Answer:
the speed of the fluid in the wider section of the pipe is 1m/s.
Explanation:
By equation of continuity we can write (for ideal (non-viscous, in-compressible).
[tex]A_1v_1 =A_2v_2[/tex]
A_1,A_2 are areas of the pipe at inlet and outlet of the pipe.
[tex]\Rightarrow \pi d_1^2v_1=\pi d_2^2v_2[/tex]_1
Here, d_1 , d_2 are diameters of inlet and outlet, also v_1, v_2 are velocities at inlet and outlet.
putting values we get
[tex]\Rightarrow \p 0.7^2\times9=\pi 2.1^2\timesv_2[/tex]
solving we get
[tex]v_2= 1m/s[/tex]
A 0.140-kg baseball is thrown with a velocity of 27.1 m/s. It is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original velocity. How long were the bat and ball in contact?
Answer:
About [tex]1.795 \times 10^{-3}[/tex] seconds
Explanation:
[tex]\Delta p=F \Delta t[/tex], where delta p represents the change in momentum, F represents the average force, and t represents the change in time.
The change of velocity is:
[tex]37-(-27.1)=64.1m/s[/tex]
Meanwhile, the mass stays the same, meaning that the change in momentum is:
[tex]64.1\cdot 0.14kg=8.974[/tex]
Plugging this into the equation for impulse, you get:
[tex]8.974=5000\cdot \Delta t \\\\\\\Delta t= \dfrac{8.974}{5000}\approx 1.795 \times 10^{-3}s[/tex]
Hope this helps!
New evidence increasingly emphasizes that __________.
The force a spring exerts on a body is a conservative force because:
a. a spring always exerts a force parallel to the displacement of the body.
b. the work a spring does on a body is equal for compressions and extensions of equal magnitude.
c. the net work a spring does on a body is zero when the body returns to its initial position.
d. the work a spring does on a body is equal and opposite for compressions and extensions of equal magnitude.
e. a spring always exerts a force opposite to the displacement of the body.
Answer:
c. the net work a spring does on a body is zero when the body returns to its initial position
Explanation:
A force is conservative when the net work done over any path that returns to the initial position is zero. Choice C matches that definition.
An ideal spring of the kind used in physics problems has the characteristic that it applies the same force at the same distance always. So any work required to extend or compress the spring is reversed when the reverse motion takes place.
How much electrical energy is used by a 75 W laptop that is operating for 12
minutes?
"1 watt" means 1 joule of energy per second.
75 W means 75 joules/sec .
Energy = (75 Joule/sec) x (12 min) x (60 sec/min)
Energy = (75 x 12 x 60) (Joule-min-sec / sec-min)
Energy = 54,000 Joules
C2B.7Suppose I drop a 60-kg anvil from rest and from such a height that the anvil reaches a speed of 10 m/s just before hitting the ground. Assume the earth was at rest before I dropped the anvil. (a) What is the earth's speed just before the anvil hits
Complete Question
C2B.7
Suppose I drop a 60-kg anvil from rest and from such a height that the anvil reaches a speed of 10 m/s just before hitting the ground. Assume the earth was at rest before I dropped the anvil.
(a) What is the earth's speed just before the anvil hits?
b) How long would it take the earth to travel [tex]1.0 \mu m[/tex] (about a bacterium's width) at this speed?
Answer:
a
[tex]|v_1| = 1.0*10^{-22} \ m/s[/tex]
b
[tex]t = 9.95 *10^{15} \approx 10 *10^{15} \ s[/tex]
Explanation:
From the question we are told that
The mass of the anvil is [tex]m_a = 60\ kg[/tex]
The speed at which it hits the ground is [tex]v = 10 \ m/s[/tex]
Generally the mass of the earth has a value [tex]m_e = 5972*10^{24} \ kg[/tex]
Now according to the principle of momentum conservation
[tex]P_i = P_f[/tex]
Where [tex]P_i[/tex] is the initial momentum which is zero given that both the anvil and the earth are at rest
Now [tex]P_f[/tex] is the final momentum which is mathematically represented as
[tex]P_f = m_a * v + m_e * v_1[/tex]
So
[tex]0 = m_a * v + m_e * v_1[/tex]
substituting values
[tex]0 = 60 * 10 + 5.972 *10^{24} * v_1[/tex]
=> [tex]v_1 = -1.0*10^{-22} \ m/s[/tex]
Here the negative sign show that it is moving in the opposite direction to the anvil
The magnitude of the earths speed is
[tex]|v_1| = 1.0*10^{-22} \ m/s[/tex]
The time it would take the earth is mathematically represented as
[tex]t = \frac{d}{|v_1|}[/tex]
substituting values
[tex]t = \frac{1.0*10^{-6}}{1.0 *10^{-22}}[/tex]
[tex]t = 10 *10^{15} \ s[/tex]
An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 60 W incandescent bulb can be replaced by a 12 W LED bulb. Both produce 800 lumens of light. Assuming the cost of electricity is $0.29 per kilowatt-hour, how much does it cost (in dollars) to run the LED bulb for one year if it runs for four hours a day?
Answer:
C = $5.08
it costs $5.08 to run the LED bulb for one year if it runs for four hours a day
Explanation:
Given;
Power of Led bulb P = 12 W
Rate r = $0.29 per kilowatt-hour
Time = 4 hours per day
The number of hours used in a year is;
time t = 4 hours per day × 365 days per year
t = 1460 hours
The energy consumption of Led bulb in a year is;
E = Pt
E = 12 W × 1460 hours
E = 17520 watts hour
E = 17.52 kilowatt-hour
The cost of the energy consumption is;
C = E × rate = Er
C = 17.52 × $0.29
C = $5.08
it costs $5.08 to run the LED bulb for one year if it runs for four hours a day
A disk between vertebrae in the spine is subjected to a shearing force of 640 N. Find its shear deformation taking it to have the shear modulus of 1.00 109 N/m2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.30 cm in diameter.
Answer:
3.08*10^-6 m
Explanation:
Given that
Total shearing force, F = 640 N
Shear modulus, S = 1*10^9 N/m²
Height of the cylinder, L = 0.7 cm
Diameter of the cylinder, d = 4.3 cm
The solution is attached below.
We have our shear deformation to be 3.08*10^-6 m
Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box of mass M. There is friction between the surfaces of blocks 2M and 3M so the three blocks accelerate together to the right.
Which block has the smallest net force acting on it?
A) M
B) 2M
C) 3M
D) The net force is the same for all three blocks Submit
Answer:
A) M
Explanation:
The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:
Box with mass M
[tex]\Sigma F = F - F' = M\cdot a[/tex]
Box with mass 2M
[tex]\Sigma F = F' - F'' = 2\cdot M \cdot a[/tex]
Box with mass 3M
[tex]\Sigma F = F'' = 3\cdot M \cdot a[/tex]
On the third equation, acceleration can be modelled in terms of F'':
[tex]a = \frac{F''}{3\cdot M}[/tex]
An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.
[tex]F' = 2\cdot M \cdot a + F''[/tex]
[tex]F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''[/tex]
[tex]F' = \frac{5}{3}\cdot F''[/tex]
Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:
[tex]F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)[/tex]
[tex]F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''[/tex]
[tex]F = 2\cdot F''[/tex]
[tex]F'' = \frac{1}{2}\cdot F[/tex]
Afterwards, F' as function of the external force can be obtained by direct substitution:
[tex]F' = \frac{5}{6}\cdot F[/tex]
The net forces of each block are now calculated:
Box with mass M
[tex]M\cdot a = F - \frac{5}{6}\cdot F[/tex]
[tex]M\cdot a = \frac{1}{6}\cdot F[/tex]
Box with mass 2M
[tex]2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F[/tex]
[tex]2\cdot M \cdot a = \frac{1}{3}\cdot F[/tex]
Box with mass 3M
[tex]3\cdot M \cdot a = \frac{1}{2}\cdot F[/tex]
As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.
A 300 g bird flying along at 6.2 m/s sees a 10 g insect heading straight toward it with a speed of 35 m/s (as measured by an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch.
Required:
What is the bird's speed immediately after swallowing?
Answer:
The velocity of the bird is [tex]v_f = 4.87 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the bird is [tex]m_1 = 300 \ g = 0.3 \ kg[/tex]
The initial speed of the bird is [tex]u_1 = 6.2 \ m/s[/tex]
The mass of the insect is [tex]m_2 = 10 \ g = 0.01 \ kg[/tex]
The speed of the insect is [tex]u_ 2 =-35 \ m/s[/tex]
The negative sign is because it is moving in opposite direction to the bird
According to the principle of linear momentum conservation
[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v_f[/tex]
substituting values
[tex](0.3 * 6.2 ) + (0.01 * (-35)) = (0.3 + 0.01 )v_f[/tex]
[tex]1.51 = 0.31 v_f[/tex]
[tex]v_f = 4.87 \ m/s[/tex]
The Final velocity of Bird = 4.87 m/s
Mass of the bird = 300 g = 0.3 kg
Velocity of bird = 6.2 m/s
Momentum of Bird = Mass of bird [tex]\times[/tex] Velocity of Bird = 0.3 [tex]\times[/tex] 6.2 = 1.86 kgm/s
Mass of the insect = 10 g = 0.01 kg
Velocity of insect = - 35 m/s
Momentum of the Insect = Mass of Insect [tex]\times[/tex] Velocity of Insect = - 0.35 kgm/s
According to the law of conservation of momentum We can write that
In the absence of external forces on the system , the momentum of system remains conserved in that particular direction.
The bird opens the mouth and enjoys the free lunch hence
Let the final velocity of bird is [tex]v_f[/tex]
Initial momentum of the system = Final momentum of the system
1.86 -0.35 = [tex]v_f[/tex] ( 0.01 + 0.3 )
1.51 = [tex]v_f[/tex] 0.31
[tex]v_f[/tex] = 4.87 m/s
The Final velocity of Bird = 4.87 m/s
For more information please refer to the link below
https://brainly.com/question/18066930
The index of refraction for a certain type of glass is 1.645 for blue light and 1.609 for red light. A beam of white light (one that contains all colors) enters a plate of glass from the air, nair≈1, at an incidence angle of 38.55∘. What is the absolute value of ????, the angle in the glass between blue and red parts of the refracted beams?
Answer:
blue θ₂ = 22.26º
red θ₂ = 22.79º
Explanation:
When a light beam passes from one material medium to another, it undergoes a deviation from the path, described by the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the incident and transmitted media refractive indices and θ are the angles in the media
let's apply this equation to each wavelength
λ = blue
in this case n₁ = 1, n₂ = 1,645
sin θ₂ = n₁/ n₂ sin₂ θ₁
let's calculate
sin θ₂ = 1 / 1,645 sint 38.55
sin θ₂ = 0.37884
θ₂ = sin⁻¹ 0.37884
θ₂ = 22.26º
λ = red
n₂ = 1,609
sin θ₂ = 1 / 1,609 sin 38.55
sin θ₂ = 0.3873
θ₂ = sim⁻¹ 0.3873
θ₂ = 22.79º
the refracted rays are between these two angles
Two metal bars experience an equal change in volume due to an equal change in temperature. The first bar has a coefficient of expansion twice as large as the second bar. How does the original volume of the first bar compare to the original volume of the second bar
Answer:
The original volume of the first bar is half of the original volume of the second bar.
Explanation:
The coefficient of cubic expansivity of substances is given by;
γ = ΔV ÷ ([tex]V_{1}[/tex]Δθ)
Given: two metal bars with equal change in volume, equal change in temperature.
Let the volume of the first metal bar be represented by [tex]V_{1}[/tex], and that of the second by [tex]V_{2}[/tex].
Since they have equal change in volume,
Δ[tex]V_{1}[/tex] = Δ[tex]V_{2}[/tex] = ΔV
For the first metal bar,
2γ = ΔV ÷ ([tex]V_{1}[/tex]Δθ)
⇒ Δθ = ΔV ÷ (2γ[tex]V_{1}[/tex])
For the second metal bar,
γ = ΔV ÷ ([tex]V_{2}[/tex]Δθ)
⇒ Δθ = ΔV ÷ ([tex]V_{2}[/tex]γ)
Since they have equal change in temperature,
Δθ of first bar = Δθ of the second bar
ΔV ÷ (2γ[tex]V_{1}[/tex]) = ΔV ÷ ([tex]V_{2}[/tex]γ)
So that;
(1 ÷ 2[tex]V_{1}[/tex]) = (1 ÷ [tex]V_{2}[/tex])
2[tex]V_{1}[/tex] = [tex]V_{2}[/tex]
[tex]V_{1}[/tex] = [tex]\frac{V_{2} }{2}[/tex]
Thus, original volume of the first bar is half of the original volume of the second bar.
What will happen to an astronaut when the jets produce these four forces
A total charge of 62 nC is uniformly distributed throughout a non-conducting sphere with a radius of 5.00 cm. The electric potential at r = 15.0 cm , relative to the potential far away, is:________
Answer:
2790 J/C
Explanation:
charge on sphere Q = 62 nC = [tex]62*10^{-9} C[/tex]
radius of the sphere r = 5.0 cm = 0.05 m
distance away from reference point d = 15.0 cm = 0.15 m
total distance of charge relative reference point R = r + d = 0.05 + 0.15 = 0.2 m
electric potential V is given as
[tex]V = \frac{kQ}{R}[/tex]
where k = Coulumb's constant = [tex]9*10^{9}[/tex] kg⋅m³⋅s⁻⁴⋅A⁻²
[tex]V = \frac{9*10^{9} * 62*10^{-9} }{0.2}[/tex] = [tex]\frac{9*62}{0.2}[/tex]
V = 2790 J/C
Identify the following as combination, decomposition, replacement, or ion exchange reactions: Al(s) + 3 Cl2(g) → 2 AlCl3(s) Ca(OH)2(aq) + H2SO4(aq) → CaSO4(aq) + 2 H2O(l
Answer:
2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)
This is a combination reaction.
Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)
This is a replacement reaction.
Explanation:
A combination reaction is a reaction in which two reagents are combined into one product. The reaction has the following general form:
A + B → AB
where A and B represent any two chemical substances.
2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)
This is a combination reaction because a single compound forms from two or more reacting species.
Double Substitution, Double Displacement or Metastasis Reactions are those in which two elements found in different compounds exchange their positions forming two new compounds. These chemical reactions do not present changes in the number of oxidation or relative load of the elements. So they are not considered redox reactions.
The solvent of the double displacement reactions usually is water and the reagents and products are usually ionic compounds (cations or anions are exchanged), although they can also be acids or bases.
In general, this type of reaction can be expressed as:
AB + CD ⇒ AD + CD
In the reaction:
Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)
This is a replacement reaction because it is a double replacement reaction in which the ions are exchanged to form new compounds.
A depiction of a famous scientific experiment is given. Consider how the beam changes when the magnet is off compared to when the magnet is on. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned off. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned off, the beam path bends toward the positively charged plate and ends at the lower half of the wide end of the tube. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned n. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned on, the beam path travels in a straight path to the center of the wide end of the tube. What type of beam was used in this experiment?
Answer:
The beam used is a negatively charged electron beam with a velocity of
v = E / B
Explanation:
After reading this long statement we can extract the data to work on the problem.
* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.
* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced
[tex]F_{e} = F_{m}[/tex]
q E = qv B
v = E / B
this configuration is called speed selector
They ask us what type of beam was used.
The beam used is a negatively charged electron beam with a velocity of v = E / B
A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a surface area of 9.50 m2. The total power it absorbs from the star is 4400 W. Assuming the surface is an ideal absorber and radiator, calculate the equilibrium temperature of the asteroid (in K).
Answer:
T = 300.6K
Explanation:
In this problem, since the asteroid is an ideal absorber, we can approximate it to a black body, and use Stefan's law
P = σ A e T⁴
where P is the absorbed power, A the area of the asteroid, and the emissivity that for a black body is worth 1 and sigma the Stefan_boltzmann constant 5,670 10⁻⁸ W / m² K⁴
they ask us for the temperature of the asteroid
T = [tex]\sqrt[4]{(P / \sigma A e)}[/tex]
let's calculate
T = (4400 / (5,670 10⁻⁸ 9.50 1)
T =(81.6857 108)
T = 3,006 102 K
T = 300.6K
In testing an automobile tire for proper alignment, a technician marks a spot on the tire 0.220 m from the center. He then mounts the tire in a vertical plane and notes that the radius vector to the spot is at an angle of 30.0° with the horizontal. Starting from rest, the tire is spun rapidly with a constant angular acceleration of 1.90 rad/s2. (Assume the spot's position is initially positive, and assume the angular acceleration is in the positive direction).A) What is the angular speed of the wheel after 1.30 s?
B) What is the tangential speed of the spot after 1.30 s?
C) What is the magnitude of the total acceleration of the spot after 1.30 s?
D) What is the angular position of the spot after 1.30 s?
Answer:
a) The angular speed of the wheel after 1.30 seconds is [tex]2.47\,\frac{rad}{s}[/tex], b) The tangential speed of the spot after 1.30 seconds is [tex]0.543\,\frac{m}{s}[/tex], c) The magnitude of the total acceleration of the spot after 1.30 seconds is [tex]1.406\,\frac{m}{s^{2}}[/tex], d) The angular position of the spot is 2.130 radians (122.011°).
Explanation:
a) Given that tire accelerates at constant rate, final angular speed can be predicted by using the following formula:
[tex]\omega = \omega_{o} + \alpha \cdot \Delta t[/tex]
Where:
[tex]\omega[/tex] - Final angular speed, measured in radians per second.
[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
[tex]\Delta t[/tex] - Time, measured in seconds.
Given that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex] (starts at rest), [tex]\alpha = 1.90\,\frac{rad}{s^{2}}[/tex] and [tex]\Delta t = 1.30\,s[/tex], the final angular speed is:
[tex]\omega = 0\,\frac{rad}{s} + \left(1.90\,\frac{rad}{s^{2}} \right) \cdot (1.30\,s)[/tex]
[tex]\omega = 2.47\,\frac{rad}{s}[/tex]
The angular speed of the wheel after 1.30 seconds is [tex]2.47\,\frac{rad}{s}[/tex].
b) The tangential speed of the spot is the product of the distance between the center of the wheel and spot. That is:
[tex]v = r \cdot \omega[/tex]
Where r is the distance between the center of the wheel and spot. The tangential speed of the spot after 1.30 seconds is:
[tex]v = (0.220\,m)\cdot \left(2.47\,\frac{rad}{s} \right)[/tex]
[tex]v = 0.543\,\frac{m}{s}[/tex]
The tangential speed of the spot after 1.30 seconds is [tex]0.543\,\frac{m}{s}[/tex].
c) The magnitude of the total acceleration of the spot is the magnitude of the vectorial sum of radial and tangential accelerations (both components are perpendicular to each other), which is determined by the Pythagorean theorem, that is:
[tex]a = \sqrt{a_{r}^{2} + a_{t}^{2}}[/tex]
Where [tex]a_{r}[/tex] and [tex]a_{t}[/tex] are the radial and tangential accelerations.
[tex]a = r\cdot \sqrt{\omega^{4} + \alpha^{2}}[/tex]
If [tex]r = 0.220\,m[/tex], [tex]\omega = 2.47\,\frac{rad}{s}[/tex] and [tex]\alpha = 1.90\,\frac{rad}{s^{2}}[/tex], then, the resultant acceleration is:
[tex]a = (0.220\,m)\cdot \sqrt{\left(2.47\,\frac{rad}{s} \right)^{4}+\left(1.90\,\frac{rad}{s^{2}} \right)^{2}}[/tex]
[tex]a \approx 1.406\,\frac{m}{s^{2}}[/tex]
The magnitude of the total acceleration of the spot after 1.30 seconds is [tex]1.406\,\frac{m}{s^{2}}[/tex].
d) Let be 30° (0.524 radians) the initial angular position of the spot with respect to center. The final angular position is determined by the following equation of motion:
[tex]\omega^{2} = \omega_{o}^{2} + 2\cdot \alpha \cdot (\theta - \theta_{o})[/tex]
Final angular position is therefore cleared:
[tex]\theta - \theta_{o} = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}[/tex]
[tex]\theta = \theta_{o} + \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}[/tex]
Given that [tex]\theta_{o} = 0.524\,rad[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 2.47\,\frac{rad}{s}[/tex] and [tex]\alpha = 1.90\,\frac{rad}{s^{2}}[/tex], the angular position of the spot after 1.30 seconds is:
[tex]\theta = 0.524\,rad +\frac{\left(2.47\,\frac{rad}{s} \right)^{2} - \left(0\,\frac{rad}{s}\right)^{2}}{2\cdot \left(1.90\,\frac{rad}{s^{2}} \right)}[/tex]
[tex]\theta = 2.130\,rad[/tex]
[tex]\theta = 122.011^{\circ}[/tex]
The angular position of the spot is 2.130 radians (122.011°).
In the rotational motion of an object, the angular acceleration is always towards the center, and the further discussion is as follows:
Rotational motion:The tangential acceleration of the object keeps changing its direction as the object rotates, always directed toward the tangent of the circle passing through the position of the object.
Radius of the spot, r = 0.220
minitial angle from the horizontal, θ = 30°
angular acceleration, α = 1.9 rad/s²
(a) from the first equation of motion we get:
ω = ω₀ + αt where
ω is the final angular speed
ω₀ is the initial angular speedand
t is the time = 1.3sω = 1.9×1.3 rad/sω = 2.47 rad/s
(b) tangential speed (v) is given by:
v = r×ωv = 0.220×2.47 m/sv = 0.5434 m/s
(c) The instantaneous tangential acceleration is given by:
[tex]a_t[/tex] = rω²so the resultant acceleration will be:
[tex]a=\sqrt{a_t^2+\alpha^2}\\\\a =\sqrt{r^2\omega^4+\alpha^2}\\\\a= \sqrt{(0.220)^2(2.47)^4+(1.9)^2}\\\\a = 1.4 \ \frac{m}{s^2}[/tex]
(d)
The angular displacement is given by:
θ = θ₀t + ¹/₂αt²θ₀ = 30° = 0.524
rad θ = 0.524×1.3 + ¹/₂×1.9×1.3²θ = 2.286 radθ = 131°
Following are the solution for points:
For a)
The angular speed is 2.47 rad/s
For b)
The tangential speed is 0.5434 m/s
For c)
Total acceleration is 1.4 m/s²
For d)
The final angular position is 131°
Learn more about rotational here:
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charged particles from the solar winds ultimately cause ___. a. the earth to maintain it's magnetic field b. the earth to change shape c. the auroras d. strong winds on earth
Answer:
The auroras C.
Explanation:
If the archerfish spits its water 30 degrees from the horizontal aiming at an insect 1.2 m above the surface of the water, how fast must the fish spit the water to hit its target? The insect is at the highest point of the trajectory of the spit water. Use g = 10 m/s2.
Answer:
The speed is [tex]v = 9.8 \ m/s[/tex]
Explanation:
From the question w are told that
The angle made is [tex]\theta = 30^o[/tex]
The distance above the surface of the water is [tex]h_{max} = 1.2 \ m[/tex]
The value of [tex]g = 10 \ m/s^2[/tex]
The maximum height attained by the fish is mathematically evaluate as
[tex]h_{max} = \frac{v^2 sin ^2 \theta }{2g }[/tex]
Making v which is the speed of the fish the subject of the formula
[tex]v = \sqrt{ \frac{2gh_{max}}{ sin^2 \theta } }[/tex]
substituting values
[tex]v = \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2 } }[/tex]
[tex]v = 9.8 \ m/s[/tex]
Location C is 0.021 m from a small sphere that has a charge of 5 nC uniformly distributed on its surface. Location D is 0.055 m from the sphere. What is the change in potential along a path from C to D?
Answer:
ΔV = -1321.73V
Explanation:
The change in potential along the path from C to D is given by the following expression:
[tex]\Delta V=-\int_a^bE dr[/tex] (1)
E: electric field produced by a charge at a distance of r
a: distance to the sphere at position C = 0.021m
b: distance to the sphere at position D = 0.055m
The electric field is given by:
[tex]E=k\frac{Q}{r^2}[/tex] (2)
Q: charge of the sphere = 5nC = 5*10^-9C
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
You replace the expression (2) into the equation (1) and solve the integral:
[tex]\Delta V=-kQ\int_a^b \frac{dr}{r^2}=-kQ[-\frac{1}{r}]_a^b[/tex] (3)
You replace the values of a and b:
[tex]\Delta V=(8.98*10^9Nm^2/C^2)(5*10^{-9}C)[\frac{1}{0.055m}-\frac{1}{0.021m}]\\\\\Delta V=-1321.73V[/tex]
The change in the potential along the path C-D is -1321.73V
What is the on ohooke benden
er ord power
What is the main difference between work, power and energy
Answer:Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.
An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination of the flight. The plane flies with an air speed of 120 m/s. If a constant wind blows at 10.0 m/s due west during the flight, what direction must the plane fly relative to north to arrive at the destination? Consider: east to the right, west to the left, north upwards and south downwards
Answer:
θ = 4.78º
with respect to the vertical or 4.78 to the east - north
Explanation:
This is a velocity compound exercise since it is a vector quantity.
The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed
v_fly² = v_nort² + v_air²
v_nort² = v_fly² + - v_air²
Let's use trigonometry to find the direction of the plane
sin θ = v_air / v_fly
θ = sin⁻¹ (v_air / v_fly)
let's calculate
θ = sin⁻¹ (10/120)
θ = 4.78º
with respect to the vertical or 4.78 to the north-east
How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for a slit width of 0.110 mm, using light of wavelength 582 nm?
Answer:
6
Explanation:
We are given that
[tex]\theta=2.12^{\circ}[/tex]
Slid width,a=0.110 mm=[tex]0.11\times 10^{-3} m[/tex]
[tex]1mm=10^{-3} m[/tex]
Wavelength,[tex]\lambda=582 nm=582\times 10^{-9}[/tex] m
[tex]1nm=10^{-9} m[/tex]
We have to find the number of diffraction maxima are contained in a region of the Fraunhofer single-slit pattern.
[tex]asin\theta=\frac{2N+1}{2}\lambda[/tex]
Using the formula
[tex]0.11\times 10^{-3}sin(2.12)=\frac{2N+1}{2}(582\times 10^{-9})[/tex]
[tex]2N+1=\frac{0.11\times 10^{-3}sin(2.12)\times 2}{582\times 10^{-9}}[/tex]
[tex]2N+1=13.98[/tex]
[tex]2N=13.98-1=12.98[/tex]
[tex]N=\frac{12.98}{2}\approx 6[/tex]
Hence, 6 diffraction maxima are contained in a region of the Fraunhofer single-slit pattern
Sophie throws a tennis ball down from a height of 1.5 m at an angle of 450 with respect to vertical. She drops another tennis ball from the same height. Use the Energy Interaction Model to predict which ball will hit the ground with greater speed.
Given that,
Height =1.5 m
Angle = 45°
We need to find the greater speed of the ball
Using conservation of energy
[tex]P.E_{i}+K.E_{f}=P.E_{f}+K.E_{f}[/tex]
[tex]mgh+\dfrac{1}{2}mv_{i}^2=mgh+\dfrac{1}{2}mv_{f}^2[/tex]
Here, initial velocity and final potential energy is zero.
[tex]mgh=\dfrac{1}{2}mv_{f}^2[/tex]
Put the value into the formula
[tex]9.8\times1.5=\dfrac{1}{2}v_{f}^2[/tex]
[tex]v_{f}^2=2\times9.8\times1.5[/tex]
[tex]v_{f}=\sqrt{2\times9.8\times1.5}[/tex]
[tex]v_{f}=5.42\ m/s[/tex]
Hence, the greater speed of the ball is 5.42 m/s.
From a height of 40.0 m, a 1.00 kg bird dives (from rest) into a small fish tank containing 50.5 kg of water. Part A What is the maximum rise in temperature of the water if the bird gives it all of its mechanical energy
Answer:
0.00185 °C
Explanation:
From the question,
The potential energy of the bird = heat gained by the water in the fish tank.
mgh = cm'(Δt)................... Equation 1
Where m = mass of the bird, g = acceleration due to gravity, h = height, c = specific heat capacity of water, m' = mass of water, Δt = rise in temperature of water.
make Δt the subject of the equation
Δt = mgh/cm'............... Equation 2
Given: m = 1 kg, h = 40 m, m' = 50.5 kg
constant: g = 9.8 m/s², c = 4200 J/kg.K
Substitute into equation 2
Δt = 1(40)(9.8)/(50.5×4200)
Δt = 392/212100
Δt = 0.00185 °C
Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12.0-V battery, the current from the battery is 1.78 A. When the resistors are connected in parallel to the battery, the total current from the battery is 11.3 A. Determine the two resistances.
Answer:
R(smaller) = 1.3 Ω and R(larger) = 5.4 Ω
Explanation:
Ohm's Law states that:
V = IR
R = V/I
where,
R = Resistance
V = Potential Difference
I = Current
Therefore, for series connection:
Rs = Vs/Is
where,
Rs = Resistance when connected in series = R(smaller) + R(larger)
Vs = Potential Difference when connected in series = 12 V
Is = Current when connected in series = 1.78 A
Therefore,
R(smaller) + R(larger) = 12 V/1.78 A
R(smaller) + R(larger) = 6.74 Ω --------------- equation 1
R(smaller) = 6.74 Ω - R(larger) --------------- equation 2
Therefore, for series connection:
Rp = Vp/Ip
where,
Rp = Resistance when connected in parallel = [1/R(smaller) + 1/R(larger)]⁻¹
Rp = [{R(smaller) + R(larger)}/{R(smaller).R(larger)]⁻¹
Rp = R(smaller).R(larger)/[R(smaller) + R(larger)]
Vp = Potential Difference when connected in parallel = 12 V
Ip = Current when connected in parallel = 11.3 A
Therefore,
R(smaller).R(larger)/[R(smaller) + R(larger)] = 12 V/11.3 A
using equation 1 and equation 2, we get:
[6.74 Ω - R(larger)].R(larger)/6.74 Ω = 1.06 Ω
6.74 R(larger) - R(larger)² = (6.74)(1.06)
R(larger)² - 6.74 R(larger) + 7.16 = 0
solving this quadratic equation we get:
R(larger) = 5.4 Ω (OR) R(larger) = 1.3 Ω
using these values in equation 2, we get:
R(smaller) = 1.3 Ω (OR) R(smaller) = 5.4 Ω
Since, it is given in the question that R(smaller)<R(larger).
Therefore, the correct answers will be:
R(smaller) = 1.3 Ω and R(larger) = 5.4 Ω
A flat roof is very susceptible to wind damage during a thunderstorm and/or tornado. If a flat roof has an area of 500 m2 and winds of speed 39.0 m/s blow across it, determine the magnitude of the force exerted on the roof. The density of air is 1.29 kg/m3.
Answer: The magnitude of the force exerted on the roof is 490522.5 N.
Explanation:
The given data is as follows.
Below the roof, [tex]v_{1}[/tex] = 0 m/s
At top of the roof, [tex]v_{2}[/tex] = 39 m/s
We assume that [tex]P_{1}[/tex] is the pressure at lower surface of the roof and [tex]P_{2}[/tex] be the pressure at upper surface of the roof.
Now, according to Bernoulli's theorem,
[tex]P_{1} + 0.5 \times \rho \times v^{2}_{1} = P_{2} \times 0.5 \rho \times v^{2}_{2}[/tex]
[tex]P_{1} - P_{2} = 0.5 \times \rho \times (v^{2}_{2} - v^{2}_{1})[/tex]
= [tex]0.5 \times 1.29 \times [(39)^{2} - (0)^{2}][/tex]
= [tex]0.645 \times 1521[/tex]
= 981.045 Pa
Formula for net upward force of air exerted on the roof is as follows.
F = [tex](P_{1} - P_{2})A[/tex]
= [tex]981.045 \times 500[/tex]
= 490522.5 N
Therefore, we can conclude that the magnitude of the force exerted on the roof is 490522.5 N.
A car travels 2500 m in 8 minutes. Calculate the speed at which the car travelled
Answer:
5.95m/s to 2 decimal places
Explanation:
In physics speed is measured in metres per second so convert 8mins to seconds
8x60=420 seconds
The formula needed:
Speed (m/s)= Distance (m)/Time (s)
2500/420=5.95m/s
A wire with mass 90.0g is stretched so that its ends are tied down at points 88.0cm apart. The wire vibrates in its fundamental mode with frequency 80.0Hz and with an amplitude of 0.600cm at the antinodes.a) What is the speed of propagation of transverse waves in the wire?b) Compute the tension in the wire.
Answer:
a) V = 140.8 m/s
b) T = 2027.52 N = 2.03 KN
Explanation:
a)
The formula for the speed of the wave is given as follows:
f₁ = V/2L
V = 2f₁L
where,
V = Speed of Wave = ?
f₁ = Fundamental Frequency = 80 Hz
L = Length of Wire = 88 cm = 0.88 m
Therefore,
V = (2)(80 Hz)(0.88 m)
V = 140.8 m/s
b)
Another formula for the speed of wave is:
V = √T/μ
V² = T/μ
T = V²μ
where,
T = Tension in String = ?
μ = Linear Mass Density of Wire = Mass of Wire/L = 0.09 kg/0.88 m
μ = 0.1 kg/m
Therefore,
T = (140.8 m/s)²(0.1 kg/m)
T = 2027.52 N = 2.03 KN
A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnitude of the impulse exerted on the ball by the floor
Answer:
9.6 Ns
Explanation:
Note: From newton's second law of motion,
Impulse = change in momentum
I = m(v-u).................. Equation 1
Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.
Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)
Substitute into equation 1
I = 2.4[2.5-(-1.5)]
I = 2.4(2.5+1.5)
I = 2.4(4)
I = 9.6 Ns
The magnitude of impulse will be "9.6 Ns".
According to the question,
Mass,
m = 2.4 kgFinal velocity,
v = 2.5 m/sInitial velocity,
u = -1.5 m/sBy using Newton's 2nd law of motion, we get
→ Impulse, [tex]I = m(v-u)[/tex]
By substituting the values, we get
[tex]= 2.4[2.5-(1.5)][/tex]
[tex]= 2.4(2.5+1.5)[/tex]
[tex]= 2.4\times 4[/tex]
[tex]= 9.6 \ Ns[/tex]
Thus the above answer is right.
Learn more about Impulse here:
https://brainly.com/question/15495020