Answer:
Acceleration, [tex]a=6.51\times 10^{10}\ m/s^2[/tex]
Explanation:
Given that,
Electric field, E = 680 N/C
Speed of the proton, v = 1.3 Mm/s
We need to find the acceleration of the proton. We know that the force due to motion is balanced by the electric force as :
[tex]qE=ma[/tex]
a and m are the acceleration and mass of the proton.
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\times 680}{1.67\times 10^{-27}}[/tex]
[tex]a=6.51\times 10^{10}\ m/s^2[/tex]
So, the acceleration of the proton is [tex]a=6.51\times 10^{10}\ m/s^2[/tex]. Hence, this is the required solution.
A person is standing in an elevator that is moving downward and slowing down. Is the magnitude of the normal force on the person greater than, less than, or equal to the magnitude of the weight force on the person?
the magnitude of the normal force is greater than the magnitude of the weight of the person
Explanation:
As the elevator moves down, there are two forces acting on the person:
- The weight of the person, [tex]W=mg[/tex], where m is the mass of the person and g the acceleration due to gravity, acting downward
- The normal reaction exerted by the floor of the lift on the person, N, acting upward
This means that Newton's second law can be written as
[tex]\sum F = N-W= ma[/tex] (1)
where we chose upward as positive direction, and a is the acceleration of the elevator.
Here we know that the elevator is moving downward and it is slowing down: this means that the velocity is negative (upward), and the acceleration is in the opposite direction (upward), so the acceleration is positive.
Eq.(1) can be rewritten as
[tex]N=W+ma[/tex]
and as we said that [tex]a>0[/tex], this means that
[tex]N>W[/tex]
So the magnitude of the normal force is greater than the magnitude of the weight of the person.
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The magnitude of the normal force on a person in a downward-slowing elevator is greater than the magnitude of the weight force.
Explanation:When a person is standing in an elevator that is moving downward and slowing down, the magnitude of the normal force on the person is greater than the magnitude of the weight force on the person.
According to Newton's third law, the normal force and the weight force are equal in magnitude and opposite in direction. In this case, since the elevator is moving downward and slowing down, the acceleration of the person is in the upward direction. Therefore, the normal force exerted by the floor of the elevator on the person must be greater than the weight force in order to cause the upward acceleration.
It is important to note that if the elevator is in free-fall and accelerating downward at the acceleration due to gravity, the normal force becomes zero and the person appears to be weightless.
You have a neutral balloon. What is its charge after 12000 electrons have been removed from it? The elemental charge is 1.6 × 10−19 C. Answer in units of µC.
Final answer:
Upon removing 12000 electrons from a neutral balloon, it will acquire a positive charge of 1.92 microcoulombs (1.92 µC). This is calculated by multiplying the number of electrons by the elemental charge and converting to the appropriate unit.
Explanation:
When 12000 electrons are removed from a neutral balloon, it obtains a positive charge because electrons carry a negative charge. To determine the charge the balloon now carries, we multiply the number of electrons removed by the elementary charge of an electron.
The charge (Q) on the balloon can be calculated using the formula:
Q = n × e
Where:
n is the number of removed electronse is the elementary charge per electron (1.6 × 10⁻¹⁹ C)Let's plug in the values into the formula:
Q = 12000 × 1.6 × 10⁻¹⁹ C
Q = 1.92 × 10⁻¹⁵ C
Converting coulombs (C) to microcoulombs (µC) by multiplying by 10¶ gives:
Q = 1.92 × 10⁻¹⁵ C × 10¶ µC/C
Q = 1.92 µC
The balloon will carry a charge of 1.92 µC after 12000 electrons have been removed.
An isolated, parallel‑plate capacitor carries a charge Q . If the separation between the plates is doubled, the electrical energy stored in the capacitor will be:a. Unchanged.b. Halved.c. Doubled.d. Quartered.e. Quadrupled.
Answer:
option C
Explanation:
Given,
Change on the capacitor = Q
Separation is doubled
Energy stored in the capacitor,E = ?
we know,
[tex]E = \dfrac{Q^2}{2C}[/tex]
and [tex]C = \dfrac{\epsilon_0A}{d}[/tex]
now,
[tex]E = \dfrac{Q^2}{2\epsilon A}\ d[/tex].......(1)
where d is the separation between the two plates.
now, when the separation is doubled
[tex]E' = \dfrac{Q^2}{2\epsilon A}\ (2d)[/tex]
[tex]E' = 2(\dfrac{Q^2}{2\epsilon_0 A}\ d)[/tex]
From equation (1)
E' = 2 E
Hence, the energy stored in the capacitor is doubled if the separation is increased.
The correct answer is option C.
charge densities such that are equal in magnitude but opposite in sign. The difference in potential between the plates is 490 V.
(a) Is the positive or the negative plate at the higher potential?
select
the positive plate
the negative plate
(b) What is the magnitude of the electric field between the plates?
____ kV/m
Answer:
a) The positive plate is at a higher potential than the negative plate.
b) Electric field between the plates = 5.76 KV/m
Explanation:
a) The positive plate is at a higher potential than the negative plate because of convention. We could define the negative plate to be "high potential." However, that convention has to be consistent. The positive plate is at the high potential, and that potential causes positive charges to flow from high potential to low potential. If we had defined the convention in the other direction, then the negative terminal would be the high potential, and that potential would cause negative charges to flow from high potential to low potential.
b) Electric field, E = potential difference across plate/distance or separation between the plates
E = V/d
V = 490V, d = 8.5cm = 0.085m
E = 490/0.085 = 5,764.706 V/m = 5.76 KV/m
Hope this helps!!!
A hot air balloon uses the principle of buoyancy to create lift. By making the air inside the balloon less dense then the surrounding air, the balloon is able to lift objects many times its own weight. A large hot air balloon has a maximum balloon volume of 2280 m3 and the required lift is 2500 N (rough estimate of the weight of the equipment and passengers). Calculate the temperature of the air inside the balloon which will produce the required lift. Assume that the outside air temperature is (exactly) 0◦ C and that air is an ideal gas under these conditions. What factor limit the maximum altitude attainable by this method for a given load?
Using the principle of buoyancy and the ideal gas law equation, we can calculate the temperature needed inside a hot air balloon to provide the required lift. The maximum altitude of the balloon is limited by air temperature, pressure, and other environmental factors.
Explanation:The principle of buoyancy used by hot air balloons to create lift is based on Archimedes' Principle, which states that the buoyant or lifting force exerted on a body immersed in a fluid equals the weight of the fluid the body displaces. In order for a hot air balloon to gain lift, the air inside the balloon must be heated to a temperature that makes it less dense than the surrounding cooler air. This difference in air density is what actually provides the uplift.
To calculate the required temperature inside the balloon, we can use the ideal gas law equation P₁V₁ / P₂V₂. Given the values in the question and the atmospheric pressure, we can derive the temperature inside the balloon required to provide the needed lift.
The maximum altitude attainable by a hot air balloon is limited by the ambient air temperature and pressure, which decrease as altitude increases. If the balloon's lift decreases (because it can no longer heat the air inside it sufficiently to provide lift), the balloon will stop ascending. Wind currents and weather conditions can also affect the maximum altitude.
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An electron moving in the y direction at right angles to a magnetic field experiences a magnetic force in the x direction. The direction of the magnetic field is in the_______.
A. x direction
B. x direction
C. y direction
D. z direction
E. z direction
Answer:
The direction of the magnetic field is in the z-direction
Explanation:
Applying right hand rule, which states that when the thumb, index finger and the middle finger are held mutually at right angle to each other, with the index finger pointing in the direction of the moving charge (y- direction), the thumb pointing in the direction of the magnetic force (x-direction) pushing on the moving charge and then the middle finger pointing in the direction of the magnetic field (z-direction).
z direction.
Explanation:Using John Ambrose Fleming's right hand rule which asks to position the middle finger, the thumb and the index finger as follows;
| thumb
|
|
index |
/
/
/
/
middle
With this arrangement shown above, if you point your index finger in the direction in which the charge is moving, and then the middle finger in the direction of the magnetic field, then your thumb will point in the direction of the magnetic force.
For example;
If a charge is moving in the x direction, and the magnetic force is moving in the y direction, then the above figure is rotated to make sure that the index finger is pointing in the direction of the moving charge so as to get the direction of the magnetic field as follows;
y| thumb (magnetic force)
|
(direction of charge) |
index x |
/
/
/
/ z
middle (direction of field)
Therefore, the magnetic field is in the z direction.
Now to the question, if the charge(electron) is moving in the y direction at right angles to a magnetic field and it experiences a magnetic field in the x direction, the diagram above can be rotated to depict this situation as follows;
y| index (direction of charge)
|
|
| x
/ thumb (direction of force)
/
/
/ z
middle (direction of field)
Therefore the magnetic field will move in the z direction.
Note:
i. For every rotation of right-hand, the index finger must always point in the direction of the charge.
ii. Since the question does not specify which direction the electron is moving in (whether positive or negative), the direction of the magnetic field (whether positive or negative) might not be determined either. But in either case, the field will move in the z direction.
A closed system consists of 0.5 kmol of ammonia occupying a volume of 6 m3. Determine (a) the weight of the system, in N, and (b) the specific volume, in m3/kmol and m3/kg. Let g 5 9.81 m/s2.
Explanation:
It is known that the molecular weight of ammonia ([tex]NH_{3}[/tex]) is as follows.
Molecular weight ([tex]NH_{3}[/tex]) = [tex]14 + 3 \times 1[/tex] = 17
(a) Therefore, we will calculate the mass as follows.
[tex]0.5 kmol \times (\frac{1000 mol}{1 kmol}) \times (\frac{17 g}{1 mol})[/tex]
= 8500 g
Now, formula to calculate weight of the system in N is as follows.
Weight = mass × g
= [tex]8500 g \times (\frac{1 kg}{1000 g}) \times (9.8 m/s^{2})[/tex]
= 83.3 N (1 [tex]kg m/s^{2}[/tex] = 1 N)
Hence, the weight of the system is 83.3 N.
(b) Relation between specific volume and number of moles is as follows.
[tex]v (m^{3}/kmol) = \frac{V}{n}[/tex]
Therefore, calculate the specific volume as follows.
[tex]V_m = \frac{6 m^{3}}{0.5 k mol}[/tex]
= 12 [tex]m^{3}/k mol[/tex]
Also,
[tex]v (m^{3}/kmol) = \frac{V}{m}[/tex]
v = [tex]\frac{6 m^{3}}{8.5 kg}[/tex]
= 0.705882 [tex]m^{3}/kg[/tex]
Therefore, we can conclude that the value of specific volume is 12 [tex]m^{3}/k mol[/tex] and 0.705882 [tex]m^{3}/kg[/tex].
Answer:
a) [tex]w=83.385\ N[/tex]
b) [tex]\bar V=12\ m^3.kmol^{-1}[/tex]
[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]
Explanation:
Given:
no. of moles of ammonia in a closed system, [tex]n=0.5\ kmol=500\ mol[/tex]
volume of ammonia, [tex]V=6\ m^3[/tex]
We know the molecular formula of ammonia: [tex]NH_3[/tex]
The molecular mass of ammonia:
[tex]M=14+3\times 1=18\ g.mol^{-1}[/tex]
Now the mass of given ammonia:
[tex]m=n.M[/tex]
[tex]m=500\times 17[/tex]
[tex]m=8500\ g=8.5\ kg[/tex]
a)
Now weight:
[tex]w=m.g[/tex]
[tex]w=8.5\times 9.81[/tex]
[tex]w=83.385\ N[/tex]
b)
Specific volume:
[tex]\bar V=\frac{6}{0.5}[/tex]
[tex]\bar V=12\ m^3.kmol^{-1}[/tex]
also
[tex]\b V=\frac{V}{m}[/tex]
[tex]\b V=\frac{6}{8.5}[/tex]
[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]
What causes a meteor shower?
Answer: Meteor showers occur when the earth in its orbit around the Sun passes through debris left over from the destruction of comets.
Explanation:A meteor is a particle broken off an asteroid or comet orbiting the Sun, it burns up as it enters the Earth's atmosphere, creating the effect called shooting star. Cosmic debris of meteor is known as meteoroids. These meteoroids, entering Earth's atmosphere, at extremely high speeds on parallel trajectories is an event known as meteor shower.
Reflected light from a thin film of oil gives constructive interference for light with a wavelength inside the film of λfilm. By how much would the film thickness need to be increased to give destructive interference?
A. 2λfilm
B. λfilm
C. λfilm/2
D. λfilm/4
The film thickness needs to be increased by C. λfilm/2 to achieve destructive interference.
Explanation:When light reflects off a thin film, it can undergo constructive or destructive interference depending on the thickness of the film and the wavelength of the light. For constructive interference, the path difference between the reflected rays should be an integer multiple of the wavelength, while for destructive interference, the path difference should be an odd multiple of half the wavelength.
Interference in thin films is a phenomenon where light waves reflect off both the top and bottom surfaces of a thin film, leading to constructive or destructive interference patterns, creating colors. Therefore, to achieve destructive interference, the film thickness needs to be increased by λfilm/2. Thus, the correct answer is option C.
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To switch from constructive to destructive interference, the film thickness must be increased by λ[tex]_f_i_l_m[/tex]/2.
In thin film interference, the thickness of the film and the wavelength of light are crucial in determining whether the interference is constructive or destructive. Constructive interference happens when the path length difference for the two reflected rays is an integral multiple of the wavelength inside the film (i.e., 2t = nλ[tex]_f_i_l_m[/tex], where n is an integer). Conversely, destructive interference occurs when this path length difference is a half-integral multiple of the wavelength (i.e., 2t = (2n + 1)λ[tex]_f_i_l_m[/tex] / 2).
Given that the film currently gives constructive interference for a wavelength of λ[tex]_f_i_l_m[/tex], the path length difference is an integral multiple of λ[tex]_f_i_l_m[/tex]. To switch to destructive interference, the thickness must be increased so the total path length difference becomes a half-integral multiple of λ[tex]_f_i_l_m[/tex]. This increase should be λ[tex]_f_i_l_m[/tex]/2.
The film thickness would need to be increased by λ[tex]_f_i_l_m[/tex]/2 to give destructive interference.
A new, previously unknown, planet Vulcan was discovered in our solar system. We measure an orbital period of 103 Earth days for the new planet. We send some astronauts to travel to the planet and they measure the planet's gravitational acceleration to be 8.2 m/s2 on its surface. They also determine that the planet's radius is half the radius of Earth.
Q1. How far away from the sun is the new planet?
(A) 8.3 x 100 m
(B) 6.4 x 100 m
(C) 1.7 x 100 m
(D) 9.5 x 1010 m
(E) 2.3 x 1010 m
Q2. What is the mass of planet Vulcan?
(A) 1.3 x 1024 kg
(B) 5.4 x 1024 kg
(C) 9.8 x 1024 kg
(D) 0.2 x 102 kg
(E) 4.8 x 1024 kg
Q3. The astronauts also discover a moon that orbits planet Vulcan with a period of 63 days. How far away is the moon from the planet Vulcan?
(A) 1 x 108 m
(B) 20 x 108 m
(C) 460 x 108 m
(D) 9 x108 m
(E) 4 x 10 m
Q4. The astronauts visit the newly discovered moon to study it. They measure the gravitational acceleration on the surface of the moon to be 2.7 m/s. When their mission is finished, they try to escape from the moon. They measure that the minimum velocity to escape the moon is 3,000 m/s. What is the mass of the moon?
(A) 1.1 x 1023 kg
(B) 5.0 x 1023 kg
(C) 7.8 x 1023 kg
(D) 29 x 104 kg
(E) 370 x 1023 kg
Answer:
Q1. corect is B, Q2. it is A, Q3. E and Q4. A
Explanation:
Q1 For this exercise we can use Newton's second law where acceleration is centripetal.
F = m a
a = v² / r.
G m M / r² = m v² / r
G M / r = v²
The velocity has a constant magnitude whereby we can divide the length of the circular orbit (2π r) between the period
G M / r = (2π r / T)²
r³ = G M T2 / 4π²
Let's calculate
T = 103 day (24 h / 1 day) (3600 s / 1h) = 8,899 10⁶ s
r³ = 6.67 10⁻¹¹ 1.99 10³⁰ (8,899 10⁶) 2 / 4π²
r = ∛ (266.25 10³⁰)
r = 6.4 10¹⁰ m
The distance matches the value in part B
Q2 Astronauts have measured the acceleration of gravity, so we can use the second law with a body on the planet's surface
F = m g
G m M_p / R_p² = m g
G M_p / R_p² = g
M_p = g R_p² / G
They indicate that the radius of the planet is half the radius of the Earth
R_p = ½ R_earth
R_p = ½ 6.37 10⁶
R_p = 3.185 10⁶ m
Let's calculate
M_p = 8.2 (3,185 10⁶)² / 6.67 10⁻¹¹
M_p = 1.25 10²⁴ kg
The correct answer is A
Q3 We use Newton's second law again, with part Q1, where M is the mass of the planet and m is the mass of the moon
r³ = G M T² / 4π²
T = 63 days (24h / 1day) (3600s / 1h) = 5.443 10⁶ s
r³ = 6.67 10⁻¹¹ 1.25 10²⁴ (5.443 10⁶)² / 4π²
r = ∛ (62.56807 10²⁴)
r = 3.97 10⁸ m
The correct answer is E
Q4 To calculate this part let's use the conservation of mechanical energy,
Starting point The surface of the moon
Em₀ = K + U = ½ m v2 - G m M / r
Final point. Infinity with zero speed
[tex]Em_{f}[/tex] = 0
Em₀ = Em_{f}
½ m v² - G m M / R = 0
v² = 2 G M / r
M = v2 r / 2G
r = 2 G M / v²
Since we don't know the radius of the moon, we will also use the equation in part 2
M = g r² / G
r = √ GM / g
Let's replace
2G M / v² = √ G M / g
4 G M / v⁴ = 1 / g
M = v⁴ / (g 4G)
M = 3000⁴ / (2.7 4 6.67 10-11)
M = 1.12 10²³ kg
corract is A
Calculate the energy of the quantum involved in the excitation of (i) an electronic oscillation of period 1.0 fs, (ii) a molecular vibration of period 10 fs, (iii) a pendulum of period 1.0 s. Express the results in joules and kilojoules per mole.
Answer:
a) E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol
b) E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol
c) E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol
Explanation:
E = hf where E = energy, H = Planck's constant = 6.62607004 × 10⁻³⁴ J.s and f = 1/time period
a) Period = 1 fs = 1 × 10⁻¹⁵ s
f = 1/(10⁻¹⁵ ) = 10¹⁵ Hz
E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁵ = 6.63 × ⁻10¹⁹ J = 6.63 × 10⁻¹⁶ KJ
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol
b) Period = 10 fs = 10 × 10⁻¹⁵ = 10⁻¹⁴ s
f = 1/period = 1/10⁻¹⁴ = 10¹⁴ Hz
E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁴ = 6.63 × 10^-20 J = 6.63 × 10⁻¹⁷ KJ
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol
c) Period = 1s
f = 1/period = 1.0 Hz
E = 6.62607004 × 10⁻³⁴ × 1 = 6.63 × 10⁻³⁴ J = 6.63 × 10⁻³¹ KJ.
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol
Answer:
(i) E = 6.626 X 10⁻¹⁹ J = 400 KJ/mol
(ii) E = 6.626 X 10⁻²⁰ J = 40 KJ/mol
(iii) E = 6.626 X 10⁻³⁴ J = 4 X 10⁻¹³ KJ/mol
Explanation:
Energy associated with excitation of a quantum is given as;
E = hf
where;
E is the energy of excitation
h is Planck's constant = 6.626 X 10⁻³⁴Js⁻¹
f is the threshold frequency in s⁻¹
Thus, E = h/t
Part (i)
E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁵)
E = 6.626 X 10⁻¹⁹ J
In (KJ/mol) = 6.626 X 10⁻²² KJ X 6.022 X10²³ = 400 KJ/mol
Part (ii)
E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁴)
E = 6.626 X 10⁻²⁰ J
In (KJ/mol) = 6.626 X 10⁻²³ KJ X 6.022 X10²³ = 40 KJ/mol
Part (iii)
E = (6.626 X 10⁻³⁴)/(1)
E = 6.626 X 10⁻³⁴ J
In (KJ/mol) = 6.626 X 10⁻³⁷ KJ X 6.022 X10²³ = 4 X 10⁻¹³ KJ/mol
(a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it in the air?
Answer
given,
height of the jump = 0.44 m
acceleration due to gravity, g = 9.8 m/s²
velocity at the height point = 0 m/s
initial speed = ?
Using equation of motion for speed calculation
v² = u² + 2 g h
0 = u² - 2 x 9.8 x 0.44
u = √8.624
u = 2.94 m/s
time taken to reach the highest point
v = u + g t
0 = 2.94 - 9.8 x t
t = 0.3 s
total time of flight will be equal to double of the time taken to reach the maximum height.
Total time = 2 x 0.3 = 0.6 s
A flea jumps straight up to a height of 0.440 m with an initial speed of 2.94 m/s and is in the air for 0.600 s.
A flea jumps straight up to a height (h) of 0.440 m. At the maximum height, the final speed (v) is zero. Given that the gravity (g) is 9.81 m/s², we can calculate the initial speed (u) using the following kinematic expression (We will assume y+ as the positive direction)
[tex]v^{2} = u^{2} + 2 \times g \times h \\\\(0m/s)^{2} = u^{2} + 2 \times (-9.81m/s^{2} ) \times 0.440m\\\\u = 2.94 m/s[/tex]
We can calculate the time (t) required to reach the maximum height using the following kinematic expression.
[tex]v = u + g \times t\\\\0m/s = 2.94m/s - 9.81m/s^{2} \times t\\\\t = 0.300 s[/tex]
The time in the air will be double the time to reach the maximum height.
[tex]t_{total} = 2 \times 0.300 s = 0.600 s[/tex]
A flea jumps straight up to a height of 0.440 m with an initial speed of 2.94 m/s and is in the air for 0.600 s.
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10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when it crosses over onto a floor that has a coefficient of kinetic friction = 0.35. How far will the puck travel across the floor before it stops?
Answer:
[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]
Explanation:
For this case we can use the second law of Newton given by:
[tex] \sum F = ma[/tex]
The friction force on this case is defined as :
[tex] F_f = \mu_k N = \mu_k mg [/tex]
Where N represent the normal force, [tex]\mu_k [/tex] the kinetic friction coeffient and a the acceleration.
For this case we can assume that the only force is the friction force and we have:
[tex] F_f = ma[/tex]
Replacing the friction force we got:
[tex] \mu_k mg = ma[/tex]
We can cancel the mass and we have:
[tex] a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}[/tex]
And now we can use the following kinematic formula in order to find the distance travelled:
[tex] v^2_f = v^2_i - 2ad[/tex]
Assuming the final velocity is 0 we can find the distance like this:
[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]
The hockey puck will travel approximately 58.5 meters across the floor before it stops.
Explanation:To solve this problem, we will use the concepts of kinetic energy and the work-energy theorem.
Initially, the hockey puck has a kinetic energy of KE1 = 0.5*m*v^2 = 0.5*0.3 kg*(20 m/s)^2 =60 J
As it travels across the floor, the friction does work on it until it stops. The work done by the friction force is W = μ*m*g*d, where: μ is the coefficient of kinetic friction (0.35), m is the mass (0.3 kg), g is the gravitational constant (9.8 m/s^2), and d is the distance traveled.
The work done by the friction is equal to the initial kinetic energy (work-energy theorem), so we can solve for d: 60 J = 0.35 * 0.3 kg * 9.8 m/s^2 * d. Simplifying this equation gives d = 60 J / (0.35 * 0.3 kg * 9.8 m/s^2) = 58.5 m
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A good-quality measuring tape can be off by 0.42 cm over a distance of 28 m. What is its percent uncertainty? (Express your answer to the correct number of significant figures and proper units.)
Answer:
0.015%
Explanation:
Data provided in the question:
Length by which the measuring tape can be off, δL = 0.42 cm
Total measured length for which Error of δL is observed, L = 28 m
Now,
we know,
1 m = 100 cm
Thus,
28 m = 28 × 100 = 2800 cm
Percent uncertainty = [δL ÷ L] × 100%
= [0.42 ÷ 2800] × 100%
= 0.015%
A 2 kg block rests on a 34o incline. If the coefficient of static friction is 0.2, how much additional force, F, must be applied to keep the block from sliding down the incline?
Answer:
Explanation:
Reaction force of inclined surface = mg cosθ
= Friction force acting in upward direction = μ x mg cosθ
If F be force required in upward direction to keep the block at rest on the plane
F + μ x mg cosθ = mg sinθ
F = mg sinθ - μ x mg cosθ
F = mg( sinθ - μ cosθ)
= 2 x 9.8 ( sin34 - 0.2 cos34 )
= 19.6 ( .559 - .1658)
= 7.7 N
This is the minimum force required .
The additional force that needs to be applied to keep a 2 kg block from sliding down a 34 degree incline, given a coefficient of static friction of 0.2, is 7.89 N.
Explanation:To find out how much additional force, F, needs to be applied to keep the 2 kg block from sliding down the 34° incline, we first need to calculate the gravitational force pulling the block down the incline. This force is given by F_gravity = m * g * sin(theta), where m = 2 kg, g = 9.8 m/s² (roughly the acceleration due to gravity on Earth), and theta = 34 degrees. Thus, F_gravity = 2 kg * 9.8 m/s² * sin(34°) = 11.10 N.
Next, we calculate the static frictional force that is preventing the block from sliding down the incline. This force is given by F_friction = m * g * cos(theta) * mu_s, where mu_s = 0.2 is the coefficient of static friction. Thus, F_friction = 2 kg * 9.8 m/s² * cos(34°) * 0.2 = 3.21 N.
The additional force, F, that must be applied to keep the block from sliding down the incline is the difference between the gravitational force and the frictional force. That's F = F_gravity - F_friction = 11.10 N - 3.21 N = 7.89 N.
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From January 26, 1977, to September 18, 1983, George Meegan of Great Britain walked from Ushuaia, at the southern tip of South America, to Prudhoe Bay in Alaska, covering 30 600 km. What was the magnitude of his average speed during that time period
Answer:
average speed = 0.146 m/s
Explanation:
given data
distance = 30600 km
time = January 26, 1977 to September 18, 1983
solution
we get here time that is January 26, 1977 to September 18, 1983
so it is = 6 year and 7 months and 22 days
and that is = 2422 days = 2.09 × [tex]10^{8}[/tex] seconds
and distance is = 30600 km = 30600 × 10³ m
so here average speed will be as
average speed = [tex]\frac{distance}{time}[/tex] ...........1
average speed = [tex]\frac{30600*10^3}{2.09*10^8}[/tex]
average speed = 0.146 m/s
A motorcycle has a velocity of 15 m/s, due south as it passes a carwith a velocity of 24 m/s, due north. What is the magnitude anddirection of the velocity of the motorcycle as seen by the driverof the car?
a. 9 m/s, north
b. 9 m/s, south
c. 15 m/s, north
d. 39 m/s, north
e. 39 m/s, south
Answer:
(b) 9m/s south
Explanation:
Case 1: A motorcycle has a velocity of 15 m/s, due south
Case 2: A car with a velocity of 24 m/s, due north.
Let the velocity of the car due south = Vs↓
Let the velocity of the car due north = Vn↑
Magnitude of the velocity of the motorcycle as seen by the driver of the car = V
V = Vn - Vs
= 24m/s - 15m/s = 9m/s↓
The magnitude and velocity of of the motorcycle as seen by the driver of the car = 9m/s south
The correct option is B
e. 39 m/s, south
Explanation:Let the velocity of the motorcycle be [tex]V_{M}[/tex]
Let the velocity of the car be [tex]V_{C}[/tex]
Let the velocity of the motorcycle relative to the car be = [tex]V_{MC}[/tex]
According to relativity of velocities in one dimension;
[tex]V_{MC}[/tex] = [tex]V_{M}[/tex] - [tex]V_{C}[/tex] --------------------------(i)
Now, take;
south to be negative (-ve)
north to be positive (+ve)
Therefore, we can say that;
[tex]V_{M}[/tex] = -15m/s [since the velocity is due south]
[tex]V_{C}[/tex] = +24m/s [since the velocity is due north]
Now, substitute the values of [tex]V_{M}[/tex] and [tex]V_{C}[/tex] into equation (i) as follows;
[tex]V_{MC}[/tex] = -15 - (+24)
[tex]V_{MC}[/tex] = -15 -24
[tex]V_{MC}[/tex] = -39 m/s
Remember that we have taken;
south to be negative (-ve)
north to be positive (+ve)
Since the result we got is negative, it means the speed is due south.
Therefore, the speed of the motorcycle as seen by the driver of the car is 3.9m/s, due south.
Consider a concentric tube heat exchanger. Assuming there is no fin attached to any of the surfaces and considering negligible fouling, what would be the appropriate equation for calculating the overall heat transfer coefficient?
Answer:
[tex]Q = T1 - T2 (KA)/L[/tex]
Explanation:
Where Q is the co-efficient of heat transfer.
T-1 is initial temperature of exchanger
T-2 is final temperature of sink where has to to be dissipated
k is the co-efficient of thermal conductivity
A is the total area of exchanger surface...
and L is the total length of exchanger...
A friend is writing a science fiction screenplay about an asteroid on a collision course with Earth. She asks you to calculate some numbers so her scenario will be correct. Astronauts will attach a rocket engine to the asteroid in an attempt to divert it. The asteroid is moving at 21 km/s. The rocket will provide an acceleration of 0.035 km/s2 at a right angle to the original motion. The rocket only has enough fuel to provide this acceleration for 40 seconds. Will this change the direction of the asteroid’s motion by at least 22°, enough to miss Earth and save civilization? (15 pts, according to Grading for problem solving, see reverse side. Your group should submit one analysis, with all group member’s names, either on the Problem solving framework or the plain white paper provided.)
Answer:
No, the deviation in the path of asteroid is not by 22°
Explanation:
Given:
velocity of asteroid, [tex]v_a=21\ km.s^{-1}[/tex]
acceleration of the rocket, [tex]a=0.035\ km.s^{-2}[/tex]
time of acceleration, [tex]t=40\ s[/tex]
Now, the final velocity of the asteroid:
using the equation of motion,
[tex]v=u+a.t[/tex]
where:
[tex]v=[/tex] final velocity
[tex]u=[/tex] initial velocity in the direction
[tex]v=0+0.035\times 40[/tex]
[tex]v=1.4\ km.s^{-1}[/tex]
Now direction of the resultant velocity:
[tex]\tan\beta=\frac{v_a}{v}[/tex]
[tex]\tan\beta=\frac{21}{1.4}[/tex]
[tex]\beta=86.186^{\circ}[/tex]
So, the deviation in the asteroid:
[tex]\theta=90-\beta[/tex]
[tex]\theta=90-86.186[/tex]
[tex]\theta=3.814^{\circ}[/tex]
A particle position in meters is given by the function x(t) = ct^4 + dt^2 + f. where c = 6m/s^4, d=8m/s^2, and f=-6m and t is in seconds.
Answer:
See explanation below.
Explanation:
For this case we have the following function:
[tex] x(t)= ct^4 +dt^2 +f[/tex]
Where [tex] c= 6 m/s^4 , d = 8m/s^2 , f=-6m [/tex]
If we replace those values we got:
[tex] x(t) = 6t^4 + 8t^2 -6[/tex]
If we want to find the position after t = 2.358 seconds for example we ust need to replace in the position function t = 2.358 and we got:
[tex] x(t=2.358) = 6(2.358)^4 + 8(2.358)^2 -6 \approx 224 m[/tex]
If we want to find the velocity we need to take the derivate of the position function and we got:
[tex] \frac{dx}{dt}=v(t) = 4ct^3 + 2dt[/tex]
[tex] v(t)= 24 t^3 + 16 t[/tex]
If we want to find the instantaneous velocity we just need to replace on v(t) a value for t
And the accelaration would be given by the second derivate of the position:
[tex] \frac{dv}{dt}= 72 t^2[/tex]
If we want to find the instantaneous acceleration we just need to replace on v(t) a value for t
Final answer:
Explaining the position function of a particle in meters at a given time using a specific function with provided values for constants.
Explanation:
Position Function: The position of a particle in meters at time 't' is described by the function x(t) = ct² + dt² + f, with given values for c, d, and f.
Explanation: To find the position of the particle using this function, substitute the values of c, d, f, and the specific time 't'. This will give you the position of the particle at that time.
Example: If c = 6m/s², d = 8m/s², f = -6m, and t = 2s, then you can plug in these values to find the position of the particle at t = 2s.
You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive the frequency as 1370 Hz. You are relieved that he is in pursuit of a different speeder when he continues past you, but now you perceive the frequency as 1330 Hz. What is the speed of the police car? The speed of sound in air is 343 m/s.
Answer:
Explanation:
Given
Apparent frequency [tex]f'=1370\ Hz[/tex]
Velocity of sound [tex]v=343\ m/s[/tex]
speed of observer [tex]v_o=35\ m/s[/tex]
Using Doppler effect Apparent frequency when source is approaching is given by
[tex]f'=f(\frac{v-v_0}{v-v_s})[/tex]
[tex]1370=f(\frac{343-35}{343-v_s})---1[/tex]
Apparent frequency when source moves away from observer
[tex]f''=f(\frac{v+v_0}{v+v_s})[/tex]
[tex]1330=f(\frac{343+35}{343+v_s})---2[/tex]
Divide 1 and 2 we get
[tex]\frac{f'}{f''}=\frac{\frac{343-35}{343-v_s}}{\frac{343+35}{343+v_s}}[/tex]
[tex]\frac{1370}{1330}=\frac{343-35}{343-v_s}\times \frac{343+v_s}{343+35}[/tex]
[tex]v_s=40\ m/s[/tex]
Thus speed of sound of police car is 40 m/s
The problem involves the Doppler effect and can be solved by applying the relevant formulas for when the source of sound is approaching and when it is receding. By setting up a system of two equations with the speed of the police car as the unknown, one can solve for it algebraically.
Explanation:This is a classic problem related to the Doppler effect, which describes the change in frequency of a wave in relation to an observer moving relative to the source of the wave. It requires us to know some principles about sound waves, including that their speed in the medium is constant, and their frequency and wavelength are inversely proportional.
To solve the problem, we need to apply the formula of the Doppler effect both when the police car is approaching and when it is moving away. The formula for the perceived frequency (f) when the source is moving towards the observer in a medium, like air, is given by f = f0*(v + vo) / (v - vs), where f0 is the emitted frequency, v is the speed of the medium (sound in this case), vo is the observer's speed, and vs is the source's speed.
When the source (police car) is moving away, the formula differs slightly and is given by f = f0 * (v - vo) / (v + vs). Here, using the frequencies when the car was approaching (1370 Hz) and moving away (1330 Hz), along with your speed (35.0 m/s) and the speed of sound (343 m/s), we have two equations with vs (the speed of the police car) as the unknown. This system of equations can be solved algebraically, yielding the speed of the police car.
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the position of a crate sliding down a ramp is given by x=0.25t^3 m, z = 6-0.75t^5/2 m, where t is in seconds. determine the magnitude of the crates velocity and acceleration when t = 2 seconds
Answer:
v = 30.15 m/s
a = 60.07 m/s2
Explanation:
Velocity is derivative of position with respect to time
[tex]v_x = x' = 3*0.25t^2 = 0.75t^2[/tex]
[tex]v_z = z' = 5*0.75t^4/2 = 1.875t^4[/tex]
Acceleration is derivative of velocity with respect to time
[tex]a_x = v_x' = 2*0.75 t = 1.5t[/tex]
[tex]a_z = v_z' = 4*1.875t^3 = 7.5t^3[/tex]
At t = 2 seconds
[tex]v_x = 0.75*2^2 = 3m/s[/tex]
[tex]v_z = 1.875*2^4 = 30m/s[/tex]
[tex]v = \sqrt{v_x^2 + v_z^2} = \sqrt{3^2 + 30^2} = \sqrt{909} = 30.15 m/s[/tex]
[tex]a_x = 1.5*2 = 3 m/s^2[/tex]
[tex]a_z = 7.5*2^3 = 60 m/s^2[/tex]
[tex]a = \sqrt{a_x^2 + a_z^2} = \sqrt{3^2 + 60^2} = \sqrt{3609} = 60.07 m/s[/tex]
The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker.
Approximately how much time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer? Select the best answer from the choices provided.
a) 1/6 seconds
b) 1/4 seconds
c) 1/2 seconds
d)1/3 seconds
e) 1/5 seconds
Answer:
1/3
Explanation:
Take the displacement and divide it by the speed of light. Meters will cancel leaving you with seconds. :)
110m / 343m/s = 0.32069971 seconds or 1/3 seconds
The time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer 1/3 second ( approx.). So, option (d) is correct.
What is velocity?The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.
Given The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker.
So, distance between the first dancer in the line and the last dancer in the line be = 120 m -10 m = 110 m
Velocity of sound in air be = 343 m/s.
Hence, The time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer = distance/velocity = 110/343 second = 1/3 second ( approx.).
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Assume that a clay model of a lion has a mass of 0.225 kg and travels on the ice at a speed of 0.85 m/s. It hits another clay model, which is initially motionless and has a mass of 0.37 kg. Both being soft clay, they naturally stick together.
What is their final velocity?
Answer:
Final velocity will be equal to 0.321 m/sec
Explanation:
We have given mass of clay model of lion [tex]m_1=0.225kg[/tex]
Its speed is 0.85 m/sec, so [tex]v_1=0.85m/sec[/tex]
Mass of another clay model [tex]m_2=0.37kg[/tex]
It is given that second clay is motionless
So its velocity [tex]v_2=0m/sec[/tex]
Now according to conservation of momentum
Momentum before collision will be equal to momentum after collision
So [tex]m_1v_!+m_2v_2=(m_1+m_2)v[/tex], here v is velocity after collision
So [tex]0.225\times 0.85+0.37\times 0+(0.225+0.37)v[/tex]
[tex]0.595v=0.191[/tex]
v = 0.321 m/sec
So final velocity will be equal to 0.321 m/sec
A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angle of θ degrees with the horizontal. Assume the spring has been compressed a distance d from its neutral position. Refer to the figure. show answer No Attempt 25%
Part (a) Set your coordinates to have the x-axis along the surface of the plane, with up the plane as positive, and the y-axis normal to the plane, with out of the plane as positive. Enter an expression for the normal force, FN, that the plane exerts on the block (in the y-direction) in terms of defined quantities and g. 25%
Part (b) Denoting the coefficient of static friction by μs, write an expression for the sum of the forces in the x-direction just before the block begins to slide up the inclined plane. Use defined quantities and g in your expression ΣFx = 25%
Part (c) Assuming the plane is frictionless, what will the angle of the plane be, in degrees, if the spring is compressed by gravity a distance 0.1 m? 25%
Part (d) Assuming θ = 45 degrees and the surface is frictionless, how far will the spring be compressed, d in meters?
Answer:
Explanation:
given
spring constant k = 880 N/m
mass m = 150 kg
Normal force will be equal to the component of weight of mass m which is perpendicular to the inclined surface
= mgcosθ
So normal force
FN = mgcosθ j , as it acts in out of plane direction .
b )
Fricrion force acting in upward direction = μs mgcosθ
component of weight acting in downward direction = mgsinθ
restoring force by spring on block in downward direction
= kd
= 880d
F( total ) = (μs mgcosθ - mgsinθ - 880d )i
c )
for balance
mgsinθ = kd
sinθ = kd / mg
= 880 x .1 / 150 x 9.8
= 88 / 1470
.0598
θ = 3.4 degree
d )
d = mgsinθ / k
150 x 9.8 sin45 / 880
= 1.18 m
The problem involves the equations of force related to a block on an inclined plane with a spring. The solution involves using concepts of statics, the restoring force of a spring, the force of gravity on an inclined plane and trigonometric functions to derive the formulas and find the answers.
Explanation:Part (a) The normal force, FN, is the product of the gravitational motion and the cosine of the angle. Therefore, FN=mgcosθ.
Part (b) Just before the block begins to slide up the plane, the sum of the forces in the x-direction is equal to the difference between the restoring force of the spring and the force due to gravity along the plane. Therefore, ΣFx = k*d - mgsinθ.
Part (c) For a frictionless plane, the angle of the plane is found by observing that the force due to gravity must be equal to the restoring force of the spring, which gives θ = arcsin(k*d/(mg)). To get the angle in degrees for d = 0.1 m, you would plug in the values: θ = arcsin(880*0.1/(150*9.81)) in radians, and convert to degrees.
Part (d) If θ = 45 degrees and the surface is frictionless, the distance the spring will be compressed, d, can be found from the equation mg*sinθ = k*d, which gives d = m*g*sinθ/k. Substituting in these values, d = 150*9.81*sin(45)/880.
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A 67 Vrms source is placed across a load that consists of a 12 ohm resistor in series with an capacitor whose reactance is 5 ohms. Compute the following: a) The average power of the load b) The reactive power of the load c) The apparent power of the load d) The power factor of the load
a) The average true power is 318.3 W
b) The reactive power is 132.6 W
c) The apparent power is 344.8 W
d) The power factor is 0.92
Explanation:
a)
For a circuit made of a resistor and a capacitor, the average (true) power is given by the resistive part of the circuit only.
Therefore, the average true power is given by:
[tex]P=I^2R[/tex]
where
I is the current
R is the resistance
In this problem, we have
V = 67 V (rms voltage)
[tex]R=12 \Omega[/tex] is the resistance of the load
[tex]X=5\Omega[/tex] is the reactance of the circuit
First we have to find the impedance of the circuit:
[tex]Z=\sqrt{R^2+X^2}=\sqrt{12^2+5^2}=13 \Omega[/tex]
Then we can find the current in the circuit by using Ohm's law:
[tex]I=\frac{V}{Z}=\frac{67}{13}=5.15 A[/tex]
Therefore, the average true power is
[tex]P=I^2R=(5.15)^2(12)=318.3 W[/tex]
b)
The reactive power of a circuit consisting of a resistor and a capacitor is the power given by the capacitive part of the circuit.
Therefore, it is given by
[tex]Q=I^2X[/tex]
where
I is the current
X is the reactance of the circuit
In this circuit, we have
[tex]I=5.15 A[/tex] (current)
[tex]X=5 \Omega[/tex] (reactance)
Therefore, the reactive power is
[tex]Q=(5.15)^2(5)=132.6W[/tex]
c)
In a circuit with a resistor and a capacitor, the apparent power is given by both the resistive and capacitive part of the circuit.
Therefore, it is given by
[tex]S=I^2Z[/tex]
where
I is the current
Z is the impedance of the circuit
Here we have
I = 5.15 A (current)
[tex]Z=13 \Omega[/tex] (impedance)
Therefore, the apparent power is
[tex]S=I^2 Z=(5.15)^2(13)=344.8 W[/tex]
d)
For a circuit with a resistor and a capacitor, the power factor is the ratio between the true power and the apparent power. Mathematically:
[tex]PF=\frac{P}{S}[/tex]
where
P is the true power
S is the apparent power
For this circuit, we have
P = 318.3 W (true power)
S = 344.8 W (apparent power)
So, the power factor is
[tex]PF=\frac{318.3}{344.8}=0.92[/tex]
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A solid sphere and a hollow sphere have the same mass and radius. The two spheres are spun with matching angular velocities. Which statement is true? I. The hollow has the greater angular momentum II. The solid sphere has the greater angular momentum. III. The angular momentum is the same for both spheres. IV. The moment of inertia is the same for both spheres.
Answer:
I. The hollow sphere has the greater angular momentum.
Explanation:
Given that the mass and radius of hollow sphere and solid sphere are same. Let the mass and radius of two spheres be m and r respectively. The two spheres are rotating having same angular velocity ω .
Moment of inertia of solid sphere, I₁ = [tex]\frac{2}{5}\times{m}r^{2}[/tex]
Moment of inertia of hollow sphere, I₂ = [tex]\frac{2}{3}\times{m}r^{2}[/tex]
Since, I₁ and I₂ are not equal. Therefore, the statement iv is wrong.
The relation between angular momentum, moment of inertia and angular velocity is :
L = Iω
Let L₁ and L₂ be the angular momentum of solid sphere and hollow sphere respectively.
L₁ = I₁ω and L₂ = I₂ω
As ω is same for both spheres but I₂ is greater than I₁, hence L₂ is greater than L₁.
Therefore, statement I is correct that the hollow sphere has the greater angular momentum.
The angular momentum of the hollow sphere is greater than that of solid sphere.
The moment of inertia of solid sphere is given as follows;
[tex]I_{ss} = \frac{2}{5} mr^2 = 0.4mr^2[/tex]
The moment of inertia of hollow sphere is given as follows;
[tex]I_{hs} = \frac{2}{3} mr^2 = 0.67 mr^2[/tex]
The angular momentum of each sphere is calculated as follows;
[tex]L = I\omega \\\\L_{ss} = 0.4mr^2 \omega \\\\L_{hs} = 0.67 mr^2 \omega[/tex]
Thus, we can conclude that the angular momentum of the hollow sphere is greater than that of solid sphere.
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In an electrically heated home, the temperature of the ground in contact with a concrete basement wall is 10.1 oC. The temperature at the inside surface of the wall is 20.8 oC. The wall is 0.18 m thick and has an area of 8.9 m2. Assume that one kilowatt hour of electrical energy costs $0.10. How many hours are required for one dollar's worth of energy to be conducted through the wall
Answer:
17 hours
Explanation:
k = Thermal conductivity of concrete = 1.1 W/m°C
A = Area = [tex]8.9\ m^2[/tex]
l = Thickness = 0.18 m
[tex]\Delta T[/tex] = Change in temperature = 20.8-10.1
Power is given by
[tex]P=\dfrac{kA\Delta T}{L}\\\Rightarrow P=\dfrac{1.1\times 8.9\times (20.8-10.1)}{0.18}\\\Rightarrow P=581.961\ W[/tex]
Time required to produce 1 kWh
[tex]t=\dfrac{3600\times 10^3}{581.961}\\\Rightarrow t=6185.98153485\ s[/tex]
For one dollar
[tex]t=\dfrac{6185.98153485}{0.1}\\\Rightarrow t=61859.8153485\ s\\\Rightarrow t=\dfrac{61859.8153485}{60\times 60}\\\Rightarrow t=17.1832820412\ hours[/tex]
The time taken is 17 hours
The total distance treaveled by a car moving in a straight line is as follows: After the first 7.0 minutes it has gone a total of 2.0 miles. After 14.0 minutes it has traveled a total of 4.5 miles. Finally at 21.0 minutes it has traveled a total of 6.0 miles. Find the average speed at: Time
Answer:
Average speed of the car will be 27.582 km/hr
Explanation:
We have given that in first 5 minutes distance traveled by car is 2 miles
After 14 minutes it has travel 4.5 miles
And finally after 21 minutes distance traveled by car is 6 miles
So total time of traveling t = 21 minutes
As we know that 1 hour = 60 minutes
So 21 minutes [tex]=\frac{21}{60}=0.35hour[/tex]
Total distance traveled = 6 miles
As 1 miles = 1.609 km
So 6 miles [tex]=6\times 1.609=9.654km[/tex]
Average speed is equal to the ratio of total distance and total time
So average speed [tex]v=\frac{9.654}{0.35}=27.582km/hr[/tex]
A mass m = 3.4 kg is at the end of a horizontal spring of spring constant k = 105 N/m on a frictionless horizontal surface. The block is pulled, stretching the spring a distance A = 6.5 cm from equilibrium, and released from rest.
(a) Write an equation for the angular frequency of the oscillation
(b) Calculate the angular frequency o of the oscillation in rad/seconds
Answer:
[tex]w=\sqrt{\frac{k}{m} }[/tex]
b. [tex]5.6rad/s[/tex]
Explanation:
a. from the spring-mass system which is explicitly describe by hooks law
from
F=-kx
which is in comparison to newtons general law of motion
F=ma
where the displacement x is expressed as
[tex]x=Asin(wt)\\[/tex]
and the acceleration is the second derivative of the displacement
[tex]a=-Aw^{2}sin(wt)\\[/tex]
hence final expression after substituting for the acceleration and the displacement is expressed as
[tex]w=\sqrt{\frac{k}{m} }[/tex]
b. for k=105N/m and m=3.4kg
we have the angular frequency to be
[tex]w=\sqrt{\frac{105}{3.4}}\\\\w=5.6rad/s[/tex]
Final answer:
The angular frequency (ω) can be found using the formula ω = sqrt(k/m). By plugging in the given values of mass (3.4 kg) and spring constant (105 N/m), the angular frequency is calculated to be approximately 5.56 rad/s.
Explanation:
The subject question pertains to simple harmonic motion (SHM) and specifically relates to the oscillatory motion of a mass attached to a horizontal spring. The question involves determining the angular frequency and calculating it in radians per second.
Part (a)
To write an equation for the angular frequency (ω) of the oscillation, we use the equation:
ω = sqrt(k/m)
where ω is the angular frequency, k is the spring constant, and m is the mass.
Part (b)
Plugging in the given values:
m = 3.4 kg
k = 105 N/m
We find ω using the formula:
ω = sqrt(105 N/m / 3.4 kg) = sqrt(30.8824 s-2) ≈ 5.56 rad/s
Therefore, the angular frequency of the oscillation is approximately 5.56 rad/seconds.