Answer:ew
Explanation:
qeeqw
Choose one. 5 points
Use the equation from week 3:
frequency =
wavespeed
wavelength
and the wavelength you found in #3 to calculate the frequency of this photon (remember the speed of
light is 3E8 m/s);
7.6E14 Hz
6.0E14 Hz
4,6E14 Hz
The frequency is 4,6E14 Hz.
What is the frequency?
Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.
Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.
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What is the maximum speed at which a car can round a curve of 25-m radius on a level road if the coefficient of static friction between the tires and road is 0.80?
I assume the curve is flat and not banked. A car making a turn on the curve has 3 forces acting on it:
• its weight, mg, pulling it downward
• the normal force from contact with the road, n, pushing upward
• static friction, f = µn, directed toward the center of the curve (where µ is the coefficient of static friction)
By Newton's second law, the net forces on the car in either the vertical or horizontal directions are
∑ F (vertical) = n - mg = 0
∑ F (horizontal) = f = ma
where a is the car's centripetal acceleration, given by
a = v ²/r
and where v is the maximum speed you want to find and r = 25 m.
From the first equation, we have n = mg, and so f = µmg. Then in the second equation, we have
µmg = mv ²/r ==> v ² = µgr ==> v = √(µgr )
So the maximum speed at which the car can make the turn without sliding off the road is
v = √(0.80 (9.80 m/s²) (25 m)) = 14 m/s
A rectangular loop of wire with sides 0.129 and 0.402 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field has a magnitude of 0.888 T and is directed parallel to the normal of the loop's surface. In a time of 0.172 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop.
Answer:
[tex]0.2677\ \text{V/m}[/tex]
Explanation:
A = Area of loop = [tex]0.129\times0.402[/tex]
B = Magnetic field = [tex]0.888\ \text{T}[/tex]
t = Time taken = [tex]0.172\ \text{s}[/tex]
Electric field is given by
[tex]E=B\dfrac{dA}{dt}\\\Rightarrow E=0.888\times\dfrac{0.129\times 0.402}{0.172}\\\Rightarrow E=0.2677\ \text{V/m}[/tex]
The emf induced is [tex]0.2677\ \text{V/m}[/tex].
A projectile is launched straight upwards at 75 m/s. Three seconds later, its velocity is...?
Answer:
V = V0 + a t
V = 75 - 9.8 * 3 = 45.6 m/s
The final velocity of the projectile after 3 seconds is equal to 45.6 m/s.
What is the equation of motion?The equations of motion can be defined as the relation of the motion of a physical system as the function of time and set up the relationship between the displacement (s), acceleration, velocity (v & u), and time of a moving system.
Given, the initial velocity of the projectile, u = 75 m/s
The time taken by the projectile, t = 3 sec
The acceleration due to gravity upward, g = - 9.8 m/s²
From the first equation of motion we can calculate the final velocity of the projectile:
v = u + at
v = u - gt
v = 75 - 9.8 ×(3)
v = 75 - 29.4
v = 45.6 m/s
Therefore, the final velocity of the projectile after three seconds is 45.6 m/s.
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Four identical balls are thrown from the top of a cliff, each with the same speed. The
first is thrown straight up, the second is thrown at 30° above the horizontal, the third
at 30° below the horizontal, and the fourth straight down. How do the speeds and
kinetic energies of the balls compare as they strike the ground? Ignore the effects of
air resistance. Explain fully using the concepts from this unit.
The comparison of the speeds and kinetic energy of the identical balls are as follows
The speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
The reason for the above comparison results areas follows;
Known parameters;
First ball is thrown straight up
Second ball is thrown 30° above the horizontal
Third ball it thrown 30° below the horizontal
The fourth ball is thrown straight down
Unknown:
Comparison of the speed and kinetic energy of the four balls
Method:
The kinetic energy, K.E. = (1/2) × m × v²
The velocity of the ball, v = u × sin(θ)
Where;
u = The initial velocity of the ball
θ = The reference angle
For the first ball thrown straight up, we have;
θ = 90°
∴ [tex]v_y[/tex] = u
The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh
Where;
h = The height of the cliff
∴ Kinetic energy of first ball, K.E.₁ = (1/2) × m × (u₁² + 2gh)²
For the second ball thrown 30° to the horizontal, we have;
K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₂ = (1/2) × m × ((0.5·u₂)² + 2·g·h)²
For the third ball thrown at 30° below the horizontal, we have;K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₃ = (1/2) × m × ((0.5·u₃)² + 2·g·h)²
For the fourth ball thrown straight down, we have;Kinetic energy K.E.₄ = (1/2) × m × (u₄² + 2gh)²
Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
u₁ = u₄, K.E₁ = K.E.₄, u₂ = u₃, K.E₂ = K.E.₃
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A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r < R) from the center of the sphere the electric field has a value E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R the magnitude of the electric field at a radius r would be equal to:__________
Answer:
Hence the answer is E inside [tex]= KQr_{1} /R^{3}[/tex].
Explanation:
E inside [tex]= KQr_{1} /R^{3}[/tex]
so if r1 will be the same then
E [tex]\begin{bmatrix}Blank Equation\end{bmatrix}[/tex] proportional to 1/R3
so if R become 2R
E becomes 1/8 of the initial electric field.
Answer:
The electric field is E/8.
Explanation:
The electric field due to a solid sphere of uniform charge density inside it is given by
[tex]E =\frac{\rho r}{3}[/tex]
where, [tex]\rho[/tex] is the volume charge density and r is the distance from the center.
For case I:
[tex]\rho = \frac{Q}{\frac{4}{3}\pi R^3}[/tex]
So, electric field at a distance r is
[tex]E = \frac { 3 Q r}{3\times 4\pi R^3}\\\\E = \frac{Q r}{4\pi R^3}[/tex]
Case II:
[tex]\rho = \frac{Q}{\frac{4}{3}\pi 8R^3}[/tex]
So, the electric field at a distance r is
[tex]E' = \frac { 3 Q r}{3\times 32\pi R^3}\\\\E' = \frac{Q r}{8\times 4\pi R^3}\\\\E' = \frac{E}{8}[/tex]
Compare the time it
takes the light to travel from your
teacher to your eye with the time
it takes sound to travel the same
distance.
Answer:
Light takes less time than sound.
Explanation:
Let's say, the teacher and the student are at a distance "d" from each other.
The medium around them would be air.
And,
The speed of light in air is approx. 3× 10⁸ m/s
while, the speed of sound in air is approx. 330 m/s
We have a formula that establishes the relation between speed, distance and time.
[tex] \boxed{ \mathsf{speed = \frac{distance}{time} }}[/tex]
Our hunt for time — Speed in both the scenarios is known to us whereas the distance is same.
Sound
[tex] \mathsf{330 = \frac{d}{time_{s}} }[/tex]
[tex] \underline{\mathsf{time _{s} = \frac{d}{330} }}[/tex]
Light
[tex] \mathsf{3 \times {10}^{8} = \frac{d}{time _{l} } }[/tex]
[tex] \underline{ \mathsf{ time _{l} = \frac{d}{3 \times {10}^{8}} }}[/tex]
The best way of comparison is finding their ratio.
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ \frac{d}{330} }{ \frac{d}{3 \times {10}^{8} } } }[/tex]
simplifying the fraction
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{d \times (3 \times {10}^{8} )}{330 \times d}}[/tex]
d gets canceled and we're left with the following expression
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ (3 \times10 \times {10}^{7} )}{330}}[/tex]
30, being a common factor in the numerator as well as denominator, gets canceled out. and in its place remains 1/ 11
(why?
=> 30÷330 = 1÷11)
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ 1\times {10}^{7} }{11}}[/tex]
taking timeₛ to the numerator on the other side.
[tex] \implies \mathsf{time_{s} = \frac{ 1\times {10}^{7} }{11}\times time_{l}}[/tex]
Therefore, we get timeₛ is approx. 10⁶ times the timeₗ.
That's a big difference, no wonder light's way much faster than sound.
As lesser the time taken to cover a distance, faster is the wave.
The sound takes about 874,000 times MORE time than the light takes.
Find the ratio of the diameter of aluminium to copper wire, if they have the same
resistance per unit length. Take the resistivity values of aluminium and copper to
be 2.65× 10−8 Ω m and 1.72 × 10−8 Ω m respectively
Answer:
1.24
Explanation:
The resistivity of copper[tex]\rho_1=2.65\times 10^{-8}\ \Omega-m[/tex]
The resistivity of Aluminum,[tex]\rho_2=1.72\times 10^{-8}\ \Omega-m[/tex]
The wires have same resistance per unit length.
The resistance of a wire is given by :
[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}[/tex]
According to given condition,
[tex]\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24[/tex]
So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.
15 . A scientist who studies the whole environment as a working unit .
Botanist
Chemist
Ecologist
Entomologist
Answer:
Ecologist.
Your answer is Ecologist.
(Ecologist) is a scientist who studies the whole environment as a working unit.
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88°C I am not 100% sure this is right but I am 98% sure this IS right
If a negatively charged particle is placed inside a uniform electric field the electric force that will act on that particle points in what direction in reference to the electric field lines?
Answer:
opposite direction
Explanation:
An electric field is defined as a physical field which surrounds the electrically charged particles that exerts force on the other particles on the field.
Now when an electron or a negatively charged particle enters a uniform electric field, the electric forces acts on the negatively charged particles and it forces the particle to move in the direction which is opposite to the direction of the field. In an uniform electric field, the field lines are parallel.
Answer:
Explanation:
negatively charged particle is placed inside uniform electric field
The force on the charge due to the electric field is
F = q E
when the charge is negative so the force on the charge is opposite to the direction of electric field.
The electric field is opposite to the force.
During take-off a 8kg model rocket is burning fuel causing its speed to increase
at a rate of 4m/s2 despite experiencing a 90N drag.
What’s is the strength of the thrust?
(Answer unit is in N)( and the answer isn’t 212)
The strength of the thrust is 122 newtons.
The motion of the rocket is described by the second Newton's law, whose model is shown below:
[tex]\Sigma F = F - D = m\cdot a[/tex] (1)
Where:
[tex]F[/tex] - Thrust, in newtons[tex]D[/tex] - Drag, in newtons[tex]m[/tex] - Mass of the rocket, in kilograms[tex]a[/tex] - Net acceleration of the rocket, in meters per square secondIf we know that [tex]D = 90\,N[/tex], [tex]m = 8\,kg[/tex] and [tex]a = 4\,\frac{m}{s^{2}}[/tex], then the strength of the thrust is:
[tex]F = D + m\cdot a[/tex]
[tex]F = 90\,N + (8\,kg)\cdot \left(4\,\frac{m}{s^{2}} \right)[/tex]
[tex]F = 122\,N[/tex]
The strength of the thrust is 122 newtons.
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An object following a straight-line path at constant speed
A.) has no forces acting on it.
B.) has a net force acting on it in the direction of motion.
C.) has zero acceleration.
D.) must be moving in a vacuum.
E.) none of the above
An object following a straight-line path at constant speed is option C.) has zero acceleration.
Are there any forces acting on a moving item traveling in a straight line at a constant speed?There are no forces operating on a body if it is travelling straight ahead at a steady speed. There are no forces operating on a body if it is travelling straight ahead at a steady speed.
Note that the physics concept of acceleration measures how quickly an object's motion is changing. An object's speed or velocity is what largely defines its motion.
Therefore, An object is considered to be accelerating when its velocity changes over time and as such since acceleration of the object is said to be zero, one can say that the net force acting on it is also zero.
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A scientist who studies the tiny microorganisms of the environment .
geologist
meteorologist
microbiologist
entomologist
Question:- A scientist who studies the tiny microorganisms of the environment
Answer:- MicrobiologistExplanation:-
Microbiologist means a person who studies micro sized living organisms
Microbiologist word is combination of two words Micro and biologist
Micro stands for objects which cannot be seen with the naked eyes and are very small in sizeBiologist a person who studies living forms.Answer:
microbiologist i think
hope it helps
please mark Brainliest if you think the answer is correct
Which technological device makes an energy conversion in the same way that a human ear makes an energy conversion?
a.) a loudspeaker
b.) a headphone
c.) a light bulb
d.) a microphone
I think it's c because of the concept of mechanical energy to electrical energy but I'm not sure
Answer:
I THINK C
Explanation:
BECAUSE A Light Emitting Diode (LED) glows even when a weak electric current passes through it.
An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. While the spring is being extended by the force, the work done by the spring is:
a. -3.6 J
b. -3.3 F
c. -3.4 times 10^-5 J
d. 3.3 J
e. 3.6 J
Answer:
b) - 3.3 J
Explanation:
Given;
mass, m = 2 kg
initial extension of the spring, x = 6 cm = 0.06 m
The weight of the mass on the spring;
W = mg
where;
g is acceleration due to gravity = 9.81 m/s²
W = 2 x 9.81
W = 19.62 N
The spring constant is calculated as;
W = kx
k = W/x
k = 19.62 / 0.06
k = 327 N/m
The work done by the spring when it is extended to an additional 10 cm;
work done = force x distance
distance = extension, x = 10 cm = 0.1 m
The work done by the spring opposes the applied force by acting in opposite direction to the force.
W = - Fx
W = - (kx) x
W = - kx²
W = - (327) x (0.1)²
W = - 3.27 J
W ≅ - 3.3 J
Therefore, the work done by the spring by opposing the applied force is -3.3 J
The spaceship Enterprise 1 is moving directly away from earth at a velocity that an earth-based observer measures to be 0.66c. A sister ship, Enterprise 2, is ahead of Enterprise 1 and is also moving directly away from earth along the same line. The velocity of Enterprise 2 relative to Enterprise 1 is 0.34c. What is the velocity of Enterprise 2, as measured by the earth-based observer
Answer:
The answer is "0.82 c".
Explanation:
Given:
Spacecraft speed 1 is [tex]u = + 0.66 \ c[/tex]
Space velocity 2 relative to spacecraft 1 is [tex]v = + 0.34\ c[/tex]
The spacecraft velocity 2 measured by the Earth observation
[tex]\to u' = \frac{u +v}{1 + ( \frac{uv}{c^2})}[/tex]
[tex]= \frac{0.66 \ c +0.34\ c}{ 1+ (\frac{0.66\ c \times 0.33\ c }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (\frac{0.2178\ c^2 }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (0.2178 )}\\\\ = \frac{1 \ c }{ 1.2178 }\\\\=0.82\ c[/tex]
Two divers, G and H, are at depths 20 m and 40 m respectively
below the water surface in lake. The pressure on G is P, while
the pressure on H is P2 if the atmospheric pressure is equivalent
to 10 m of water, then the value of P2/P1 is.
A. 1.67.
B. 2.00.
C. 0.50.
D. 0.60.
Answer:
B
Explanation:
P1/P1 = 40/20
=2
what effect does the force of gravity have on a stone thrown vertically upwards
Answer:
rock go down
Explanation:
what comes up must come down.
A bicycle tire with a volume of 0.00210 m^3 is filled to its recommended absolute pressure of 495 kPa on a cold winter day when the tire's temperature is -14°C. The cyclist then brings his bicycle into a hot laundry room at 32°C.
a. If the tire warms up while its volume remains constant, will the pressure increase be greater than, less than, or equal to the manufacturer's stated 10% overpressure limit?
b. Find the absolute pressure in the tire when it warms to 32 degrees Celcius at constant volume.
(A) The pressure will be greater than 10% overpressure limit.
(B) The final pressure will be "582.915 kPa".
Given:
Volume,
[tex]V = 0.0021 \ m^3[/tex]Initial pressure,
[tex]P_o= 495 \ kPa[/tex]Initial temperature,
[tex]T_o = -14^{\circ} C[/tex][tex]= 259 \ K[/tex]
Final temperature,
[tex]T = 32^{\circ} C[/tex](B)
Number of moles,
→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]
then,
The final absolute pressure,
→ [tex]P = \frac{nRT}{V}[/tex]
[tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]
[tex]=(\frac{T}{T_o} )P_o[/tex]
[tex]= (\frac{305}{259} )\times 495[/tex]
[tex]= 582.915 \ kPa[/tex]
Thus the above approach is correct.
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A hot-air balloon plus cargo has a mass of 308 kg and a volume of 2910 m3 on a day when the outside air density is 1.22 kg/m3. The balloon is floating at a constant height of 9.14 m above the ground.
Required:
What is the density of the hot air in the balloon?
9514 1404 393
Answer:
1.114 kg/m³
Explanation:
The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...
2910d +308 = 2910×1.22
Solving for d, we find ...
2910d = 2919(1.22) -308
d = 1.22 -308/2910
d ≈ 1.114 . . . kg/m³
The density of the hot air is about 1.114 kg/m³.
What is the de Broglie wavelength of a red blood cell with a mass of 1.00 * 10-11 g that is moving with a speed of 0.400 cm/s? Do we need to be concerned with the wave nature of the blood cells when we describe the flow of blood in the body?
Answer:
The wavelength is "[tex]=16.5675\times 10^{-18} \ m[/tex]".
Explanation:
Given:
Mass,
m = [tex]1\times 10^{-11} \ g[/tex]
Speed,
V = [tex]0.400 \ cm/s[/tex]
or,
= [tex]0.4\times 10^{-2}[/tex]
According to De Broglie,
The wavelength will be:
⇒ [tex]\lambda = \frac{h}{mV}[/tex]
[tex]=\frac{6.627\times 10^{-34}}{1\times 10^{-11}\times 10^{-3}\times 0.4\times 10^{-2}}[/tex]
[tex]=16.5675\times 10^{-18} \ m[/tex]
So, blood cells move these wavelength.
A 40 kg boy is standing on the edge of a stationary 30-kg platform that is free to rotate without friction. The boy tries to walk around the platform in a counterclockwise direction. As he does:
a. the platform doesn't rotate.
b. the platform rotates in a clockwise direction just fast enough so that the boy remains stationary relative to the ground.
c. the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground.
d. both go around with equal angular velocities but in opposite directions
Answer:
the correct one is C
Explanation:
To find the answer, let's propose the solution of the problem
We create a system formed by the child and the platform so that all the forces have been internal and the angular momentum is conserved.
Initial instant. Before starting to walk
L₀ = 0
Final moment. After the child is walking
L_f = I₁ w₁ + m r v₂
where index 1 is used for the platform and index 2 for the child
linear and angular velocity are related
v₂ = w₂ r
angular momentum is conserved
0 = I₁ w₁ + m r (w₂ r)
w₁ = [tex]- \frac{m r^2}{I1} \ w_2[/tex]
the moment of inertia of the platform bringing it closer to a disk or cylinder
I₁ = [tex]\frac{1}{2}[/tex] M r²
sustitute
w₁ = [tex]- \frac{2 m }{M} \ w_2[/tex]
W₁ = - [tex]- \frac{2 40}{30} \ w_2 = - \frac{8}{3} \ w_2[/tex]
from here we can see that the platform and the child rotate in the opposite direction and with different angular speeds
when examining the answers the correct one is C
Answer:
Option C (the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground)Explanation:
relative to the ground the boy moves in a counter clockwise motion , now the boy and the wheel are one system
so by conservation of angular momentum their net sum of angular momentum relative to a point outside the system(say ground) should be zero
so the wheel moves in a clockwise direction , their angular velocity may or may not be same depending on I. so option D is wrong
option B is wrong because relative to ground their angular momentum should be equal and opposite
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Help me in my hw,A train starts from rest.Its velocity becomes 90km/hr after 1 min,Calculate the acceleration of train and distance covered by the train.Answer it ASAP
Answer:
I am serious about that
Explanation:
.......
A block with a mass of 0.26 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.2 N on the block. When the block is released, it oscillates with a frequency of 1.4 Hz. How far was the block pulled back before being released?
Answer:
2
Explanation:
pulling force because of it force
Answer:
5.9 cm
Explanation:
f: frequency of oscillation
frequency of oscillationk: spring constant
frequency of oscillationk: spring constantm: the mass
[tex]f = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } [/tex]
in this problem we know,
F= 1.4 Hz
m= 0.26 kg
By re-arranging the formula we get
[tex]k = {(2\pi \: f )}^{2} m = {(2\pi(1.4hz))}^{2} 0.26kg = 20.1 \frac{n}{m} [/tex]
The restoring force of the spring is:
F= kx
where
F= 1.2 N
k= 20.1 N/m
x: the displacement of the block
[tex]x = \frac{f}{k} = \frac{1.2 \: n}{20.1 \frac{n}{m} } = 0.059m \: = 5.9 \: cm[/tex]
Which of the following is an ethical question? A. How do we clone humans? B. Can we clone humans? C. Will cloning technology be useful? D. Do clones have the same rights as humans?
Answer:
D.) Do clones have the same rights as humans?
Explanation:
This is the only option choice that relates to ethics, more specifically, bioethics. The rest of the answer choices look either seek for scientific explanations on cloning itself or the implications of clothing. Only answer choice D relates to ethics.
Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing
Answer:
The correct answer is "4.443 sec".
Explanation:
Given:
Mass of child,
= 34 kg
Mass of swing,
= 18 kg
Length,
= 4.9 m
The time period of pendulum will be:
T = [tex]2 \pi \sqrt{4g}[/tex]
= [tex]2 \pi \sqrt{\frac{4.9}{9.8} }[/tex]
= [tex]4.443 \ sec[/tex]
Answer:
The time taken to back and forth is 4.4 s .
Explanation:
Length, L = 4.9 m
let the time period is T.
Acceleration due to gravity, g = 9.8 m/s^2
Use the formula of time period
[tex]T = 2 \pi\sqrt{L}{g}\\\\T = 2 \times 3.14\sqrt{4.9}{9.8}\\\\T = 4.4 s[/tex]
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to 109 rad/sec in 1.87 seconds. The blades of the fan have a diameter of 6.7 meters and their deceleration rate is 4.7 rad/sec2.
What was the initial angular speed of the fan in rev/sec?
ωi =
Answer:
wo = 18.75 rev / s
Explanation:
This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²
w_f = w₀ - a t
w₀ = w_f + a t
w₀ = 109 + 4.7 1.87
w₀ = 117.8 rad / s
let's reduce to revolutions / s
w₀ = 117.8 rad / s (1 rev / 2pi rad)
w₀ = 18.75 rev / s
A cyclist rides in a circle with speed 8.1 m/s. What is his centripetal
acceleration if the circle has a radius of 27 m?
Explanation:
We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:
Vt = ωr with ω being the angular velocity.
We now calculate ω:
ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s
Now that we have ω we can calculate the centripetal aceleration:
a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2
Question 3 of 10
What has the same value no matter where it is located in the universe?
A. Volume
B. Weight
C. Mass
D. Density
Reset Selection
Answer:
C. Mass
Explanation:
