Answer:
Surface tension is, the surface where the water meets the air, water molecules cling even more tightly to each other.
g A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 39.0kg and diameter 78.0cm. The power is off for 34.0s, and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 170 complete revolutions.At what rate is the flywheel spinning when the power comes back on?
Answer:
[tex]10.54\ \text{rad/s}[/tex]
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = 500 rpm = [tex]500\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]
[tex]\omega_f[/tex] = Final angular velocity
t = Time = 34 s
[tex]\theta[/tex] = Angular displacement = 170 revs = [tex]170\times 2\pi\ \text{rad}[/tex]
[tex]\alpha[/tex] = Angulr acceleration
From the kinematic equations of angular motion we have
[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \alpha=\dfrac{\theta-\omega_it}{\dfrac{1}{2}t^2}\\\Rightarrow \alpha=\dfrac{170\times 2\pi-500\times \dfrac{2\pi}{60}\times 34}{\dfrac{1}{2}\times 34^2}\\\Rightarrow \alpha=-1.23\ \text{rad/s}^2[/tex]
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=500\times \dfrac{2\pi}{60}+(-1.23)\times 34\\\Rightarrow \omega_f=10.54\ \text{rad/s}[/tex]
The rate at which the wheel is spinning when the power comes back on is [tex]10.54\ \text{rad/s}[/tex].
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 15.0 km ? Take the mass of the sun to be ms = 1.99×1030 kg , the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be g = 9.8 m/s2 . Express your weight wstar in newtons.
Answer:
W' = 1.66 x 10¹⁴ N
Explanation:
First, we will calculate the mass:
[tex]W = mg[/tex]
where,
W = weight on earth = 690 N
m = mass = ?
g = acceleration due to gravity on earth = 9.8 m/s²
Therefore,
[tex]m = \frac{W}{g} = \frac{690\ N}{9.8\ m/s^2}\\\\m = 70.4\ kg[/tex]
Now, we will calculate the value of g on the neutron star:
[tex]g' = \frac{GM}{R^2}[/tex]
where,
g' = acceleration due to gravity on the surface of the neutron star = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of the Neutron Star = 1.99 x 10³⁰ kg
R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m
Therefore,
[tex]g' = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{(7500\ m)^2}\\\\g' = 2.36\ x\ 10^{12}\ m/s^2[/tex]
Therefore, the weight on the surface of the neutron star will be:
[tex]W' = mg'\\W' = (70.4\ kg)(2.36\ x\ 10^{12}\ m/s^2)[/tex]
W' = 1.66 x 10¹⁴ N
If acceleration is zero what statement about velocity is
true *
A)Velocity is zero
B)Velocity is constant
C)Velocity cannot be determined
D) Velocity is changing
Answer: A
Velocity is zero because the acceleration isn't affected, and velocity is the rate of change, so it can't be any other options.
Answer:
B)Velocity is constantExplanation:
If an object moves with a velocity and there is no acceleration, then the velocity remains constant. His velocity after five second will be equal to his initial velocity.#keeplearning dude:)6. If an object accelerates at 3m/s/s, how long does it take for the object to travel at a speed of 12 m/s.
Answer:
4 seconds
Explanation:
Assuming that the object started from rest,
v = at
--> t = v/a = (12 m/s) / (3 m/s^2)
= 4 seconds
9. Mr. Smith went skiing in Maine last weekend. He traveled 523 kilometers to Sugarloaf from
Leominster. His average speed was 109 km/hr. How long did it take Mr. Smith to hit the slopes?
Answer:
Time taken by Mr. smith = 4.80 hour (Approx.)
Explanation:
Given:
Distance travel by Mr. smith = 523 kilometer
Average speed of Mr. smith = 109 km/hr
Find;
Time taken by Mr. smith
Computation:
Time taken = Distance cover / Speed
Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.
smith
Time taken by Mr. smith = 523 / 109
Time taken by Mr. smith = 4.798 hr
Time taken by Mr. smith = 4.80 hour (Approx.)
You and a friend are playing with a Coke can that you froze so it's solid to demonstrate some ideas of Rotational Physics. First, though, you want to calculate the Rotational Kinetic Energy of the can as it rolls down a sidewalk without slipping. This means it has both linear kinetic energy and rotational kinetic energy. [The freezing only matters because if there is liquid inside, the calculation for the Moment of inertia becomes more complicated]. A Coke can can be modeled as a solid cylinder rotating about its axis through the center of the cylinder. This can has a mass of 0.33 kg and a radius of 3.20 cm. You'll need to look up the equation for the Moment of Inertia in your textbook. It is rotating with a linear velocity of 6.00 meters / second in the counter-clockwise (or positive) direction. You can use this to determine the angular velocity of the can (since it is rolling without slipping). What is the Total Kinetic Energy of the Coke can
Answer:
K_{total} = 8.91 J
Explanation:
In this exercise you are asked to find the kinetic energy of the can of coca-cola
K_total = K_ {Translation} + K_ {rotation}
the translational kinetic energy is
K_ {translation} = ½ m v²
the kinetic energy of rotation is
K_ {rotation} = ½ I w²
The moment of inertia of a cylinder is
I = ½ m r²
we substitute
K_ {total} = ½ m v² + ½ (½ m r²) w²
angular and linear velocity are related
v = w r
we substitute
K_ {total} = ½ m v² + ¼ m r² v² / r²
K_ {total} = m v² (½ + ¼)
K_ {total} = ¾ m v²
let's calculate
K_ {total} = ¾ 0.33 6.00²
K_{total} = 8.91 J
How much work will a 500 watt motor do in 10 seconds?
Answer:
50j
Explanation:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Answer:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Explanation:
Plutonium-238 has a half life of 87.7 years. What percentage of a 5 kilogram (kg) sample remains after 50 years?
Answer:
i dont know but i should know try g o o g l e
Explanation:
Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. It emerges from the second filter with intensity 132 W/m2 . You may want to review (Pages 897 - 898) . Part A What is the angle from vertical of the axis of the second polarizing filter
Answer:
θ = 32.4º
Explanation:
For this exercise let's use Malus's law
I = Io cos² θ
in this case it indicates that the incident intensity is 370 W/m², when the first polarization passes, only the radiation with the same polarization of the polarizer emerges, that is, vertical
I₀ = 370/2 = 185 W / m²
this is the radiation that affects the second polarizer, let's apply the expression of Maluz
θ = cos⁻¹ ([tex]\sqrt{\frac{I}{I_o} }[/tex])
θ = cos⁻¹ ([tex]\sqrt{132/185}[/tex])
θ = cos⁻¹ (0.844697)
θ = 32.4º
What happens when Earth rotates on its axis and how long does it take
Answer:
You get Day and Night
It takes 24 hour
Answer:
Explanation:
The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.
what is force,momentum,and velocity.
Answer:
A force is a push or pull upon an object resulting from the object's interaction with another object.
Momentum is force or speed of movement.
Velocity defines the path of the motion of the frame or the object
What is the magnitude of the gravitational force acting on a
1.0 kg object which is 1.0 m from another 1.0 kg object?
Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!
bit.[tex]^{}[/tex]ly/3a8Nt8n
Increasing the telescope diameter beyond the value found in part (a) will increase the light-gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of 10-m diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope (5-m diameter) on Palomar Mountain in California
Answer:
Ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
Explanation:
If increasing the Diameter of a Telescope beyond a given value will increase the ability of the telescope to capture more light and also capture astronomical objects located in a very distant position without improving resolution.
Hence the superiority of Keck telescope atop Mauna Kea over Hale Telescope atop Palomar mountain in California is the ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plane, if the coefficient of the static friction is 0.20 and kinetic friction is 0.15 (1) find the value of P to cause motion up the plane (2) find P to prevent motion down the plane. (3) Find P to cause continuous motion up the plane.
Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If a charge concentration of 25.0 C is above the plane at a height of 3000 m within the cloud and a charge concentration of -40.0 C is at height 850 m, what is the electric field at the aircraft
Answer:
[tex]523269.9\ \text{N/m}[/tex]
Explanation:
q = Charge
r = Distance
[tex]q_1=25\ \text{C}[/tex]
[tex]r_1=3000\ \text{m}[/tex]
[tex]q_2=40\ \text{C}[/tex]
[tex]r_2=850\ \text{m}[/tex]
The electric field is given by
[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]
The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]
Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thruster consists of a propeller mounted in a duct; the unit is then mounted below the waterline in the bow or stern of the ship. The duct runs completely across the ship. Calculate the thrust developed by a 1900 kW unit supplied to the propeller if the duct is 2.6 m in diameter and the ship is stationary.
Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
Determine the Thrust developed
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : calculate the area of the duct
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
next : calculate the velocity of propeller
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
Finally determine the thrust developed
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN
The universe cooled after the Big Bang.At some point hydrogen atoms combined to form helium.What is this process called?
Answer:
Nuclear fusion
Explanation:
A radioactive material produces 1160 decays per minute at one time, and 4.0 h later produces 170 decays per minute. whats the half life
Answer:
Half life is 3.23 hours
Explanation:
Given
Decay rate at starting = 1160 decays per minute
Decay rate after 4 hours = 170 decays per minute
As we know know
[tex]N = N_0 *e ^{\Lambda *T}[/tex]
Substituting the given values, we get -
[tex]170 = 1160 *e ^{-4*\Lambda}\\0.1465 = e ^{-4*\Lambda}\\-0.834 = -4 * \Lambda\\\Lambda = 0.834/4\\\Lambda = 0.2085[/tex]
Also
[tex]t_{1/2} = \frac{ln2}{\Lambda}[/tex]
Substituting the given values we get -
[tex]t_{1/2} = =0.693/0.2085\\= 3.23[/tex]hours
1.What is the Kinetic energy of a 3 kg object moving at 4 m/s?
Plz help I’ll give points
Answer:
24 J
Explanation:
[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]
whem completing an emergency Roaside stop,it is necessary to put on your parking brake
A. True
B. False
Answer:
trueeeeeeee..........mmmm...........
A sound wave travels with a velocity of 1.5 m/s and has a frequency of 500 Hz. What is its wavelength?
Scientists are constantly exploring the universe, looking for new planets that support life similar to the life on
Earth. A new planet that supports life would have all of the following characteristics except -
A. a gaseous atmosphere.
B. an orbiting moon.
C. liquid water.
D. protection from radiation.
A new planet that supports life would have all the following characteristics except an orbiting moon. Hence, option B is correct.
What is a Planet?An enormous, spherical celestial object known as a planet is neither a star nor its remains. The nebular hypothesis, which states how an interstellar cloud falls out of a nebula to produce a young protostar encircled by a protoplanetary disk, is now the best explanation for planet formation.
By gradually accumulating material under the influence of gravity, or accretion, planets develop in this disk.
The rocky planets Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. These planets all revolve around axes that are inclined relative to their respective polar axes.
To know more about Planet:
https://brainly.com/question/14581221
#SPJ2
Which statement best explains why objects are pulled toward Earth’s center?
Answer:
Earth has a much greater mass than objects on its surface
Can someone take there time and answer this :)
Answer: I think B.)
Explanation:
What type of force is F-?
Answer:
Frictional Force
Explanation:
i think so
03: A mass with a 60 g vibrate at the end of a spring. The amplitude of the motion is 0.394 ft
and a frequency is 0.59 HZ. Find the perind and spring constant, the maximum speed and
acceleration of the mass, the speed and acceleration when the displacement is 6 cm, compute the
kinetic and the potential energy when the position is 6 cm
Answer:
a) T = 1.69 s, b) k = 0.825 N / m, c) v = 1.46 feet/s, d) a = 5.41 ft / s²,
e) v = - 1,319 ft / s, a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J
Explanation:
In a mass-spring system with simple harmonic motion, the angular velocity is
w = [tex]\sqrt{\frac{k}{m} }[/tex]
a) find the period
angular velocity, frequency, and period are related
w = 2π f = 2π / T
f = 1 / T
T = 1 / f
T = 1 / 0.59
T = 1.69 s
b) the spring constant
w = 2π f
w = 2π 0.59
w = 3.70 rad / s
w² = k / m
k = w² m
k = 3.70² 0.060
k = 0.825 N / m
c) the maximum speed
simple harmonic movement is described by the expression
x = A cos (wt + Ф)
speed is defined by
v =[tex]\frac{dx}{dt}[/tex]
v = -A w sin (wt + fi)
the speed is maximum when the cosine is ± 1
v = A w
v = 0.394 3.70
v = 1.46 feet/s
d) maximum acceleration
a = [tex]\frac{dv}{dt}[/tex]
a = - A w² cos wt + fi
the acceleration is maximum when the cosine is ±1
a = A w²
a = 0.394 3.70²
a = 5.41 ft / s²
e) velocity and acceleration for x = 6 cm
let's reduce the cm to feet
x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot
Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s
let's use the expression for the velocity
v = -A w sin (0 + Фi)
0 = - A w sin Ф
so sin Ф = 0 which implies that Фi = 0
the equation of motion is
x = A cos wt
x = 0.394 cos 3.70t
we substitute
0.1969 = 0.394 cos 370t
3.70 t = cos⁻¹ (0.1969 / 0.394)
let's not forget that the angle is in radians
3.70, t = 1.047
t = 1.047 / 3.70
t = 0.2826 s
we substitute this time in the equation for velocity and acceleration
v = - Aw sin wt
v = - 0.394 3.70 sin 3.70 0.2826
v = - 1,319 ft / s
a = - A w² cos wt
a = - 0.394 3.70² cos 3.70 0.2826
a = - 2.70 ft / s²
f) the kinetic and potential energy at this point
K = ½ m v²
let's slow down to the SI system
v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s
K = ½ 0.060 0.402²
K = 4.8 10⁻³ J
U = ½ k x²
U = ½ 0.825 0.06²
U = 1.49 10⁻³ J
1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20
Answer:
It will go up to 93.75 m before it is moving at 20 m/s
Explanation:
As we know that
[tex]v^2 - u^2 = 2aS[/tex]
here v is the final speed i.e 20 m/s
u is the initial speed i.e 5 m/s
a is the acceleration due to gravity i.e 2 m/s^2
Substituting the given values in above equation, we get -
[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters
If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity or the boy skater
his mass is 50 kg?
Question: Two people stand facing each other at a roller-skating rink then push off each other. If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity of the boy skater if his mass is 50 kg?
Answer:
3 m/s
Explanation:
Applying,
The Law of conservation of momentum
Momentum of the girl skater = momentum of the boy skater
MV = mv...................... Equation 1
Where M = mass of the girl skater, V = velocity of the girl skater, m = mass of the boy skater, v = velocity of the boy skater
From the question, we were asked to calculate v
v = MV/m.................. Equation 1
Given: M = 30 kg, V = 5 m/s, m = 50 kg
Substitute these values into equation 1
v = (30×5)/50
v = 3 m/s
Hence the velocity of the the boy skater is 3m/s
A truck is traveling on a level road. The driver suddenly applies the brakes, causing the truck to decelerate by an amount g/2. This causes a box in the rear of the truck to slide forward. If the coefficient of sliding friction between the box and the truckbed is 2/5, find the acceleration of the box relative to the truck and relative to the road.
Answer:
Truck [tex]\dfrac{g}{10}[/tex]
Road [tex]-\dfrac{g}{10}[/tex]
Explanation:
[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]
[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]
Frictional force is given by
[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]
Net acceleration is given by
[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]
The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.
An ideal horizontal spring-mass system has a mass of 1.0 kg and a spring with constant 78 N/m. It oscillates with a period of 0.71 seconds. When this same spring-mass system oscillates vertically instead, the period is _______ seconds. Enter 2 significant figures (a total of three digits) and use g = 10.0 m/s2 if necessary.
Answer:
T = 0.71 seconds
Explanation:
Given data:
mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.
We have to calculate time period when this same spring-mass system oscillates vertically.
As we know
[tex]T = 2\pi \sqrt{\frac{m}{K} }[/tex]
This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.
Therefore, T = 0.71 seconds
