the distance, which means that if the distance is doubled, the force will decrease by a factor of four.
In this case, if the electron moves 0.5 m away from the proton, the new distance becomes 1.5 m. Using Coulomb's Law, we can calculate the new force:
F = k * (q1 * q2) / r^2
Where F is the force, k is the Coulomb's constant (approximately 9 x 10^9 Nm^2/C^2), q1 and q2 are the charges (in this case, both 1 C), and r is the distance (1.5 m). Plugging in these values, we get:
F = (9 x 10^9 Nm^2/C^2) * (1 C * 1 C) / (1.5 m)^2
F ≈ 4 x 10^9 N
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.61 and standard deviation 0.82. (a) If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? Between 2.61 and 3.00
Answer: The probability that the sample average sediment density is at most 3.00 = 0.9913
The probability that the sample average sediment density is between 2.61 and 3.00 = 0.4913
Explanation:
Given : Mean : [tex]\mu=2.61 [/tex]
Standard deviation : [tex]\sigma =0.82[/tex]
Sample size : [tex]n=25[/tex]
The value of z-score is given by :-
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
a) For x= 3.00
[tex]z=\dfrac{3.00-2.61}{\dfrac{0.82}{\sqrt{25}}}=2.38[/tex]
The p-value : [tex]P(z\leq2.38)=0.9913[/tex]
b) For x= 2.61
[tex]z=\dfrac{2.61-2.61}{\dfrac{0.82}{\sqrt{25}}}=0[/tex]
The p-value : [tex]P(0<z\leq2.38)=P(2.38)-P(0)=0.9913-0.5=0.4913[/tex]
A uniform thin rod is hung vertically from one end and set into small amplitude oscillation. If the rod has a length of 3.6 m, this rod will have the same period as a simple pendulum of length ____ cm. Round your answer to the nearest whole number.
Answer:
The length of the simple pendulum is 2.4 meters.
Explanation:
Time period of simple pendulum is given by :
[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]
L is the length of pendulum
The time period of the rope is given by :
[tex]T=2\pi\sqrt{\dfrac{2L'}{3g}}[/tex]
L' is the length of the rod, L' = 3.6 m
It is given that, the rod have the same period as a simple pendulum and we need to find the length of simple pendulum i.e.
[tex]2\pi\sqrt{\dfrac{L}{g}}=2\pi\sqrt{\dfrac{2L'}{3g}}[/tex]
On solving the above equation as :
[tex]\dfrac{L}{g}=\dfrac{2L'}{3g}[/tex]
L = 2.4 m
So, the length of the thin rod that is hung vertically from one end and set into small amplitude oscillation 2.4 meters. Hence, this is the required solution.
A 4m^3 container is being filled at a rate of 0.051m^3/s. What is the time required to completely fill this container?
Answer:
Time required to completely fill this container = 78.43 seconds
Explanation:
Volume of container = 4 m³
Rate of filling of container = 0.051 m³/s
We have the equation
[tex]\texttt{Time required}=\frac{\texttt{Volume of container}}{\texttt{Rate of filling of container}}[/tex]
Substituting
[tex]\texttt{Time required}=\frac{4}{0.051}=\frac{4000}{51}=78.43 s[/tex]
Time required to completely fill this container = 78.43 seconds
While entering a freeway, a car accelerates from rest at a rate of 2.1 m/s^2 for 14.5 s. How far does the car travel in those 14.5 s, in meters?
What is the car's final velocity, in meters per second?
Answer:
220.8 m
30.5 m/s
Explanation:
a = acceleration of the car = 2.1 m/s²
t = time of travel = 14.5 s
x = displacement of the car
v₀ = initial velocity of the car = 0 m/s
Displacement of the car is given as
x = v₀ t + (0.5) a t²
Inserting the values
x = (0) (14.5) + (0.5) (2.1) (14.5)²
x = 220.8 m
v = final velocity of the car
Final velocity of the car is given as
v = v₀ + at
v = 0 + (2.1) (14.5)
v = 30.5 m/s
the density of zinc is 7.14 g/cm^3 and has an atomic radius of 1.332A. Calculate the height of its unit cell. the atomic weight is 65.37 grams/mole.
Answer : The height of the unit cell is, [tex]4.947\AA[/tex]
Explanation : Given,
Density of zinc = [tex]7.14g/cm^3[/tex]
Atomic radius = [tex]1.332\AA[/tex]
Atomic weight of zinc = 65.37 g/mole
As we know that, zinc has haxagonal close packed crystal structure. The number of atoms in unit cell of HCP is, 6.
Formula used for density :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex] .............(1)
where,
[tex]\rho[/tex] = density of zinc
Z = number of atom in unit cell = 6 atoms/unit cell (for HCP)
M = atomic mass
[tex](N_{A})[/tex] = Avogadro's number = [tex]6.022\times 10^{23}atoms/mole[/tex]
a = edge length of unit cell
[tex]a^3[/tex] = volume of unit cell
Now put all the values in above formula (1), we get
[tex]7.14g/cm^3=\frac{(6\text{ atoms per unit cell})\times (65.37g/mole)}{(6.022\times 10^{23}atoms/mole)\times a^{3}}[/tex]
[tex]V=a^{3}=9.122\times 10^{-23}cm^3[/tex]
[tex]V=9.122\times 10^{-29}m^3[/tex]
Now we have to calculate the height of the unit cell.
Formula used :
[tex]V=6\sqrt {3}r^2h[/tex]
where,
V = volume of unit cell
r = atomic radius = [tex]1.332\AA=1.332\times 10^{-10}m[/tex]
conversion used : [tex](1\AA=10^{-10}m)[/tex]
h = height of the unit cell
Now put all the given values in this formula, we get:
[tex]9.122\times 10^{-29}m^3=6\sqrt {3}\times (1.332\times 10^{-10}m)^2h[/tex]
[tex]h=4.947\times 10^{-10}m=4.947\AA[/tex]
Therefore, the height of the unit cell is, [tex]4.947\AA[/tex]
To find the height of the zinc unit cell, we apply the geometric relationship of the hexagonal close-packed structure to the given atomic radius. After performing the calculation, we get the height of the unit cell. Density provides context for the metal's packing efficiency but is not directly used in the calculation.
Explanation:To calculate the height of the unit cell of zinc, we must first determine the type of crystal structure zinc has. Zinc crystallizes in a hexagonal close-packed (hcp) structure. Given that its atomic radius is 1.332 Å (or 0.1332 nm), we can use this information along with the geometry of the hcp structure to calculate the height (c) of the unit cell.
In an hcp structure, the height (c) can be found using the relationship between the atomic radius (r) and the height, which is c = 2\sqrt{\frac{2}{3}}\cdot r. Substituting the given value for the atomic radius:
c = 2\sqrt{\frac{2}{3}}\cdot (0.1332 \text{ nm})
Calculating this gives us the height of the unit cell for zinc.
While this question does not require use of the density directly, it's worth noting that the density of zinc is 7.14 g/cm³, which indicates a relatively high atomic packing efficiency typical for metals. Using the atomic weight (65.37 grams/mole) and the Avogadro's number, we could further explore the relationship between the density and the unit cell parameters if needed.
A person steps off the end of a 3.45 � high diving board and drops to the water below. (a) How long does it take for the person to reach the water? (b) What is person’s speed on entering the water? (c) What is person’s speed on entering the water if they step off a 12.0 � diving
(a) It takes approximately 0.841 seconds to reach the water from a 3.45 m diving board, with a speed of about 8.24 m/s.
(b) If stepping off a 12.0 m diving board, it takes approximately 1.565 seconds to reach the water, with a speed of about 15.34 m/s.
To solve this problem, we can use the kinematic equations of motion to calculate the time taken to reach the water and the speed upon entering the water.
(a) To find the time taken to reach the water, we can use the kinematic equation:
Height = (1/2) * gravity * time^2
Where:
Height = 3.45 m (height of the diving board)
Gravity = 9.8 m/s² (acceleration due to gravity)
Time is the time taken to reach the water.
Rearranging the equation to solve for time:
time = sqrt((2 * height) / gravity)
time = sqrt((2 * 3.45 m) / 9.8 m/s²)
time ≈ sqrt(0.7061)
time ≈ 0.841 s
(b) To find the speed upon entering the water, we can use the equation:
Speed = gravity * time
Where:
Speed is the speed upon entering the water.
Substituting the known values:
Speed = 9.8 m/s² * 0.841 s
Speed ≈ 8.24 m/s
(c) If the person steps off a 12.0 m diving board, we can repeat the above calculations using height = 12.0 m:
time = sqrt((2 * 12.0 m) / 9.8 m/s²)
time ≈ sqrt(2.449)
time ≈ 1.565 s
Speed = 9.8 m/s² * 1.565 s
Speed ≈ 15.34 m/s