Answer:
30J
Explanation:
Given parameters:
Mass of hamster = 0.104kg
Velocity = 24m/s
Unknown:
Kinetic energy = ?
Solution:
Kinetic energy is the energy due to the motion of a body. It is mathematically derived by;
Kinetic energy = [tex]\frac{1}{2}[/tex] m v²
m is the mass
v is the velocity
Kinetic energy = [tex]\frac{1}{2}[/tex] x 0.104 x 24² = 30J
Which element has the same number of valence electrons as krypton (Kr)?
fluorine (F)
chlorine (Cl)
neon (Ne)
selenium (Se)
The element that has the same number of valence electrons as krypton (Kr) is known as Neon (Ne). Thus, the correct option for this question is C.
What are Valence electrons?Valence electrons may be defined as the number of electrons present in the outermost shell of any atom during the time of electronic configuration. For example, Sodium has an atomic number of 11, its electronic configuration is 2, 8, and 1. It means that the valence electron on sodium is 1.
Neon and Krypton being the members of the same family which is known as noble gases or inert gases occupy the same valence electrons of eight in their outermost shell. It is group 8A of the periodic table. Apart from Ne, and Kr, other elements like argon, Xenon, and Radon also belong to the same family and occupy the same valence electron except for helium.
Therefore, the element that has the same number of valence electrons as krypton (Kr) is known as Neon (Ne). Thus, the correct option for this question is C.
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A turbofan operates at 25,000 ft and moves at 815 ft/s. It ingests 1.2 times the amount of air into the fan than into the core, which all exits through the fan exhaust. The fuel-flow-to-core airflow ratio is 0.0255. The exit densities of the fan and core are 0.00154 and 0.000578 slugs/ft3, respe~tively. The exit pressures from the fan and core are 10.07 and 10.26 psia, respectively. The developed thrust is 10,580 !bf, and the exhaust velocities from the fan and core are 1147 and 1852 ft/s, respectively. (a) Find the ingested air mass flow rate for the core and TSFC. (b) What are the exit areas of the fan and core nozzles
Answer:
a)
Mass flow rate of core = [tex]m_{e}[/tex] = 60.94 Kg/s
Mass flow rate of fan = [tex]m_{s}[/tex] = 73.12 kg/s
TSFC = 3.301 x [tex]10^{-5}[/tex]
b)
Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]
Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]
Explanation:
Data Given:
Height = 25000 ft
Vehicle velocity = [tex]u_{a}[/tex] = 815 ft/s = 248.41 m/s
[tex]m_{s} = 1.2m_{e}[/tex]
[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]
Where,
[tex]m_{s}[/tex] = Mass flow rate of fan
[tex]m_{e}[/tex] = Mass flow rate of core
F = Thrust
Density of core = [tex]D_{e}[/tex] = 0.000578 slugs/[tex]ft^{3}[/tex] = 0.2979 kg/[tex]m^{3}[/tex]
Density of fan = [tex]D_{s}[/tex] = 0.00154 slugs/[tex]ft^{2}[/tex] = 0.7937 kg/[tex]m^{3}[/tex]
Ambient Pressure of Fan = [tex]P_{s}[/tex] = 10.07 Psi = 69430.21 Pa
Ambient Pressure of core = [tex]P_{e}[/tex] = 10.26 Psi = 70740.2 Pa
Thrust = F = 10580 lbf = 47062.2 N
Velocity of fan = [tex]u_{s}[/tex] = 1147 ft/s = 349.6 m/s
Velocity of core = [tex]u_{e}[/tex] = 1852 ft/s = 564.5 m/s
At the height of 25000 ft, P = 37600 [tex]P_{a}[/tex]
Now,
we have:
[tex]m_{e}[/tex] = [tex]u_{e}[/tex] x [tex]D_{e}[/tex] x [tex]A_{e}[/tex]
Plugging in the values, we get:
[tex]m_{e}[/tex] = 168.16 [tex]A_{e}[/tex] Equation 1
And,
[tex]m_{s}[/tex] = [tex]D_{s}[/tex] x [tex]A_{s}[/tex] x [tex]u_{s}[/tex]
[tex]m_{s}[/tex] = 277.5 [tex]A_{s}[/tex] Equation 2
As, we know,
[tex]m_{s} = 1.2m_{e}[/tex]
[tex]m_{s}[/tex] = 277.5 [tex]A_{s}[/tex]
And now for Thrust, we have:
F = [tex]A_{e}[/tex] x ([tex]P_{e}[/tex] - [tex]P_{a}[/tex] ) + [tex]A_{s}[/tex] x ([tex]P_{s}[/tex] - [tex]P_{a}[/tex] ) + [tex]m_{e}[/tex]x ([tex]u_{e}[/tex] - [tex]u_{a}[/tex] ) + [tex]m_{s}[/tex] x ([tex]u_{s}[/tex] - [tex]u_{a}[/tex] ) Equation 3
Now, substitute equation 1 and 2 in equation 3, we get:
Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]
Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]
Mass flow rate of core = [tex]m_{e}[/tex] = 60.94 Kg/s
Mass flow rate of fan = [tex]m_{s}[/tex] = 73.12 kg/s
TSFC = Thrust Specific Fuel Consumption = fuel mass flow rate / Thrust
TSFC = [tex]m_{f}[/tex]/F
And,
[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]
[tex]m_{e}[/tex] = 60.94
[tex]m_{f}[/tex] = 0.0255 x 60.94
[tex]m_{f}[/tex] = 1.55397
TSFC = [tex]m_{f}[/tex]/F
TSFC = 1.55397/47062.2
TSFC = 3.301 x [tex]10^{-5}[/tex]
Low TSFC = High efficiency
High TSFC = Low efficiency
a)
Mass flow rate of core = [tex]m_{e}[/tex] = 60.94 Kg/s
Mass flow rate of fan = [tex]m_{s}[/tex] = 73.12 kg/s
TSFC = 3.301 x [tex]10^{-5}[/tex]
b)
Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]
Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]
Cite an example of an issue showing the power of Media and information to affect change? sentence pls :)
Answer:
The influence of Media and Information over elections, politics and, governance.
Explanation:
Once we used to have politicians travel to interact with citizens and meet them face to face. That is slowly fading away in place of townhalls that take place online.
Video conferencing, video calls, social media apps, and technology now make it easy to connect with a lot of people at once, across various timezones, without changing location.
Once upon a time, the Television media, Radio Media, Cable Television (all categorized as traditional media) used to be the custodian of information, hence, the chief influencers of politics and governance.
Today, the internet has made it possible for individuals or a group of people without much tech-spend to set up and garner a following that has enough power to influence governance.
Online petitions and Wikileaks, the proliferation of terrorist organizations that have strong information and technology as well as social media skills have modified forever, the way public office holders and other stakeholders involved used to look at governance.
How does the magnetic force move throughout the magnet
Answer:
its always active
Explanation:
If a cyclist travels 30 km in 2 h, What is her average speed?
Answer:
15km/h
Explanation:
→ Speed = Distance ÷ Time
30 ÷ 2 = 15km/h
Sam heaves a 16lb shot straight upward, giving it a constant upward acceleration from rest of 35 m/s^2 for 64.0 cm. He releases it 2.20m above the ground. You may ignore air resistance.
(a) What is the speed of the shot when Sam releases it?
(b) How high above the ground does it go?
(c ) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?
Answer:
6.69 m/s
4.483 m
1.42s
Explanation:
Given that:
Initial Velocity, u = 0
Final velocity, v =?
Acceleration, a = 35m/s²
1.) using the relation :
v² = u² + 2as
v² = 0 + 2(35) * 64*10^-2m
v² = 70 * 0.64
v = sqrt(44.8)
v = 6.693
v = 6.69 m/s
B.) height from the ground, h0 = 2.2
How high ball went , h:
Using :
v² = u² + 2as
Upward motion, g = - ve
0 = 6.69² + 2(-9.8)*(h - 2.2)
0= 6.69² - 19.6(h - 2.2)
44.7561 + 43.12 - 19.6h = 0
19.6h = 44.7561 - 43.12
h = 87.8761 / 19.6
h = 4.483 m
C.)
vt - 0.5gt² = h - h0
6.69t - 0.5(9.8)t²
6.69t - 4.9t² = 1.83 - 2.2
-4.9t² + 6.69t + 0.37 = 0
Using the quadratic equation solver :
Taking the positive root:
1.4185 = 1.42s
kid shoots a BB gun directly upward at a ock of birds. The initial velocity of the BB is21 m/s. If the BB hits a bird at less than 1 m/s then it will not harm the bird. Calculate the minimumheight above the gun at which the birds may safely y. Use a magnitude of acceleration of 9.8m
Answer:
22.5 m
Explanation:
Using v² = u² - 2gy where u = initial velocity of BB = 21 m/s, v = final velocity of BB = 1 m/s (since this is the required speed of BB in which it will not harm the birds), g = acceleration due to gravity = 9.8 m/s² and y = minimum height of BB above the gun at which the birds may safely fly.
Substituting the values of the variables into the equation, we have
v² = u² - 2gy
(1 m/s)² = (21 m/s)² - 2(9.8 m/s²)y
collecting like terms, we have
(1 m/s)² - (21 m/s)² = - 2(9.8 m/s²)y
1 m²/s² - 441 m²/s² = -(19.6 m/s²)y
simplifying, we have
- 440 m²/s² = -(19.6 m/s²)y
dividing through by -19.6 m/s², we have
y = - 440 m²/s² ÷ -19.6 m/s²
y = 22.45 m
y ≅ 22.5 m
One of the disadvantages of experimental research is that __________.
A.
it isn’t easily replicated
B.
it doesn’t often reflect reality
C.
the results aren’t generalizable
D.
conditions are not controllable
Please select the best answer from the choices provided
Answer:
B
Explanation:
An air-track glider attached to a spring oscillates between the 10 cm mark and the 60 cm mark on the track. The glider completes 10 oscillations in 33 s. What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider
Answer:
a) Time period is 3.3 seconds
b) The frequency is 0.3030 Hz
c) amplitude is 0.25 m
d) maximum speed is 0.476 m/s
Explanation:
Given the data in the question;
a) Period
Time Period T = Time taken for one oscillation
T = 33s / 10 = 3.3 seconds
Therefore, Time period is 3.3 seconds
b) Frequency
we know that frequency is the inverse of time period
so;
Frequency f = 1/T = 1 / 3.3 s
Frequency f = 0.3030 Hz
Therefore, The frequency is 0.3030 Hz
c) amplitude
amplitude A = [tex]\frac{1}{2}[/tex]( 60 cm - 10 cm )
A = [tex]\frac{1}{2}[/tex] × 50 cm
A = 25 cm
A = 0.25 m
Therefore, amplitude is 0.25 m
d) maximum speed of the glider
maximum speed [tex]V_{max}[/tex] = ωA
and ω = 2π/T
so maximum speed [tex]V_{max}[/tex] = [tex]\frac{2\pi }{T}[/tex]A
so we substitute
so maximum speed [tex]V_{max}[/tex] = [tex]\frac{2\pi }{3.3}[/tex] × 0.25 m
so maximum speed [tex]V_{max}[/tex] = 0.476 m/s
Therefore, maximum speed is 0.476 m/s
The time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.
Based on the given information,
• The air-track glider connected with a spring oscillates between the 10 cm mark and the 60 cm mark.
• In 33 seconds, the glider completes 10 oscillations.
There is a need to find, the period, frequency, amplitude, and maximum speed of the glider.
a) Time period (T) for one oscillation is,
[tex]\frac{33s}{10} = 3.3 s[/tex]
b) The frequency (f) is the reciprocal of the time period,
[tex]f = \frac{1}{T} =\frac{1}{3.3S} = 0.303 Hz[/tex]
c) The amplitude (A) is,
[tex]A = \frac{1}{2} (60 cm-10cm) = 25cm[/tex]
[tex]A = 0.25 m[/tex]
d) Maximum speed of the glider is,
[tex]Vmax = \frac{2\pi }{T} (0.25 m)[/tex]
[tex]Vmax = 0.47575 m/s[/tex]
Thus, time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.
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a car accidently rolls off a cliff. as it leaves the cliff it has a horizontal velocity of 13m/s it hits the ground 60m from the shoreline. calculate the height of the cliff
Answer:
104.59 m
Explanation:
From the question given above, the following data were obtained:
Horizontal velocity (u) = 13 m/s
Horizontal distance (s) = 60 m
Height of cliff =?
Next, we shall determine the time taken for the car to hit the ground. This can be obtained as follow:
Horizontal velocity (u) = 13 m/s
Horizontal distance (s) = 60 m
Time (t) =?
s = ut
60 = 13 × t
Divide both side by 13
t = 60 /13
t = 4.62 s
Finally, we shall determine the height of the cliff. This can be obtained as follow:
Time (t) = 4.62 s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) =?
h = ½gt²
h = ½ × 9.8 × 4.62²
h = 4.9 × 21.3444
h = 104.59 m
The, the height of the cliff is 104.59 m
Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.
a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?
Answer:
a) ω₁ = ω₂ = 3.7 rad/sec
b) Δθ₁ = Δθ₂ = 18.5 rad
c) d₁ = 14.5 m d₂ = 57.5 m
d) Fc1 = 273.9 N Fc2 = 1069.8 N
e) The boy near the outer edge.
Explanation:
a)
Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.However, linear speeds of points at different distances from the center, are different.Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:[tex]v = \omega*r (1)[/tex]
Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:[tex]\omega = \frac{v_{out} }{r_{out} } = \frac{11.5m/s}{3.14m} = 3.7 rad/sec (2)[/tex]
As we have already said, ωout = ωin = 3.7 rad/secb)
Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:[tex]\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)[/tex]⇒ Δθ₁ = Δθ₂ = 18.5 rad.
c)
The linear distance traveled by each child, will be related with the linear speed of them.Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:[tex]v_{inn} = \omega * r_{inn} = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)[/tex]
vout is a given of the problem ⇒ vout = 11. 5 m/s
Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:[tex]d_{inn} = v_{inn} * t = 2.9m/s* 5.0 s = 14.5 m (5)[/tex]
[tex]d_{out} = v_{out} * t = 11.5 m/s* 5.0 s = 57.5 m (6)[/tex]
d)
The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:[tex]F_{c} = m*\frac{v^{2}}{r} (7)[/tex]
Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:[tex]F_{cin} = m*\frac{v_{in}^{2}}{r_{in}} = 25.4 kg* \frac{(2.9m/s)^{2} }{0.78m} = 273.9 N (8)[/tex]
In the same way, we get Fcout (the force on the boy near the outer edge):[tex]F_{cout} = m*\frac{v_{out}^{2}}{r_{out}} = 25.4 kg* \frac{(11.5m/s)^{2} }{3.14m} = 1069.8 N (9)[/tex]
e)
The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.The maximum friction force is given by the product of the coefficient of static friction times the normal force.Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.As both boys have the same mass, the normal force is also equal.This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:[tex]F_{frs} = \mu_{s} * m* g (10)[/tex]If this force is greater than the centripetal force, the boy will be able to hold on.So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.An RC car is carrying a tiny slingshot with a spring constant of 85 N/m at 0.2 m off the ground at 5.6 m/s. The sling shot is pulled back 3.5 cm from a relaxed state and shoots a 25 g steel pellet in the same direction the car is moving. What is the velocity of the steel pellet relative to the ground as it leaves the sling shot
Answer:
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
Explanation:
Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:
[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]
[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]
[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)
Where:
[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.
[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.
[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.
[tex]x[/tex] - Initial deformation of the spring, measured in meters.
If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:
[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]
[tex]v \approx 5.960\,\frac{m}{s}[/tex]
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
A child uses one hand to charge a balloon by rubbing it against her shirt. She then holds a rod in her other hand and finds that the balloon and rod, when brought close to one another, repel. Which one of the following is true?
a. the rod must be a conductor
b. the rod must be an insulator
c. it could be either
Answer:
B. The rod must be an insulator
Explanation:
We have that this rod must be an insulator. After this child rubbed the balloon, the balloon acquired static charge. So holding a rid against it is going to cause it to repel, this is to say it is repelling because the rod also is carrying some static charges. If this rid was a conductor, there would be no charge in its surface. The charge would have passed through her hand as it comes in contact.
What industry uses extremely large equipment?
O mining
O harvesting
O food packaging
O welding
Answer:
mining
Explanation:
because they use the most heavy duty machines
A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Answer:
cos 0 = 1.
Fs = 7×8 = 56 J
Explanation:
I NEED HELP please answer it
A wheel starts from rest and has an angular acceleration that is given by α (t) = (6.0 rad/s4)t2. After it has turned through 10 rev its angular velocity is:
Answer:
75 rad/s
Explanation:
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)]
Hence: ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dtdt
Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:
θ(t) = ∫∫α(t) dtdt
θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³
ω(3.348) = 2(3.348)³ = 75 rad/s
The required value of angular velocity of wheel is 75 rad/s.
Given data:
The angular acceleration of the wheel is, [tex]\alpha (t) = 6.0 t^{2} \;\rm rad/s^{4}[/tex].
The turning rate of wheel is, n = 10 rev.
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)
Clearly, the angular velocity is the single integral of angular acceleration. Then,
ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dt dt
Since,
α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴.
Then solve by substituting the values as,
θ(t) = ∫∫α(t) dt dt
θ(t) = ∫∫6t² dt dt
θ(t) =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
Also,
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³ ......................................................(1)
Substitute the value of time in equation (1) as,
ω( t = 3.348) = 2(3.348)³ = 75 rad/s
Thus, we can conclude that the required value of angular velocity of wheel is 75 rad/s.
Learn more about the angular velocity here:
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A hi-lo lifts an 25 N skid to top of a pallet rack. The pallet rack is 3.6 meters tall. The hi-lo takes 12 seconds to get the skid on top. Calculate the power output of the hi-lo.
Answer:
7.5Watts
Explanation:
Given parameters:
Force of lift = 25N
Height = 3.6m
Time = 12s
Unknown:
Power output = ?
Solution:
Power is the rate at which work is done ;
Power = [tex]\frac{force x height }{time}[/tex]
Power = [tex]\frac{25 x 3.6}{12}[/tex] = 7.5Watts
Suppose it takes 1.4 s for the monkey to catch the coconut and the initial upward speed of the coconut is 2.9 m/s. Assume the acceleration of gravity is 9.8 m/s 2 . Determine the y coordinate of the location where the monkey catches the coconut. Answer in units of m.
Answer:
y-coordinate where the monkey catches the coconut is 13.664 m.
Explanation:
Given;
time taken for the monkey to catch the coconut, t = 1.4 s
initial upward speed of the coconut, Uy = 2.9 m/s
acceleration due to gravity, g = 9.8 m/s²
The y-coordinate where the monkey catches the coconut is calculated as;
[tex]h_y = U_yt +\frac{1}{2} gt^2\\\\h_y = (2.9\times 1.4) +\frac{1}{2} (9.8)(1.4^2)\\\\h_y = 4.06 + 9.604\\\\h_y = 13.664 \ m[/tex]
Therefore, y-coordinate where the monkey catches the coconut is 13.664 m.
Briefly describe factors and characteristics that influence conflict resolution.
Answer:
Personal attitudes play a role in conflict resolution and include prejudices, biases, prior experience, and the level of importance of a topic to the parties involved. It is important to note that you can control your own attitude, but you cannot control the attitudes of others; it is necessary to focus on your attitudes and assumptions before you address the other party's. Next, you need to know what the nature of the conflict is, or what the conflict is about. It is necessary to know how severe the conflict is, how easily it can be resolved, how much sacrifice it will take to overcome, and how heated it is likely to get. Lastly, one needs to understand the capabilities of the people involved. Is everyone on the same level and do they have what they need to work on the conflict, i.e. money, time, and support?
Explanation:
Answer:
The guy above is correct I'm just trying to make the post more reliable if two people say it then it's good plus I checked also Thanks other guy.
Explanation:
CAN YOU GUYS PLEASE ANSWER THIS QUICKLY THIS IS DUE IN AN HOUR AND IM GETTING WORRIED
I'm pretty sure I the third option C.
Explanation:
sorry if I'm wrong
A 2028 kg Oldsmobile traveling south on Abbott Road at 14.5 m/s is unable to stop on the ice covered intersection for a red light at Saginaw Street. The car collides with a 4146 kg truck hauling animal feed east on Saginaw at 9.7 m/s. The two vehicles remain locked together after the impact. Calculate the velocity of the wreckage immediately after the impact. Give the speed for your first answer and the compass heading for your second answer. (remember, the CAPA abbreviation for degrees is deg) -1.75
Answer:
v = 8.1 m/s
θ = -36.4º (36.4º South of East).
Explanation:
Assuming no external forces acting during the collision (due to the infinitesimal collision time) total momentum must be conserved.Since momentum is a vector, if we project it along two axes perpendicular each other, like the N-S axis (y-axis, positive aiming to the north) and W-E axis (x-axis, positive aiming to the east), momentum must be conserved for these components also.Since the collision is inelastic, we can write these two equations for the momentum conservation, for the x- and the y-axes:We can go with the x-axis first:[tex]p_{ox} = p_{fx} (1)[/tex]
⇒ [tex]m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)[/tex]
Replacing by the givens, we can find vfx as follows:[tex]v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)[/tex]
We can repeat the process for the y-axis:[tex]p_{oy} = p_{fy} (4)[/tex]
⇒[tex]m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)[/tex]
Replacing by the givens, we can find vfy as follows:[tex]v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)[/tex]
The magnitude of the velocity vector of the wreckage immediately after the impact, can be found applying the Pythagorean Theorem to vfx and vfy, as follows:[tex]v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)[/tex]
In order to get the compass heading, we can apply the definition of tangent, as follows:[tex]\frac{v_{fy} }{v_{fx} } = tg \theta (8)[/tex]
⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)
⇒ θ = tg⁻¹ (-0.738) = -36.4º
Since it's negative, it's counted clockwise from the positive x-axis, so this means that it's 36.4º South of East.A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 6.5 w/m^2 at a distance of 2.5 m from the speaker. At what rate does this speaker produce energy?
Answer:
the rate at which the speaker produces energy is 510.51 J/s
Explanation:
Given;
sound intensity of the speaker, I = 6.5 W/m²
distance, r = 2.5 m
The rate at which the speaker produces energy on a spherical surface is given as;
P = IA
P = I(4πr²)
P = 6.5 (W/m²) x 4π x (2.5 m)²
P = 510.51 J/s
Therefore, the rate at which the speaker produces energy is 510.51 J/s
A primary coil has 360 turns, while secondary has 120 turns in a transformer. The output voltage is 220V. What is the input voltage, and what type of transformer is it
Answer:
550V
Step - Down transformer
Explanation:
Given parameters:
Number of turns in primary coil = 300 turns
Secondary turns = 120 turns
Output voltage = 220V
Unknown:
Input voltage = ?
Type of transformer = ?
Solution:
To solve this problem, we use the expression below:
[tex]\frac{V_{out} }{V_{in} }[/tex] = [tex]\frac{Ns}{Np}[/tex]
So insert the parameters and find Vin;
[tex]\frac{220}{Vin}[/tex] = [tex]\frac{120}{300}[/tex]
120Vin = 220 x 300
Vin = [tex]\frac{220 x 300}{120}[/tex] = 550V
Since the input voltage is greater than the output voltage, this is step - down transformer.
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Answer:
it will option option A hope it helps
Question 7 of 25
A sound wave with a wavelength of 26 m passes through glass, and then it
passes into air, where its wavelength changes to 1.6 m. If the sound wave has
a constant frequency of 220 Hz, what is its approximate speed as it passes
through the air? (The equation for the speed of a wave is v= f *1.)
O A. 3,500 m/s
B. 350 m/s
O C. 5,700 m/s
O D. 570 m/s
Answer: 350 ms
Explanation:
Just took the quiz:)
Answer:
B. 350
Explanation:
Magnification of lens is 1. What does it mean?
Answer:
It means when you look into the lens your vision magnifies by x1
Explanation:
All types of mass movement are caused by the force of
Answer:
Mass and Acceleration
Explanation:
The typical Force equation is:
F = ma
where m = mass, and a=acceleration.
Answer:
Gravity is the main force responsible for mass movements. Gravity is a force that acts everywhere on the Earth's surface, pulling everything in a direction toward the center of the Earth
Explanation:
The _______ changes light energy into nerve signals using receptors called rods and cones. A. retina B. lens C. iris D. pupil
Answer:
A. Retina
Explanation:
In which number are the zeros not significant?
100.0
O 0.0003
O 4.00005
O 1.0004
Answer:
0.0003
Explanation:
In the rules of Sig Figs, all zeros before with decimals are not sigificant. I.E. 0.00000000000000009. Despite how many 0's there are, only the 9 is significant. Zeros before a number is not significant. In 100, only the one is signficant in 100. with a dot at the end, the one and the two zeros are significant. hope this helps.
Answers:
the second option
Explanation: