Answer:
[tex]heigth=2.86m[/tex]
Explanation:
Given data
time=0.19 s
distance=1.6 m
To find
height
Solution
First we need to find average velocity
[tex]V_{avg}=\frac{distance}{time}\\V_{avg}=\frac{1.6m}{0.19s}\\V_{avg}=8.42m/s[/tex]
Also we know that average velocity
[tex]V_{avg}=(V_{i}+V_{f})/2\\[/tex]
Where
Vi is top of window speed
Vf is bottom of window speed
Also we now that
[tex]V_{f}=V_{i}+gt\\V_{f}=V_{i}+(9.8)(0.19)\\V_{f}=V_{i}+1.862[/tex]
Substitute value of Vf in average velocity
So
[tex]V_{avg}=(V_{i}+V_{f})/2\\where\\V_{f}=V_{i}+1.862\\and\\V_{avg}=8.42m/s\\So\\8.42m/s=(V_{i}+V_{i}+1.862)/2\\2V_{i}+1.862=16.84\\V_{i}=(16.84-1.862)/2\\V_{i}=7.489m/s\\[/tex]
Vi is speed of balloon at top of the window
Now we need to find time
So
[tex]V_{i}=gt\\t=V_{i}/g\\t=7.489/9.8\\t=0.764s[/tex]
So the distance can be found as
[tex]distance=(1/2)gt^{2}\\ distance=(1/2)(9.8)(0.764)^{2}\\ distance=2.86m[/tex]
To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight. Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can solve for the initial height to be approximately 0.01485 m or 14.85 cm above the top of the window.
Explanation:To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight.
Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can plug in these values into the equation to solve for the initial height:
1.6 = 0.5*9.8*(0.19)^2
Simplifying the equation, we find that the balloon was dropped from a height of approximately 0.01485 m or 14.85 cm above the top of the window.
Learn more about height calculation here:https://brainly.com/question/30400807
#SPJ3
The mass of the Sun is 2x1030 kg, and the mass of Mars is 6.4x1023 kg. The distance from the Sun to Mars is 2.3X1011 m. Calculate the magnitude of the gravitational force exerted by the Sun on Mars. N Calculate the magnitude of the gravitational force exerted by Mars on the Sun. N
The magnitude of the gravitational force exerted by the Sun on Mars is approximately 1.49x10^22 N, while the magnitude of the gravitational force exerted by Mars on the Sun is approximately 5.92x10^15 N.
To calculate the magnitude of the gravitational force exerted by the Sun on Mars, we can use Newton's law of gravitation. The formula is given by F = G * (m1 * m2) / r^2, where F is the magnitude of the force, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between them.
Plugging in the values for the Sun's mass (2x10^30 kg), Mars' mass (6.4x10^23 kg), and the distance between them (2.3x10^11 m), we get
F = (6.673x10^-11 N·m²/kg²) * ((2x10^30 kg) * (6.4x10^23 kg)) / (2.3x10^11 m)^2
Simplifying the equation and performing the calculations, the magnitude of the gravitational force exerted by the Sun on Mars is approximately 1.49x10^22 N.
Similarly, to calculate the magnitude of the gravitational force exerted by Mars on the Sun, we can use the same formula with the masses and distance reversed. Plugging in the values, we get
F = (6.673x10^-11 N·m²/kg²) * ((6.4x10^23 kg) * (2x10^30 kg)) / (2.3x10^11 m)^2
Simplifying and calculating the equation, the magnitude of the gravitational force exerted by Mars on the Sun is approximately 5.92x10^15 N.
For more such questions on gravitational force, click on:
https://brainly.com/question/18961003
#SPJ3
A metal cylinder is measured to have a length of 5.0 cm and a diameter of 1.26 cm. Compute the volume of the cylinder. Write your final answer with the correct number of significant figures.
Answer:
V = 6.23 cm³
Explanation:
given,
Length of the cylinder, h = 5 cm
Diameter of the cylinder, d = 1.26 cm
r = 0.63 cm
Volume of the cylinder = ?
We know,
[tex]V = \pi r^2 h[/tex]
[tex]V = \pi \times 0.63^2\times 5[/tex]
V = 6.23 cm³
Volume of the cylinder is equal to V = 6.23 cm³
The volume of a metal cylinder with a length of 5.0 cm and a diameter of 1.26 cm (-approximately 4.9 cm^3- when calculated using the formula: Volume = π * (d/2)^2 * h.
Explanation:To calculate the volume of a cylinder, you use the formula: Volume = π * (d/2)^2 * h. Here, d represents the diameter of the cylinder, h represents the height, and π (pi) is a constant approximately equal to 3.14159. Plugging the values into the formula, we get:
Volume = π * (1.26 cm/2)^2 * 5.0 cm
After performing the above computations, the volume of the metal cylinder is approximately 4.9 cm^3 when rounded to two significant figures.
https://brainly.com/question/32822827
#SPJ3
Collapse question part Part 4 (d) What is the unit vector in the direction of the spacecraft's velocity? (Express your answer in vector form.)
Answer:
unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]
Explanation:
Given:
v = (-23.2, -104.4, 46.4) m/s
Above expression describes spacecraft's velocity vector v.
Find:
Find unit vector in the direction of spacecraft velocity v.
Solution:
Step 1: Compute magnitude of velocity vector.
mag (v) = sqrt ( 23.2^2 + 104.4^2 + 46.4^2)
mag (v) = 116.58 m/s
Step 2: Compute unit vector unit (v)
unit (v) = vec (v) / mag (v)
unit (v) = [ -23.2 i -104.4 j + 46.4 k ] / 116.58
unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]
The unit vector in the direction of the spacecraft’s velocity is found by dividing the velocity vector by its magnitude. Accordingly, if one knows the component values of the spacecraft's velocity, one can calculate the components for the unit vector by dividing each component by the magnitude of the velocity.
Explanation:The unit vector in the direction of the spacecraft's velocity can be determined from its velocity vector. The unit vector is a vector of length 1 that points in the same direction as the given vector. It can be found by dividing the velocity vector of the spacecraft by its magnitude. If V represents the velocity vector, then the unit vector U is calculated as U = V / |V| where |V| is the magnitude of the vector V. Thus, if you know the component values of the spacecraft's velocity, you can calculate the respective components for the unit vector by dividing each component of the velocity vector by the velocity's magnitude.
Learn more about Unit Vector here:https://brainly.com/question/34814688
#SPJ3
1. A hollow conductor carries a net charge of +3Q. A small charge of -2Q is placed inside the cavity in such a way that it is isolated from the conductor. How much charge is on the outer surface of the conductor?
Answer:
+Q
Explanation:
As no electric field can exist (in electrostatic condition) inside a conductor, if we apply Gauss 'Law to a spherical gaussian surface with a radius just a bit larger than the distance of the inner surface to the center (but less tah the distance of the outer surface), the net flux through this surface must be zero, due to E=0 at any point of the gaussian surface.
Therefore, as the net flux must be proportional to the charge enclosed by the surface, it follows that Qenc = 0.
⇒ Qenc = Qc + Qin = -2Q + Qin = 0 ⇒ Qin = +2Q
So, if the net charge of the conductor is + 3Q (which must remain the same due to the conservation of charge principle) and no charge can exist within the conductor (in electrostatic conditions), we have the following equation:
Qnet = Qin + Qou = +3Q ⇒ +2Q + Qou = +3Q
⇒ Qou = +Q
An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance between right and left wheels is 1.5 m. What are the rotating speeds of each driving wheel as fractions of the drive shaft speed?
Explanation:
The given data is as follows.
Inner wheel Radius = 20 m,
Distance between left and right wheel = 1.5m,
Let us assume speed of drive shaft is N rpm.
Formula to calculate angular velocity is as follows.
Angular velocity of automobile = w = [tex]\frac{V}{R}[/tex]
where, V = linear velocity of automobile m/min,
R = turning radius from automobile center in meter
In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.
Now, we assume that
u = linear velocity of inner wheel
and, u' = linear velocity of outer wheel.
Formula for angular velocity of inner wheel w = ,
Formula for angular velocity of outer wheel w =
Now, for inner wheels
w =
= [tex]\frac{u}{(R - d)}[/tex]
u = [tex]V \times \frac{(R - d)}{R}[/tex]
= [tex]V \times (1 - \frac{d}{R})[/tex]
If radius of wheel is r it will cover distance in one min.
Since, velocity of wheel is u it will cover distance u in unit time(min)
Thus, u = [tex]2\pi rn[/tex] = [tex]V \times (1 - \frac{d}{R})[/tex]
Now, rotation per minute of inner wheel is calculated as follows.
n = [tex]\frac{V}{2 \pi r \times (1 - \frac{d}{R})}[/tex]
= [tex]\frac{V}{2 \pi r \times (1 - \frac{0.75}{20})}[/tex] (since 2d = 1.5m given, d = 0.75m),
= [tex]\frac{V}{r} \times 0.1532[/tex]
So, rotation per minute of outer wheel; n' =
= [tex]\frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}[/tex]
= [tex]\frac{V}{r} \times 0.1651[/tex]
In a sharp left turn, the car's right wheel rotates 1.0375 times the speed of the drive shaft, while the left wheel rotates at the speed of the drive shaft. This difference is due to the right wheel covering a larger distance than the left wheel.
The principle in operation here is that in a sharp turn, the outer wheel (in this case, the right wheel) has to cover a larger distance than the inner wheel (left wheel). Therefore, the right wheel will rotate faster than the left one.
Now, calculating the difference in rotation speeds of the wheels, we consider the radii of the paths of the two wheels. For the left wheel, the radius is 20 m and for the right wheel, the radius is greater by half the wheelbase distance, that is, 20.75 m.
The ratio of the speeds is equal to the ratio of the radii of the two paths. Therefore, the speed of the right wheel as a fraction of the drive shaft speed is 20.75/20 = 1.0375 and the speed of the left wheel as a fraction of the drive shaft speed is 20/20 = 1.
So, in conclusion, during a sharp turn to the left, the right wheel rotates 1.0375 times the speed of the drive shaft, while the left wheel rotates at the same speed as the drive shaft.
For more such questions on speed, click on:
https://brainly.com/question/32891458
#SPJ6
A water droplet of mass ‘m’ and net charge ‘-q’ remains stationary in the air due to Earth’s Electric field.
(a) What must be the direction of the Earth’s electric field?
(b) Find an expression for the Earth’s electric field in terms of the mass and charge of the droplet.
Answer:
(a) the electric field of the Earth will be directed towards the negatively charged water droplet.
(b) E = (9.8*m)/q
Explanation:
Part (a) the direction of the Earth’s electric field
Electric field is always directed towards negatively charged objects.
The water droplet has a negative charge ‘-q’, therefore the electric field of the Earth will be directed towards the negatively charged water droplet.
Part (b) an expression for the Earth’s electric field in terms of the mass and charge of the droplet
The magnitude of Electric field is given as;
E = F/q
where;
f is force and q is charge
Also from Newton's law, F = mg
where;
m is mass and g is acceleration due to gravity = 9.8 m/s²
E = F/q = mg/q
E = (9.8*m)/q, Electric field in terms of mass and charge of the droplet.
Facilitated diffusion and simple diffusion are processes that decrease the potential energy stored across a membrane true or false
Answer:
It's true. Potential energy is actually the concentrations of different elements. Diffusion is the process of moving the elements or materials from the area of high concentration (high potential energy) to the low concentration. So in bot facilitated and simple types of diffusion the level of potential energy decreases across the membrane.
Explanation:
The specific heat capacity of lead is 0.13 J/g-°C. How much heat (in J) is required to raise the temperature of of 91 g of lead from 22°C to 37°C?
Answer:
Q = 177J
Explanation:
Specific heat capacity of lead=0.13J/gc
Q=MCΔT
ΔT=T2-T1,where T1=22degrees Celsius and T2=37degree Celsius.
ΔT=37 - 22 = 15
Q = Change in energy
M = mass of substance
C= Specific heat capacity
Q = (91g) * (0.13J/gc) * (15c)= 177.45J
Approximately, Q = 177J
What will be the induced magnetic field strength 7.5 cm radially outward from the center of the plates?
Answer:
B = 9.867 * 10^-8 T
Explanation:
Given:
- Rate of accumulating charge I = 37.0 mC/s
- radial distance from center of the plate r = 7.5 cm
- magnetic constant u_o = 4*pi *10^-7 H/m
Find:
The induced magnetic field strength 7.5 cm radially outward from the center of the plates?
Solution:
- Apply Gaussian Law on the surface:
B.(dA) = u_o*I
- The surface integral is dA = (2*pi*r):
B.(2*pi*r) = u_o*I
B = u_o*I / (2*pi*r)
- Plug values in:
B = (4*pi *10^-7)*(37*10^-3) / (2*pi*0.075)
B = 9.867 * 10^-8 T
Answer:
answer is B on Plato if you have it
Explanation:
If a jumping frog can give itself the same initial speed regardless of the direction in which it jumps (forward or straight up), how is the maximum vertical height to which it can jump related to its maximum horizontal range Rmax = v20/g?
Answer:
Explanation:
If u is the initial velocity at an angle \theta with horizontal then
Horizontal range of Frog can be given by
[tex]R=ut+\frac{1}{2}at^2[/tex]
where u=initial velocity
a=acceleration
t=time
Here initial horizontal velocity
[tex]u_x=u\cos \theta [/tex]
and there is no acceleration in the horizontal motion
Therefore
[tex]R=u\cos \theta \times t+0[/tex]
Considering vertical motion
[tex]Y=ut+\frac{1}{2}at^2[/tex]
here Initial vertical velocity [tex]u_y=u\sin \theta [/tex]
acceleration [tex]a=g[/tex]
for complete motion Y=0 i.e.displacement is zero
[tex]0=u\sin \theta \times t-\frac{1}{2}gt^2[/tex]
[tex]t=\frac{2u\sin \theta }{g}[/tex]
Therefore Range is
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
Range will be maximum when [tex]\theta =45[/tex]
[tex]R=\frac{u^2}{g}----1[/tex]
and Maximum height [tex]h_{max}=\frac{u^2\sin ^2 \theta }{2g}[/tex]
for [tex]\theta =45[/tex]
[tex]h_{max}=\frac{u^2}{4g}----2[/tex]
Divide 1 and 2
[tex]\frac{R_{max}}{h_{max}}=\frac{\frac{u^2}{g}}{\frac{u^2}{4g}}[/tex]
[tex]\frac{R_{max}}{h_{max}}=4[/tex]
The number of confirmed exoplanets is (a) less than 10; (b) roughly 50; (c) more than 500; (d) more than 5000.
Answer:
(c) more than 500
Explanation:
Until 2019, more than 3000 planetary systems have been discovered that contain more than 4000 exoplanets, since some of these systems contain multiple planets. Most known extrasolar planets are gas giants equal to or more massive than the planet Jupiter, with orbits very close to its star.
Final answer:
The number of confirmed exoplanets is more than 5000. Through missions like Kepler, the catalog of exoplanets includes systems with a variety of planet arrangements and types, indicating an incredibly diverse universe with potentially billions of Earth-size planets. The correct answer is option is d) more than 5000.
Explanation:
The number of confirmed exoplanets is (d) more than 5000. As of July 2015, NASA's Kepler mission had detected a total of 4,696 possible exoplanets, and 1,030 of those candidates had been confirmed as planets. Advancing to 2018, astronomers had data on nearly 3,000 exoplanet systems.
By 2022, this knowledge expanded to include data on over 800 systems, many with multiple planets, and an understanding that one quarter of stars may have exoplanet systems. This implies the existence of at least 50 billion planets in our Galaxy alone. Recent studies have shown that planets like Earth are the most common type of planet, leading to an estimated 100 billion Earth-size planets around Sun-like stars in the Galaxy.
It's important to note that detections have been made possible using the Doppler and transit techniques, and while most of the exoplanets found are more massive than Earth, the shortage of small rocky planets detected is an observational bias, as they are more difficult to detect.
The ensemble of exoplanets discovered is incredibly diverse, suggesting a variety of planet formations and arrangements in these systems. Some systems have been observed to have rocky planets closer to their stars than in our solar system, and others have large gas giants, known as 'hot Jupiters', very close to their stars.
Say for a particular population body temperature measured in Celsius has a mean of 37 and the standard deviation is equal to 0.5. What is the mean and standard deviation of the population when temperature is measured in Fahrenheit units?
Answer
given,
Mean of temperature in Celsius = 37 Standard deviation in Celsius = 0.5
relation between Fahrenheit and Celsius
[tex]F = \dfrac{9}{5}C + 32[/tex]
Mean of temperature in Fahrenheit
[tex]Mean_F =\dfrac{9}{5}\times Mean_C + 32[/tex]
[tex]Mean_F =\dfrac{9}{5}\times 37 + 32[/tex]
[tex]Mean_F = 98.6\ F[/tex]
Standard deviation of Fahrenheit.
[tex]SD_F = \dfrac{9}{5}\times SD_C [/tex]
addition of 32 will not change the Standard deviation.
[tex]SD_F = \dfrac{9}{5}\times 0.5[/tex]
[tex]SD_F = 0.9[/tex]
Air temperature in a desert can reach 58.0°C (about 136°F). What is the speed of sound (in m/s) in air at that temperature?
The speed of sound in air increases with temperature. By taking the difference between the given temperature and 0°C, then multiplying by the rate of speed increase per degree Celsius, you can find the speed of sound. This comes out to approximately 365.8 m/s at 58.0°C.
Explanation:The speed of sound in air varies depending on the temperature of the air. In general, the speed of sound increases by approximately 0.6 m/s for each degree Celsius increase in temperature. Therefore, we need to find the difference between the given temperature (58.0°C) and 0°C, which is 58, and then multiply that by 0.6 to find the increase in speed due to temperature.
Step 1: Find the temperature difference = 58.0°C - 0°C = 58°C
Step 2: Multiply the temperature difference by the rate of speed increase, which is 0.6 m/s/°C. We get: (58°C) x (0.6 m/s/°C) = 34.8 m/s.
Step 3: To find the speed of sound at the higher temperature, add this increase to the speed of sound at 0°C, which is 331 m/s.
Step 4: So, the speed of sound in air at 58.0°C is 331 m/s + 34.8 m/s = 365.8 m/s (approximately).
Learn more about Speed of Sound here:https://brainly.com/question/35989321
#SPJ11
If the humidity in a room of volume 410 m3 at 25 ∘C is 70%, what mass of water can still evaporate from an open pan? Express your answer using two significant figures.
Answer:
[tex]m=1864.68\ g[/tex]
Explanation:
Given:
volume of air in the room, [tex]V=410\ m^3[/tex]
temperature of the room, [tex]T=25+273=298\ K[/tex]
Saturation water vapor pressure at any temperature T K is given as:
[tex]p_{sw}=\frac{e^{(77.3450 + 0.0057\times T - \frac{ 7235}{T} )}}{T^{8.2}}[/tex]
putting T=298 K we have
[tex]p_{sw}=3130\ Pa[/tex]
The no. of moles of water molecules that this volume of air can hold is:
Using Ideal gas law,
[tex]P.V=n.R.T[/tex]
[tex]n=\frac{P_{sw}.V}{R.T}[/tex]
[tex]n=\frac{3130\times 410}{8.314\times 298}[/tex]
[tex]n=518\ moles[/tex] is the maximum capacity of the given volume of air to hold the moisture.
Currently we have 80% of n, so the mass of 20% of n:
[tex]m=(20\%\ of\ n)\times M}[/tex]
where;
M= molecular mass of water
[tex]m=0.2\times 518\times 18[/tex]
[tex]m=1864.68\ g[/tex] is the mass of water that can vaporize further.
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a 23.0-V potential difference is applied to these plates, calculate the following.
(a) Calculate the electric field between the plates.
(b) Calculate the surface charge density.
(c) Calculate the capacitance.
(d) Calculate the charge on each plate.
Answer:
a. 11.5kv/m
b.102nC/m^2
c.3.363pF
d. 77.3pC
Explanation:
Data given
[tex]area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v[/tex]
to calculate the electric field, we use the equation below
V=Ed
where v=voltage, d= distance and E=electric field.
Hence we have
[tex]E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m[/tex]
b.the expression for the charge density is expressed as
σ=ξE
where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2
If we insert the values we have
[tex]8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2} \\102nC/m^{2}[/tex]
c.
from the expression for the capacitance
[tex]C=eA/d[/tex]
if we substitute values we arrive at
[tex]C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF[/tex]
d. To calculate the charge on each plate, we use the formula below
[tex]Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC[/tex]
For the air-filled capacitor:
(a) The electric field is 11.5 kV/m
(b) The surface charge density is 1.02×10⁻⁷C/m²
(c) The capacitance is 3.36pF
(d) Charge on the plate: 76pC
Capacitor:Given that an air-filled capacitor consists of two parallel plates such that the:
Area of the plates, A = 7.6 cm² = 7.6×10⁻⁴ m²
distance between the plates, d = 2mm = 2×10⁻³m
voltage applied, V = 23V
(a) Electric field is given by:
E = V\d
E = 23/2×10⁻³
E = 11.5 kV/m
(b) The electric field for a parallel plate capacitor in terms of the surface charge density is given by:
E = σ/ε₀
where σ is the surface charge density:
σ = ε₀E
σ = 8.85×10⁻¹²×11.5×10³
σ = 1.02×10⁻⁷C/m²
(c) the capacitance is given by:
C = ε₀A/d
C = (8.85×10⁻¹²)×(7.6×10⁻⁴)/2×10⁻³
C = 3.36×10⁻¹²F
C = 3.36 pF
(d) The charge (Q) on the plate is given by:
Q = σA
Q = 1.02×10⁻⁷× 7.6×10⁻⁴
Q = 76 pC
Learn more about capacitors:
https://brainly.com/question/17176550?referrer=searchResults
After a great many contacts with the charged ball, how is the charge on the rod arranged (when the charged ball is far away)?
a. There is positive charge on end B and negative charge on end A.
b. There is negative charge spread evenly on both ends.
c. There is negative charge on end A with end B remaining neutral.
d. There is positive charge on end A with end B remaining neutral.
Answer: Option (b) is the correct answer.
Explanation:
Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.
Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.
Thus, we can conclude that negative charge spread evenly on both ends.
Option (b) is the correct answer.
b. There is negative charge spread evenly on both ends.
The following information should be considered:
Since, there should be the negative charge present on the ball and a positive charge present on the rod. Due to this, at the time when the negatively charged metal ball will come in contact with the rod so positive charges from rod get conducted towards the metal ball. Therefore, the rod gets neutralized. However towards the metal ball there is a continuous supply of negative charges.Learn more: https://brainly.com/question/10024737?referrer=searchResults
How much energy is required to raise the temperature of 11.4 grams of gaseous helium from 24.2 °C to 38.3 °C ?
Answer:
Q= 835J
Explanation:
Specific heat capacity of Helium=5.193J/gk
Q=msΔT
ΔT=T2-T1
Q=change in energy
m=mass of substance
S= Specific heat capacity
Q= Change in energy
T2= 38.3 degrees Celsius =(38.3+273)k= 311.3k
T1=24.2 degree Celsius =24.2+273k
T1=297.2k
ΔT=T2-T1=(311.3-297.2)k=14.1K
Q=(11.4g)*(5.193J/gk)*(14.1k)
Q=834.72282J
Approximately, Q= 835J
On a dry day, the temperature in Boulder (altitude: 5330’) is 40F. What is the temperature (in F) on nearby Bear Peak (altitude: 8460’)?
Answer: °C = 4.44°C
Explanation: The centigrade scale and Fahrenheit scale are related by the formulae below
9 *°C = 5(°F - 32)
Where °C = measurement of temperature in centigrade scale.
°F = measurement of temperature in Fahrenheit scale =40°F
By substituting the parameters, we have that
9 * °C = 5 (40 - 32)
9* °C = 5(8)
9 * °C = 40
°C = 40/9
°C = 4.44°C
Temperature decreases by approximately 3.5°F for every 1000 feet increase in altitude. Given the altitude difference of 3130 feet from Boulder to Bear Peak and Boulder's temperature of 40°F, we can calculate that the temperature at Bear Peak is expected to be approximately 29°F.
Explanation:The question asks for the temperature at a higher altitude given the temperature at a lower one. This involves understanding how temperature changes with altitude. It is stated that, on average, temperature decreases by about 3.5°F for every 1000 feet you climb in altitude, a phenomenon known as the lapse rate.
Given that, if the starting temperature in Boulder (altitude: 5330’) is 40F, and we need to find the temperature on Bear Peak (altitude: 8460’), we need to calculate the difference in altitude and the corresponding temperature decrease.
The altitude difference is 8460 - 5330 = 3130 feet. From this height difference, we can determine the corresponding temperature change by multiplying 3130 feet by 3.5°F per 1000 feet, yielding a temperature decrease of about 11°F.
To find the temperature at the higher altitude (Bear Peak), we then subtract this temperature change from the starting temperature. Thus, 40°F - 11°F gives 29°F as the expected temperature on Bear Peak.
Learn more about Temperature change with altitude here:https://brainly.com/question/27681105
#SPJ12
A certain substance has a heat of vaporization of 37.51 kJ / mol. At what Kelvin temperature will the vapor pressure be 3.50 times higher than it was at 307 K?
Answer:
T2=336K
Explanation:
Clausius-Clapeyron equation is used to determine the vapour pressure at different temperatures:
where:
In(P2/P1) = ΔvapH/R(1/T1 - 1/T2)
p1 and p2 are the vapour pressures at temperatures
T1 and T2
ΔvapH = the enthalpy of vaporization of the liquid
R = the Universal Gas Constant
p1=p1, T1=307K
p2=3.50p1; T2=?
ΔvapH=37.51kJ/mol=37510J/mol
R=8.314J.K^-1moL^-1
In(3.50P1/P1)= (37510J/mol)/(8.314J.K^-1)*(1/307 - 1/T2)
P1 and P1 cancelled out:
In(3.50)=4511.667(T2 - 307/307T2)
1.253=14.696(T2 - 307/T2)
1.253=(14.696T2) - (14.696*307)/T2
1.253T2=14.696T2 - 4511.672
Therefore,
4511.672=14.696T2 - 1.253T2
4511.672=13.443T2
So therefore, T2=4511.672/13.443=335.61
Approximately, T2=336K
One of the primary visible emissions from a distant planet occurs at 425 nm. Calculate the energy of a mole of photons of this emission.]
Answer:
Explanation:
Given
Wavelength of incoming light [tex]\lambda =425\ nm[/tex]
We know
[tex]speed\ of\ wave=frequency\times wavelength[/tex]
[tex]frequency=\frac{speed}{wavelength}[/tex]
[tex]\mu =\frac{3\times 10^8}{425\times 10^{-9}}[/tex]
[tex]\mu =7.058\times 10^{14}\ Hz[/tex]
Energy associated with this frequency
[tex]E=h\mu [/tex]
where h=Planck's constant
[tex]E=6.626\times 10^{-34}\times 7.058\times 10^{14}[/tex]
[tex]E=46.76\times 10^{-20}\ Hz[/tex]
Energy of one mole of Photon[tex]=N_a\times E[/tex]
[tex]=6.022\times 10^{23}\times 46.76\times 10^{-20}[/tex]
[tex]=281.58\times 10^{3}[/tex]
[tex]=281.58\ kJ[/tex]
To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.
Explanation:To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).
Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.
Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).
Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).
Learn more about energy of photons here:https://brainly.com/question/34240307
#SPJ3
Substance A has a heat capacity that is much greater than that of substance B. If 10.0 g of substance A initially at 30.0 ∘C is brought into thermal contact with 10.0 g of B initially at 80.0 ∘C, what can you conclude about the final temperature of the two substances once the exchange of heat between the substances is complete?
When substances with different heat capacities are brought into thermal contact, heat transfers until they reach thermal equilibrium. In this case, the final temperature will be closer to the initial temperature of substance A.
When two substances with different heat capacities are brought into thermal contact, heat will transfer from the substance with a higher initial temperature to the substance with a lower initial temperature until they reach thermal equilibrium. In this case, since substance A has a much greater heat capacity than substance B, it can absorb and transfer a larger amount of heat. Therefore, the final temperature of the two substances will likely be closer to the initial temperature of substance A and lower than the initial temperature of substance B.
Learn more about heat capacity here:https://brainly.com/question/28302909
#SPJ3
A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s relative to the water. The local ocean current is 1.53 m/s in a direction 40° north of east. What is the velocity of the ship in meters per second relative to the Earth in degrees north of east?
Answer:
8.07 m/s, 81.7º NE.
Explanation:
The ship, due to the local ocean current, will be deviated from its original due north bearing.In order to find the magnitude of the velocity of the ship, we need to convert a vector equation, in an algebraic one.If we choose two axes coincident with the N-S and W-E directions, we can find the components of the velocity along these directions.Clearly, the velocity of the ship, relative to water, is only due north, so it has no component along the W-E axis.The local ocean current, as it is directed at an angle between both axes, has components along these axes.These components can be found from the projections of the velocity vector along these axes, as follows:[tex]vocx = voc* cos 40 = 1.53 m/s * 0.766 = 1.17 m/s\\vocy = voc* sin 40 = 1.53 m/s * 0.643 = 0.98 m/s[/tex]
The component along the N-S axis (y-axis) of the velocity of the ship will be the sum of the velocity relative to water, plus the component of the ocean current along this same axis:[tex]vshy = vsw + vwy = 7.00 m/s + 0.98 m/s = 7.98 m/s[/tex]
The component along the W-E axis, is just the component of the local ocean current in this direction:vshx = 1.17 m/s
We can find the magnitude of the velocity vector, applying the Pythagorean theorem, as follows:[tex]v = \sqrt{vshx^{2} + vshy^{2} } =\sqrt{(7.98m/s)^{2} +(1.17m/s)^{2} } =8.07 m/s[/tex]
The direction of the vector relative to the W-E axis (measured in counterclockwise direction) is given by the relative magnitude of the x and y components, as follows:[tex]tg \theta = \frac{vshy}{vshx} = \frac{7.98}{1.17} = 6.82 \\ \theta = tg^{-1} (6.82)\\ \theta= 81.7\deg[/tex]
The velocity of the ship, relative to Earth, is 8.07 m/s, 81.7º North of East.A 40-kg rock is dropped from an elevation of 10 m. What is the velocity of the rock when it is 5-m from the ground?
Answer:
Explanation:
Given
mass of rock [tex]m=40\ kg[/tex]
Elevation of Rock [tex]h=10\ m[/tex]
Distance traveled by rock with time
[tex]h=ut+\frac{1}{2}at^2[/tex]
where, u=initial velocity
t=time
a=acceleration
here initial velocity is zero
when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m
[tex]5=0\times t+\frac{1}{2}(9.8)(t^2)[/tex]
[tex]t^2=\frac{10}{9.8}[/tex]
[tex]t=1.004\approx 1\ s[/tex]
velocity at this time
[tex]v=u+at[/tex]
[tex]v=0+9.8\times 1.004[/tex]
[tex]v=9.83\ m/s[/tex]
Final answer:
To calculate the velocity of a 40-kg rock when it is 5 meters from the ground, after being dropped from 10 meters, we use the kinematic equation, leading to a final velocity of approximately 9.9 m/s.
Explanation:
The question concerns calculating the velocity of a 40-kg rock when it is 5 meters from the ground, after being dropped from an elevation of 10 meters. This problem is fundamentally based on the principles of kinematic equations that describe the motion of objects under the influence of gravity. Assuming the acceleration due to gravity (g) is 9.8 m/s2, and ignoring air resistance, we can use the equation of motion v² = u² + 2as, where v is the final velocity, u is the initial velocity (0 m/s, since the rock is dropped), a is the acceleration due to gravity, and s is the distance covered.
To find out how fast the rock is moving when it is 5 meters from the ground, we first calculate the distance it has fallen, which is 10 meters - 5 meters = 5 meters. Plugging the values into the equation, we have v² = 0 + 2(9.8)(5) = 98. Therefore, v = √98 = approximately 9.9 m/s.
A spherical shell of radius 9.7 m is placed in a uniform electric field with magnitude 1310 N/C. Find the total electric flux through the shell.
The electric flux through a spherical shell in a uniform electric field is calculated using Gauss's Law. The physical principles of uniform electric fields and spherical symmetry are applied to determine the flux, emphasizing the concept of electric flux through closed surfaces.
Explanation:The question involves calculating the electric flux through a spherical shell placed in a uniform electric field. According to Gauss's Law, the electric flux (ΦE) through a closed surface surrounding a charge is proportional to the enclosed charge (ΦE = q/ε0), regardless of the shape of the surface. In a uniform electric field, the electric flux through a closed surface, like a spherical shell, can be derived from the formula ΦE = E⋅A, where E is the electric field strength and A is the area of the spherical shell.
For a spherical shell of radius 9.7 m in a uniform electric field of 1310 N/C, the area of the shell (A) is 4πr2, resulting in A = 4π(9.72) square meters. Substituting the values into ΦE = E⋅A gives us the total electric flux through the shell.
A rod of mass M = 146 g and length L = 47 cm can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 19 g, moving with speed V = 5 m/s, strikes the rod at angle θ = 29° a distance D = L/2 from the end and sticks to the rod after the collision.What is the angular speed ωf of the system immediately after the collision, in terms of system parameters and I?
Answer:
[tex]\omega = \frac{mv\frac{L}{2}\sin(29^\circ)}{\frac{1}{3}ML^2 + m(\frac{L}{2})^2} = 0.91~{\rm rad/s}[/tex]
Explanation:
The angular speed of the system can be found by conservation of angular momentum. Since the ball and the rod are stick together, the collision is completely inelastic, ergo kinetic energy is not conserved.
[tex]L_{initial} = L_{final}\\m\vec{v}\times \vec{r} = I\omega[/tex]
The moment of inertia of the combined objects is equal to the sum of moment of inertia of the separate objects.
[tex]I = \frac{1}{3}ML^2 + m(\frac{L}{2})^2[/tex]
The cross product in the left-hand side can be written as a sine of the angle.
Therefore;
[tex]mv\frac{L}{2}\sin(29^\circ) = (\frac{1}{3}ML^2 + m(\frac{L}{2})^2)\omega\\(19\times 10^{-3})(5)(\frac{47\times 10^{-2}}{2})\sin(29^\circ) = (\frac{1}{3}(146\times 10^{-3})(47\times 10^{-2})^2 + (19\times 10^{-3})(\frac{47\times 10^{-2}}{2})^2)\omega\\\omega = 0.91~{\rm rad/s}[/tex]In terms of system parameters:
[tex]\omega = \frac{mv\frac{L}{2}\sin(29^\circ)}{\frac{1}{3}ML^2 + m(\frac{L}{2})^2}[/tex]
A man drops a baseball from the top of a building. If the ball is held at a height of 1m before it is dropped, and takes 6.8 seconds to hit the ground, how high is the building in meters? (Neglect air resistance)
To solve this problem we will apply the linear motion kinematic equations.
The equation that describes the position as a function of the initial velocity, acceleration and time is given by the relation
[tex]s = v_0 t +\frac{1}{2} at^2[/tex]
Here,
[tex]v_0 =[/tex] Initial velocity
t = Time
a = Acceleration, at this case due to gravity
There is not initial velocity then we have that the equation to the given time is
[tex]s = \frac{1}{2} (9.8)(6.8)^2[/tex]
[tex]s = 226.8m[/tex]
If the ball is held at a height of 1m before it is dropped, we have that the Building height is
[tex]h = 226.8-1[/tex]
[tex]h = 225.8m[/tex]
In the absence of air resistance, a ball is thrown vertically upward with a certain initial KE. When air resistane is a factor affecting the ball, does it return to its original level with the same, less or more KE? Does your answer contradict the law of energy of conservation?
Answer:
Explanation:
A ball is thrown vertically upward with a certain Kinetic Energy in the absence of air resistance and while returning it experiences air resistance.
Air resistance causes the ball to lose its kinetic energy as it provides resistance which will convert some of its kinetic energy to heat energy.
So in a way total energy is conserved but not kinetic energy as some portion of it is lost in the form of heat.
Final answer:
A ball thrown upward with air resistance will return with less kinetic energy due to the negative work done by air resistance, which transforms some of its kinetic energy into heat. This does not violate the conservation of energy as the energy is still conserved but in different forms. Mechanical energy decreases but is compensated by the increase in thermal energy in the air.
Explanation:
When a ball is thrown vertically upward in the absence of air resistance, it will return to its original level with the same kinetic energy (KE) because energy is conserved in a system where no external forces do work. However, when air resistance is a factor, it will return with less kinetic energy. This is because air resistance does negative work on the ball, converting some of its Kinetic Energy into thermal energy as it dissipates heat into the air.
The conservation of energy principle is not contradicted by this scenario. The total energy (kinetic plus potential plus any energy converted into heat due to air resistance) is still conserved. However, the mechanical energy of the ball (the sum of its kinetic and potential energy) decreases due to the work done by air resistance. This reduction in mechanical energy is exactly balanced by the increase in thermal energy of the air.
If an object like a feather is thrown upward, it will experience significant air resistance, and it will definitely return with less kinetic energy than it had when it was thrown up. Again, this doesn't violate the conservation of energy; instead, the energy is simply transformed into different forms, primarily heat, due to interactions with the air.
Use your observations to determine qualitatively how the strength of the electric interaction between charged objects depends on the distance between them. Explain your reasoning (this is trickier than it might seem at first).
Answer:
Explanation:
The experiment confirm the inverse square law which relates the force between two charged particles to the product of the charges and the distance between the charges.
From the general equation, we notice the force is inversely related to the distance between the charges, when the distance is halved, the force increase by a factor of 4, hence a decrease in distance leads to a corresponding increase in the force value.
Also the electric field intensity a charge exert on another charge within the region of its field is independent on the distance because distance has no effect on electric field strength.
A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward with a force FS whose direction makes an angle of 30.0° with the ramp (Fig. E4.4). (a) How large a force FS is necessary for the component Fx parallel to the ramp to be 90.0 N?(b) How large will the component Fy perpendicular to the ramp be then?
Answer:
(a) 104 N
(b) 52 N
Explanation:
Given Data
Angle of inclination of the ramp: 20°
F makes an angle of 30° with the ramp
The component of F parallel to the ramp is Fx = 90 N.
The component of F perpendicular to the ramp is Fy.
(a)
Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.
Resolve F into its x-component from Pythagorean theorem:
Fx=Fcos30°
Solve for F:
F= Fx/cos30°
Substitute for Fx from given data:
Fx=90 N/cos30°
=104 N
(b) Resolve r into its y-component from Pythagorean theorem:
Fy = Fsin 30°
Substitute for F from part (a):
Fy = (104 N) (sin 30°)
= 52 N
To find the force FS necessary for the component Fx parallel to the ramp to be 90.0 N, we can use the equation Fx = FS * cos(30°) = 90.0 N. To find the component Fy perpendicular to the ramp, we can use the equation Fy = FS * sin(30°).
Explanation:To find the force FS necessary for the component Fx parallel to the ramp to be 90.0 N, we can use the equation Fx = FS * cos(30°) = 90.0 N. Rearranging the equation, FS = 90.0 N / cos(30°). Therefore, FS ≈ 103.9 N.
To find the component Fy perpendicular to the ramp, we can use the equation Fy = FS * sin(30°). Substituting the value of FS, Fy ≈ 103.9 N * sin(30°). Therefore, Fy ≈ 51.9 N.
Two water waves meet at the same point, one having a displacement above equilibrium of 60 cm and the other having a displacement above equilibrium of 80 cm. At this moment, what is the resulting displacement above equilibrium?
To solve this problem it will be necessary to apply the interference principle. Under this principle interference is understood as a phenomenon in which two or more waves overlap to form a resulting wave of greater, lesser or equal amplitude. In this case, if both are at the same point, the result of the total displacement will be the sum of the individual displacements, therefore
[tex]x = \sum h_i[/tex]
[tex]x = 60cm + 80cm[/tex]
[tex]x =140cm[/tex]
Therefore the resulting displacement above equilibrium is 140cm
Final answer:
When two water waves meet at the same point, the resulting displacement above equilibrium can be calculated by adding the individual displacements together. In this case, the resulting displacement above equilibrium is 140 cm.
Explanation:
The resulting displacement above equilibrium when two water waves meet at the same point can be determined by adding the individual displacements together. In this case, one wave has a displacement above equilibrium of 60 cm and the other wave has a displacement above equilibrium of 80 cm. The resulting displacement is found by adding 60 cm and 80 cm, which gives a resulting displacement above equilibrium of 140 cm.