Answer:
rotational kinetic energy is 1156.81 J
Explanation:
Given the data in the question;
first we find the moment of inertia of the rod as axis passes through the middle or center;
[tex]I = \frac{1}{12}ml^2[/tex]
where m is the mass of the road( 3.0 kg)
[tex]l[/tex] is the length of the rod ( 2.6 m )
so we substitute
[tex]I = \frac{1}{12}[/tex] × 3 × (2.6)²
[tex]I[/tex] = 1.69 kg.m²
now, to calculate the rotational kinetic energy,
kinetic energy due to rotation is;
KE[tex]_R[/tex] = [tex]\frac{1}{2}Iw^2[/tex]
where [tex]I[/tex] is moment of inertia( 1.69 kg.m² ),
w is angular frequency ( 37 radians/s )
we substitute
KE[tex]_R[/tex] = [tex]\frac{1}{2}[/tex] × 1.69 × ( 37 radians/s )²
KE[tex]_R[/tex] = 1156.81 J
Therefore, rotational kinetic energy is 1156.81 J
The rotational kinetic energy will be 1156.81 J. It is half of the product of the moment of inertia and angular speed.
What is rotational kinetic energy?The rotational kinetic energy is the kinetic energy generated by an object's rotation and It is a component of its overall kinetic energy.
The given data in the problem is;
m is the mass = 3.0 kg
l is the length = 2.6 m
[tex]\rm \omega[/tex] is the angular speed = 37 radians/s
v is the speed = 13 m/s
[tex]\rm E_k[/tex] is the rotational kinetic energy=?
The momentum of inertia is found by;
[tex]\rm I = \frac{1}{12} ml^2 \\\\ \rm I = \frac{1}{12} \times 3 \times (2.6)^2 \\\\\ \rm I = 1.69 \kg.m^2[/tex]
The rotational kinetic energy is found by;
[tex]\rm E_k= \frac{1}{2}I (\omega)^2 \\\\ \rm E_k= \frac{1}{2}\times 1.69 (37)^2 \\\\ \rm E_k=1156.81 \ J[/tex]
Hence the rotational kinetic energy will be 1156.81 J.
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how can you Make different objects using blocks
SCIENCE
The growth of algae in ocean water is limited by their need for
a.
warm ocean currents.
b.
carbon dioxide and sunlight.
c.
dissolved oxygen.
d.
low salinity.
Answer:
c is trueeeeeeeeeeeeeee
The growth of algae in ocean water is limited by their need for dissolved oxygen. Thus, the correct answer is option C.
What is algae?The term "algae" refers to a large and diverse group of photosynthetic eukaryotic organisms. It is a polyphyletic grouping of species from several distinct clades. Most are aquatic and autotrophic, lacking many of the distinct cell and tissue types found in land plants, such as stomata, xylem, and phloem.
Algae, like all organisms, normally grow in balance with their ecosystems, with the amount of nutrients in the water limiting their growth. Deep ocean water contains fewer algae because algae require sunlight and carbon dioxide to thrive.
Therefore, due to need for dissolved oxygen the growth of algae in ocean water is limited.
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Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance between the centers of the balls. Construct a problem in which you calculate the electric field (magnitude and direction) due to the balls at various points along a line running through the centers of the balls and extending to infinity on either side. Choose interesting points and comment on the meaning of the field at those points. For example, at what points might the field be just that due to one ball and where does the field become negligibly small? Among the
things to be considered are the magnitudes of the charges and the distance between the centers of the balls. Your instructor may wish for you to consider the electric field off axis or for a more complex array of charges, such as those in a water molecule.
Answer:
interest point:
1) Point on the left side
2) Point within the radius r₁ of the first sphere
3) Point between the two spheres
4) point within the radius r₂ of the second sphere
5) Right side point
Explanation:
In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres
We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is
E_ {total} = E₁ + E₂
E_{ total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}[/tex]
the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d
x₂ = x₁ -d
E total = [tex]k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}[/tex]
Let's analyze the field for various points of interest.
1) Point on the left side
in this case
E_ {total} = [tex]k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )[/tex]
E_ {total} = [tex]k \frac{Q}{x_1^2}[/tex] [tex]( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )[/tex]
We have several interesting possibilities:
* We can see that as the point is further away the field is more similar to the field created by two point charges
* there is a point where the field is zero
E_ {total} = 0
x₁² = (x₁ + d)²
2) Point within the radius r₁ of the first sphere.
In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point
E_ {total} = [tex]-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}[/tex]
this expression holds for the points located at
-r₁ <x₁ <r₁
3) Point between the two spheres
E_ {total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}[/tex]
This champ is always different from zero
4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes
E_ {total} = [tex]+ k \frac{Q}{(d-x_1)^2}[/tex]+ k Q / (d-x1) 2
point range
-r₂ <x₂ <r₂
5) Right side point
E_ {total} = [tex]k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}[/tex]
E_ {total} = [tex]- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )[/tex]- k Q / x22 (1- 1 / (x1 + d) 2)
we have two possibilities
* as the distance increases the field looks more like the field created by two point charges
* there is a point where the field is zero
difference between ferromagnetic and antiferromagnetic ..
Answer:
ferromagnetic materials are materials that are highly attracted to magnets while antiferromagnetic materials are materials that are not attracted by magnets
1. A person kicks a rock off a cliff horizontally with a speed of 20 m/s. It takes 7.0 seconds to hit the
ground, find:
a. height of the cliff
b. final vertical velocity
C. range
D.speed and angle of impact
This problem involved half projectile.
initial velocity, vo = 20 m/s
time of flight, t = 7 s
(a) Simply use the formula to get the height, h:
h = vo*t - (1/2)gt^2
(b) To get the final vertical velocity or terminal velocity (vf), use the formula:
(vf)^2 - (vo)^2 = 2gh
(c) Use the formula find the horizontal distance traveled, R:
R = vo * cos(θ) * t
But since the angle involved with respect to horizontal is zero, and cos(0) = 1, we have
R = vo * t
Hope this helps~ `u`
Jai
Anyone know how to do this???
Answer:
World War 1 was caused by entangled alliances, nationalism, imperialism, and major
advancements in military technology. Does the Treaty of Versaille address those issues?
Explain your answer using facts. (5 points)
A 1200kg car is moving down a 30° hill. The driver applies the
brakes at a time that the car's speed is 12m/s. What constant force F
must result if the car is to stop after travelling 100m?
Two type of microscopes used to view cells are optical and__ microscopes
options:
laser
Electron
A projectile was fired horizontally from a cliff 20m above the ground. If
the horizontal range of the projectile is 40m, calculate the initial velocity
of the projectile.
The initial velocity of the projectile is 19.8m/s
Explanation:
First, find time.
From our kinematics equations:
delta y = Vi•t + (1/2)at^2
rearrange,
t = sqrt[(2•delta y)/a]
t = sqrt[(2•20m)/9.8m/s^2]
t = 2.02s
Next, plug time into new kinematics equation to solve for the Vi in the x direction (horizontal)
delta x = Vi•t + (1/2)at^2
delta x = Vi•t
Rearrange:
Vi = delta x/t
Vi = 40m/2.02s
Vix = 19.8m/s
A potter's wheel is a uniform disk of mass 4.50 kg and radius 0.650 m and can spin freely around a vertical axis through its center. With the wheel spinning at an angular speed of 4.70 rad/s, a small piece of clay of mass 0.870 kg is dropped at the outer edge of the wheel and sticks to it. Find the final angular speed of the wheel clay. Treat the piece of clay as a point particle. Group of answer choices
Answer:
3.39 rad / s.
Explanation:
Given data:
mass of disk = 4.50 Kg
radius of wheel = 0.650 m
mass of the clay = 0.870 kg
The moment of inertial of the wheel = I = 4.5 kg x ( 0.65 m )2 / 2 = 0.95 kg . m2.
Now, applying the principle of angular momentum conservation :
Iω_i = ( I + mr2 )ω_f.
where ω_i = initial angular speed= 4.70 rad/s, ω_f = final angular speed
Hence, ω_f = Iω_i / ( I + mr2 )
= ( 0.95 kg . m2 x 4.7 rad / s ) / [ 0.95 kg . m2 + 0.87 kg x ( 0.65 m )2 ]
= 3.39 rad / s.
Hence, correct answer is : 3.39 rad / s.
Please help me!!!!!!!!!
Hi there! :)
[tex]\large\boxed{17.32 m/s}[/tex]
Use the following equation in solving for kinetic energy:
KE = 1/2mv² where:
KE = kinetic energy (J)
m = mass (kg)
v = velocity (m/s)
Plug in the given values:
12,000 = 40v²
Divide both sides by 40:
12,000 / 40 = v²
300 = v²
Take the square root of both sides:
√300 = v
v ≈ 17.32 m/s
Surface tension is often calculated using a machine that lifts a wire ring from the surface of a liquid. In this case the ring and liquid have some cohesive forces and attract rather than repel. In order to lift a ring of radius 2.75 cm off of the surface of a pool of blood plasma, a vertical force of 2.00*10-2 N greater than the weight of the ring is required. Consider the situation just before the ring breaks contact with the blood plasma where the blood plasma makes a contact angle of approximately zero degrees along the circumference of the ring and is stretched down vertically on both sides of the ring.
Required:
Calculate the surface tension of blood plasma from this information.
Answer:
0.116 N/m
Explanation:
Since the net force acting on the ring must be greater than 2.00 × 10⁻² N, and the surface tension T = F/L where F = net force = 2.00 × 10⁻² N and L = circumference of ring = 2πr where r = radius of ring = 2.75 cm = 2.75 × 10⁻² m.
So, T = F/L
= F/2πr
= 2.00 × 10⁻² N ÷ 2π(2.75 × 10⁻² m)
= 1/2.75π N/m
= 1/8.64 N/m
= 0.116 N/m
You are a member of an alpine rescue team and must get a box of supplies, with mass 3 kg , up an incline of constant slope angle 30.0, so that it reaches a stranded skier who is a vertical distance 4 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 0.05. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s2 .What is the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier
Answer:
v = 9.04 m / s
Explanation:
For this exercise we can use the relation that the work of the non-conservative force (friction) is equal to the variation of the mechanical energy of the system.
W = Em_f - Em₀ (1)
Starting point. Lower slope
Em₀ = K = ½ m v²
highest point. Where is the skier at a height h
Em_f = U = m g h
The work of rubbing
W = -fr x
the negative sign is because the friction force opposes the movement.
Let's set a reference system where the x axis is parallel to the slope and the y axis is perpendicular
let's use trigonometry to break down the weight
cos θ = W_y / W
sin θ = Wₓ / W
W_y = W cos θ
Wₓ = W sin θ
Y axis
N - Wₓ = 0
N = mg sin θ
X axis
fr = m a
the friction force has the expression
fr = μ N
fr = μ mg sin θ
we look for the job
W = - μ mg sin θ x
where x is the distance along the slope
we substitute in 1
-μ mg sin θ x = mg h - ½ m v²
let's use trigonometry to find the distance x
tan 30 = h / x
x = h / tan 30
we substitute
- [tex]\mu \ mg \ sin \theta \ \frac{h}{tan 30} \ x[/tex] = m gh - ½ m v²
we use
tan 30 = sin30 / cos30
v² = 2g h + 2 μ g h cos 30
v = [tex]\sqrt{ 2gh \ (1+ cos 30}[/tex]
let's calculate
v = [tex]\sqrt{ 2 \ 9.8 \ 4 \ (1 + 0.05 \ cos \ 30)}[/tex]
v = 9.04 m / s
A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 520 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tensions in the rope to the left and to the right of the mountain climber.
Answer:
The answer is "892.90 N"
Explanation:
Following are the solution to these question:
Calculating the vertical force of the summation that is equal to zero:
[tex]\to TL \cos 65 + TR \cos 80 -520 = 0\\\\\to 0.4226\ TL + 0.1736\ TR = 520\\[/tex]
Calculating the sum of horizontal forces that is equal to zero:
[tex]\to TL\sin 65 - TR \sin 80 = 0 \\\\\to 0.9063TL - 0.9846TR = 0\\\\\to TL = (\frac{0.9846}{0.9063})\ TR \ \ = 1.0866\ TR\\\\\to 0.4226(1.0866) \ TR +0.1736\ TR =520 \ N\\\\\to 0.6328 \ TR = 520 \\\\\to TR = 821.74 \ N \\\\\to TL = 1.0866 \times 821.74 = 892.90\ N[/tex]
A graduated beaker with 375 mL of water is sitting on a scale which measures the weight of the glass and water to be 7.60 N. When a rock is put into the glass, the volume level of the water changes to 450 mL and the scale reading changes to 9.22 N. What is the specific gravity of the rock
Answer:
Volume of water displaced = 450 - 375 = 75 ml
Vr = volume of rock = 75 ml
Wr = 9.22 - 7.60 = 1.62 N weight of 75 ml of rock
Density of rock = 1.62 N / 75 ml = .0216 N / ml
Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml
Density of rock / density of water = .0216 / .0098 = 2.20
The specific gravity of the rock in the given water volume is 0.2.
The given parameters;
initial volume of the water, = 375 mlweight of the water, = 7.6 Nfinal volume of water = 450 mlchange in scale reading = 9.22 NThe specific gravity of the rock is calculated as follows;
[tex]S.G = \frac{weight \ in \ air}{Weight \ in \ water} \\\\S.G = \frac{450 - 375}{375} \\\\S.G = 0.2[/tex]
Thus, the specific gravity of the rock in the given water volume is 0.2.
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Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is |v with arrow| = 230 km/s and the orbital period of each is 15.5 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 1030 kg.)
Answer:
[tex]1.554\times 10^{32}\ \text{kg}[/tex]
Explanation:
M = Mass of each star
T = Time period = 15.5 days
v = Orbital velocity = 230 km/s
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
Radius of orbit is given by
[tex]R=\dfrac{vT}{2\pi}[/tex]
We have the relation
[tex]\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}[/tex]
The mass of each star is [tex]1.554\times 10^{32}\ \text{kg}[/tex]
I need help please .
Answer:
option 5
Explanation:
because all u do is have to add them up
A vessel having a capacity of 0.05 m³ contains a mixture of saturated water an saturated steam at a temperature 245°C the mass of the liquid present is 10 kg. find the following: i- The pressure. ii- The mass. iii- The specific volume. iv- The specific enthalpy. v- The specific internal energy.
When the temperature of a certain solid, rectangular object increases by AT, the length of one
side of the object increases by 0.010% = 1.0 x 10-4 of the original length. The increase in volume
of the object due to this temperature increase is
Explanation:
iausiaolalalLosjjskskakOaokasksoososoapapozosiosoa
A farmer plants the same crop in a field year after year. Every year there are months during which the field is left without any plants. The farmer notices a decline in soil
quality during these months. What causes this decrease in soil quality?
Erosion
Drought
Desertification
Consumption of nutrients by the crops
Answer:
Drought
Explanation:
Please help I have no idea how to do this
Answer:
So an object with mass is attracted to another object with mass, and the gravitational force is directly proportional to the masses of the two objects, and inversely proportional to the square of the distance between the two objects.
If distance were to increase, than the gravitational force would decrease. If mass were to increase, so would the gravitational force.
Explanation:
Explanation:
Lets take gravitational force(F) and mass(m) and distance (r)
now for a body in contact with the surface of the earth, its mass is also considered(m‘),now the mass of the earth(m") is also considered,then the distance from the body to the center of the earth (r).ie it r because its practically the radius of the earth. is also considered
So by using dimensional analysis ....
we get F a m'•m"/r² ,where a is proportional to.
now since F is directly increaseproportional to m ie. F a m, then an increase in mass of the body increases it's gravitational force(and clearly that makes sense because the bigger you are the stronger you get pulled to the ground)
then we also see that F is inversely proportional to r ie.F a 1/r ,then an increase in the distance between the ground an the object decrease it's gravitational force ( meaning as any object on earth keeps on moving away from the ground the gravitational force between the object and the center of the earth is weak, when it reaches space then the force becomes virtually negligible!)
So to answer the second question, we clearly see that doubling the mass of the body increases the gravitational force between it and the earth
and doubling the distance on the other hand will decrease the attraction between the body and the earth
So a body forcefully projected into the air fights against gravity but its easier as it keeps on getting higher, If it has a greater mass like that of a trail or truck , it will not even probably stay in the air for long , unless its projected with a very high velocity
I hope this helps, and you can ask me any question concerning this via the comments platform.
A train crosses 650m long bridge and 800m long platform in 20sec and 30 sec respectively. what is speed of train?
Answer:
29 m/s
Explanation:
Assuming the bridge and the platform are back to back,
Average speed = Total Distance/ Total Time
Avg S = 1,450m/50s = 29 m/s
The average speed of the train is 29 m/s
The average speed is defined as the ratio of total distance traveled and the total time taken.
Total distance = 650 m + 800 m = 1450 m
Total Time = 20 s + 30 s = 50 s
Average speed = total distance/total time
= 1450 / 50
= 29 m/s is the average speed of the train.
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HELP ITS DUE IN 4 MINUTES
Answer:
igneous = melted rocks formed by cooled magma
sedimentary is brken rocks, kayers with fossil...
metamophic is rocks formed by pressure and heat
What is the force of gravity between two 40.0kg masses that are separated by 3.00m?
Answer:
[tex]f = g \times \frac{m1 \times m2}{ {d}^{2} } [/tex]
[tex]f = 6.67 \times {10}^{ - 11} \times \frac{40 \times 40}{9} [/tex]
F=1.2x 10^-82.3 The motion of an object is accelerated
when its speed:
a
decreases
b remains constant
increases.
Answer:
The motion of an object is accelerated when its speed increases.
1. Describe the components of the reflex arc
Please help !!!!!!!!!!!!!!!
Answer:
9v
Explanation:
What kind of weather would MOST LIKELY lead to hurricane formation?
Answer:
The recipe for a hurricane is a combination of warm, humid wind over tropical waters. The temperature of tropical waters must be at least 80 degrees F for up to 165 feet below the ocean’s surface. As this warm water meets the wind that blows west from Africa across the ocean, it causes the water to vaporize. The water vapor then rises into the atmosphere, where it cools and liquefies.
Explanation:
I hope this helps
Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.40 cm. If the potential difference across the plates was 23.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.
Answer:
E = 1.64 x 10⁶ V/m
Explanation:
The electric field in the region between the plates can be given by the following formula:
[tex]E = \frac{\Delta V}{d}[/tex]
where,
E = Electric Field = ?
ΔV = Poetential Difference across the plates = 23 KV = 23000 V
d = distance between plates = 1.4 cm = 0.014 m
Therefore, using these values in the equation, we get:
[tex]E = \frac{23000\ V}{0.014\ m}[/tex]
E = 1.64 x 10⁶ V/m
How much power is required to carry a 35N
package a vertical distance of 18 m if the work on
the package is accomplished in 30 s?
Explanation:
force=35
distance=18
time=30
power=f×d/t
p=35×18/30
p=21