Answer:
Centripetal acceleration = 83.77m/s²
Explanation:
Given the following data;
Radius, r = 0.13m
Velocity, v = 3.3m/s
To find centripetal acceleration;
Centripetal acceleration is given by the formula;
[tex] Acceleration, a = \frac {v^{2}}{r}[/tex]
Substituting into the equation, we have;
[tex] Centripetal \; acceleration, a = \frac {3.3^{2}}{0.13}[/tex]
[tex] Centripetal \; acceleration, a = \frac {10.89}{0.13}[/tex]
Centripetal acceleration = 83.77m/s²
Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s².
Answer:
the answer is 84
Explanation:
A solid cylinder is released from the top of an inclined plane of height 0.50 m. From what height, in meters, on the incline should a solid sphere of the same mass and radius be released to have the same speed as the cylinder at the bottom of the hill?
Answer:
Explanation:
for rolling motion down the plane acceleration is given by the following expression
a = g sinθ / (1 + k² / R²)
here k is radius of gyration and R is radius of the object rolling down .
for cylinder I = 1/2 m R²
so k² = R² / 2
k² / R² = 1/2
a = g sinθ /( 1 + 1 / 2 )
= 2 / 3 x g sinθ
v = √ 2 a s
= √ (2 x 2 / 3 x g sinθ s )
= √ (4 / 3 x g h )
= √ (4 / 3 x g x .5 )
= √ 2g / 3
for sphere I = 2/5 m R²
so k² = 2/5 R²
k² / R² = 2 / 5
a = g sinθ / (1 + 2 / 5)
= 5 / 7 x g sinθ
v = √ 2 a s
= √ (2 x 5 / 7 x g sinθ s )
= √ (10/7 x g h )
Given
√ (10/7 x g h ) = √ 2g / 3
10/7 x g h = 2g / 3
h = 14 / 30 m
= .47 m .
Which of the following is true?
A
The Atlantic, Pacific, Indian, Arctic, and Southern Oceans are completely separate
from each other.
B
The ocean covers about half of the Earth's surface.
с
Scientists have studied most of the ocean, but a tiny bit remains unexplored.
D
Scientists know more about the moon than they do the ocean.
Answer:
options B,C,D are true
Explanation:
Question 25
The drawings below show four objects that are acted upon by different forces The direction of the forces are represented by the arrows, and the units of the forces are in newtons (N) Based on Newton's law of
inertia, which object is moving at a steady speed, assuming that all four objects are moving on a frictionless surface?
Answer:
The one with equal forces on both sides, D is your answer
Explanation:
A, 5 and 10 No
B 13 and 8 no
C 5 and 8 no
D 6 and 6 YES
Object 4 have steady speed. Object 4 have same force on both the sides of object. Same force is applied on both the sides.
What is steady speed?Uniform or constant velocity describes a body's motion when there is no acceleration. A body's velocity is its rate of movement in a specific direction.
When running steadily, your exhaustion should be caused by the length of the run, not by your speed. A steady-state run pace, according to the McMillan Running website, is halfway between your current 30k and half-marathon pace. This speed should allow you to run for 25 to 75 minutes.
Therefore, Object 4 have steady speed. Object 4 have same force on both the sides of object. Same force is applied on both the sides.
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A 80 kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 5 meters starting from rest, its speed is 6.0 m/s. Find the magnitude of the net force on the bobsled.
How do you solve this question?
Answer:
F = 288 [N]
Explanation:
To solve this problem we must use the following equation of kinematics and find the value of acceleration.
[tex]v_{f}^{2} =v_{o}^{2} +2*a*x[/tex]
where:
Vf = final velocity = 6 [m/s]
Vo = initial velocity = 0 (starting from rest)
a = acceleration [m/s²]
x = distance = 5 [m]
Now replacing, we have:
[tex](6)^{2}=0+(2*a*5)\\36=10*a\\a = 3.6 [m/s^{2}][/tex]
Since we already have the value of acceleration, we can use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
ΣF = m*a
[tex]F =80*3.6\\F = 288 [N][/tex]
45.0g of a sample placed in a graduated cylinder causes the water level to rise from
25.0mL to 40.0mL. Density of the sample is?
Answer:
3g/mL
Explanation:
The density of a substance can be calculated using the formula;
Density = mass/volume
Where;
Density = g/mL
mass = grams (g)
volume = (mL)
According to this question, 45.0g of a sample placed in a graduated cylinder causes the water level to rise from
25.0mL to 40.0mL. This means that the volume of the sample is 40mL - 25mL = 15mL
Using D = m/v
D = 45/15
D = 3g/mL
Hence, the density of the sample is 3g/mL
Which of the following is not an example of work being done on an object?
Pushing on a rock that will not move
Paddeling a canoe down a river
Lifting a bag of groceries
Throwing a ball across a field
Answer:
Lifting a bag of groceries
Answer:
paddeling a canoe down a river :D or throwing a ball across a field
Explanation:
if a person has a mass of 60 kg and a velocity of 2 m/s what is the magnitude of his momentum
Answer:
120 kg m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 60 × 2
We have the final answer as
120 kg m/sHope this helps you
PLZ HELP ILL MARK BRAINLEIST!!!!
Why did the bowling ball make a bigger
splash than the ping pong ball?
What kind of energy made that splash
happen?
Answer:
a. The bowling ball would have more kinetic energy because of its greater mass.
b. Potential energy
Explanation
a. Bowling ball has higher mass, self explanatory.
b. A high diver has lots of stored energy when they are on the diving platform. When they dive this stored energy helps make the splash when they hit the water. Stored energy is also called potential energy.
Was the Big Bang a loud explosion? Why?
Answer:
bc it was a universal explosion and It started the future
Explanation:
FACTS
Answer:
i wouldn't believe so.
Explanation:
because there was no room or air for the sound to move through. this is because of immense heat and the amount of hyperactive neutrons, electrons and protons clouding everywhere. This would mean that even if there was sound it would a. not travel far or b. go in a completely different direction than expected.
I need help understanding this question, so I know the arrow is traveling 80 meters per second, but it was launched from a starting point of 32 meters. I know for a fact an arrow does not have any thrust left at around 3 seconds of being in the air.
I just need someone to explain the questions and provide an answer to each.
Answer:
a) h(g) = 358,53 m
b) t = 8,16 s
c) t(t) = 16,71 s
Explanation:
Equations for vertical shooting are:
Vf = V₀ - g * t ; h = V₀*t - (1/2)*g*t² ; Vf² = V₀² - 2*g*h
And at maximum heigt Vf = 0 then
0 = V₀ - g * t
t = V₀/g V₀ = 80 m/s and g = 9,8 m/s²
t = 80 / 9,8 (s)
t = 8,16 s
Then 8,16 s is the time to get maximum height
If we plug t = 8,16 (s) in equation h = V₀*t - (1/2)*g*t²
we get: h (max) = (80)*8,16 - 0,5*9,8*(8,16)² (m)
h (max) = 652,8 - 326,27 m
h (max) = 326,53 m
Then relative to ground that height becomes
h(g) = 326,53 + 32
h(g) = 358,53 m
In order to get the time the arrow is in the air we proceed as follows:
a) for the arrow to be at the launched point will take the same time that from the launched point to the maximum height, and after that we have to find out the time the arrow takes from 32 m down to the ground level
Then
t(t) = 8,16 + 8,16 + tₓ (2)
Where tₓ is the time from 32 m height to ground
h = V₀*tₓ - (1/2)*g*tₓ² but since the arrow now is going down then we change the sign of the second term on the right side of the equation
32 = (80)*tₓ + 0,5 * 9,8 * tₓ² Note that when the arrow is at 32 m height the speed is again V₀ = 80 m/s
32 = 80*tₓ + 4,9*tₓ²
A second-degree equation for tₓ, solving it
4,9*tₓ² + 80*tₓ - 32 = 0
t₁,₂ = -80 ± √ 6400 + 627,2 / 9,8
t₁,₂ =( - 80 ± 83,8 ) / 9,8
there is not a negative time therefore we dismiss such solution and
t₁ = 3,8 / 9,8
t₁ = 0,39 s
And
t(t) = 8,16 + 8,16 + 0,39 s
t(t) = 16,71 s
The students look through the side of the aquarium.
They notice that the image of the tongs appears to break as the tongs enter the water.
Which property of light are the students observing in this situation?
Answer:
light refraction
Explanation:
The coulombic force between two ions is reduced to ______ of its original strength when the distance between them is quadrupled.
Answer:
1/16Explanation:
According to the coulombs law, the force existing vetween the ions is expressed as;
F = kQq/r² .... 1
Q and q are the ions
r is the distance between the ions
If the distance between the ion is quadrupled, then;
F2 = kQq/(4r)²
F2 = kQq/16r² ... 2
Divide equation 2 by 1;
F2/F = kQq/16r² ÷ kQq/r²
F2/F = kQq/16r² × r²/kQq
F2/F = 1/16
F2 = 1/16 F
Therefore the coulombic force between two ions is reduced to 1/16 of its original strength when the distance between them is quadrupled.
A (10.0+A) g ice cube at -15.0oC is placed in (125 B) g of water at 48.0oC. Find the final temperature of the system when equilibrium is reached.
Specific heat of ice: 2.090 J/g K
Specific heat of water: 4.186 J/gK
Latent heat of fusion for water: 333 J/g
Answer: Final temperature is 34.15°C.
Explanation: When two objects have different temperature, they will exchange heat energy until there is no more net energy transfer between them. At that state, the objects are in thermal equilibrium.
So, when in equilibrium, the total heat flow must be zero, i.e.:
[tex]Q_{1}+Q_{2}=0[/tex]
In our case, there will be a change in state of ice into water, so total heat flow will be:
[tex]m_{1}c_{1}(T_{f}-T_{i})+m_{2}c_{2}(T_{f}-T_{i})+mL=0[/tex]
where
m₁ is mass of ice
m₂ is mass of water
c₁ is specific heat of ice
c₂ is specific heat of water
[tex]T_{f}[/tex] is final temperature
[tex]T_{i}[/tex] is initial temperature
L is latent heat fusion
Temperature is in Kelvin so the transformation from Celsius to Kelvin:
For ice:
T = -15 + 273 = 258K
For water:
T = 48 + 273 = 321K
Solving:
[tex]21(2.09)(T_{f}-258)+158(4.186)(T_{f}-321)+21(333)=0[/tex]
[tex]43.89T_{f}-11323.62+661.4T_{f}-212305.55+6993=0[/tex]
[tex]705.3T_{f}=216636.17[/tex]
[tex]T_{f}=[/tex] 307.15K
In Celsius:
[tex]T_{f}=[/tex] 34.15°C
Final temperature of the system when in equilibrium is 34.15°C
If Jack weighs more that Jill, and they run up the same hill, who has done more work?
If Jack and Jill weigh the same, and Jill runs up the hill in half the time as Jack, who had more power?
Answer:
jack has done more work pulling more weight and Jill has more power.
Explanation:
PLEASE ans The question's in the pictures, please don't answer what already has answers. Only answer if you can finish both pages completely PLEASE I NEED HELP :(( if ur ans is relevant I will mark brainliest
The fact that our preconceived ideas contribute to our ability to process new information best illustrates the importance of: the serial position effect. O repression iconic memory . semantic encoding . retroactive interference .
Answer:
It’s a
Explanation:
Don’t actually put that i needed the points mb
What is the rms current flowing through a light bulb that uses an average power of 60.0 W when it is plugged into a wall receptacle supplying an rms voltage of 120.0 V?
Answer:
I = 0.5 A
Explanation:
Given that,
Average power of a light bulb, P = 60 W
The rms voltage supplied is 120 V
We need to find the value of rms current flowing through the light bulb. The relation between the power, voltage and current is given by :
P = VI
[tex]I=\dfrac{P}{V}\\\\I=\dfrac{60}{120}\\\\I=0.5\ A[/tex]
So, the rms value of current is 0.5 A.
Any living thing is called an organism,no matter if it is one-celled or many-celled. True or False?.
Answer:
I think it's most likely true.
Explanation:
any organism has the properties of a living thing, which includes cells, whether it has one cell or many
Answer:
False
Explanation:
An organism is a living thing that is a single-celled life form
While getting buff at the gym you lift a bunch of weights applying 1000N of force to lift them from the ground to a height of 2m. How much work did you do?
A. 2000 J
B. 1000J
C. -2000 J
D. -1000 J
A rotating heavy wheel is used to store energy as kinetic energy. If it is designed to store 1.00 x 106 J of kinetic energy when rotating at 64 revolutions per second, find the moment of inertia (rotational inertia) of the wheel. (Hint: Start with the expression for rotational kinetic energy.)
We know, [tex]1\ rpm = \dfrac{2\pi}{60} \ rad/s[/tex] .
[tex]64\ rpm\ is = \dfrac{2\pi}{60}\times 64\ rad/s\\\\= \dfrac{32\pi}{15}\ rad/s[/tex]
We know, kinetic energy is given by :
[tex]K.E = \dfrac{I\omega^2}{2}\\\\I = \dfrac{2(K.E)}{\omega^2}\\\\I = \dfrac{2\times 10^6}{\dfrac{32}{15}\times \pi}\\\\I = 298415.52 \ kg \ m^2[/tex]
Hence, this is the required solution.
A child pulls a wagon across the grass so that it accelerates using a force of 50 N at an angle of 42 degrees above the ground. The loaded wagon has a mass of 12 kg. If the coefficient of friction between the wagon and grass is 0.64. What is the acceleration of the wagon? Describe the motion of the wagon.
Answer:
[tex]-1.398\ \text{m/s}^2[/tex]
Decelerating or slowing down
Explanation:
F = Force = 50 N
[tex]\theta[/tex] = Angle force is being applied = [tex]42^{\circ}[/tex]
[tex]\mu[/tex] = Coefficient of friction = 0.64
m = Mass of wagon = 12 kg
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Normal force is given by
[tex]N=mg-F\sin\theta[/tex]
Frictional force is given by
[tex]f=\mu N\\\Rightarrow f=\mu (mg-F\sin\theta)[/tex]
The force balance is given by
[tex]F\cos\theta-f=ma\\\Rightarrow \dfrac{F\cos\theta-\mu (mg-F\sin\theta)}{m}=a\\\Rightarrow a=\dfrac{50\times \cos42^{\circ}-0.64(12\times 9.81-50\times\sin42^{\circ})}{12}\\\Rightarrow a=-1.398\ \text{m/s}^2[/tex]
The acceleration of the wagon is [tex]-1.398\ \text{m/s}^2[/tex]. The negative sign indicates that the wagon is decelerating or slowing down.
The acceleration of the wagon is [tex]-1.398 \;\rm m/s^{2}[/tex].
Given data:
The magnitude of pulling force is, F = 50 N.
The angle of inclination is, [tex]\theta = 42^{\circ}[/tex].
The mass of wagon wheel is, m = 12 kg.
Coefficient of friction between wagon and grass is, [tex]\mu =0.64[/tex].
The given problem is based on the concept of frictional force. The standard expression for the frictional force is,
[tex]f= \mu \times N[/tex]
Here, N is the normal force and its value is,
[tex]N=mg-Fsin \theta[/tex]
And the net force acting on wagon is,
[tex]F' = Fcos\theta -f\\\\ma = Fcos\theta -(\mu(mg-Fsin \theta))\\\\a = \dfrac{Fcos\theta -(\mu(mg-Fsin \theta))}{m}[/tex]
Here, a is the acceleration of wagon.
Solving as,
[tex]a = \dfrac{50 \times cos42 -(0.64(12 \times 9.8-(50 \times sin42)))}{12}\\\\a=-1.398 \;\rm m/s^{2}[/tex]
Thus, we can conclude that the acceleration of the wagon is [tex]-1.398 \;\rm m/s^{2}[/tex].
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What is the flow sensitivity of a biosensor?
Answer:
Sensitivity of biosensor
The biosensor showed good linear correlation in the wide detection range of 0.001–2000 ng/mL with good sensitivity. In addition, it retained its biosensing property for seven days with high reproducibility
Explanation:
Which of the following is the recommended amount of fats per meal for male clients
Answer:
44 grams- 55 grams through the whole day. Probably about 14.6 grams per meal.
Explanation:
Answer:
2 thumbs (ISSA Guide)
Explanation:
which of the following to all food chains depend on in an ecosystem
Answer:
The sun is the ultimate source of energy for all food chains. Through the process of photosynthesis, plants use light energy from the sun to make food energy. Energy flows, or is transferred through the system as one organism consumes another.
*Urgent* I WILL GIVE BRAINLIEST
Select the answer that helps conserve the most energy.
O walking to school
O driving a car to school
Otaking the bus to school
Answer:
walking to school
Explanation:
Driving a car to school
, and taking the bus to school both take up energy, unlike walking to school.
unless ur talking about energy, counting energy you produce and use to complete things, then it would be the 3rd one, taking the bus to school.
Please help true or false
Answer:
the answer is true.......
A wave travels at 175 m/s along the x-axis.If the period of the periodic vibrations of the wave is 3.00 milliseconds,then what is the wavelength of the wave?
A) 25.5 cm
B) 35.6 cm
C) 42.9 cm
D) 49.5 cm
E) 52.5 cm
Answer:
E) 52.5 cmExplanation:
Step one:
given data
period T= 3 milliseconds= 0.003
velocity v= 175m/s
wave lenght λ=?
Step two:
we know that f=1/T
the expression relating period and wave lenght is
v=λ/T
λ=v*T
λ=175*0.002
λ=0.525m
to cm= 0.525*100
=52.5cm
The wavelength of the wave is E) 52.5 cm
Thorium^+2
Chemical symbol:
Atomic Number:
Mass: 232
# of protons
# of neutrons
Group #
Period #
Answer:
chemical symbol: Th
atomic number:90
protrons :90
neutrons:142
group#:4
period#: 9
Explanation:
you take the atomic weight (232.038)and subtract the atomic number to get (90) which is your neutrons
Two blocks of masses 1.0 kg and 2.0 kg, respectively, are pushed by a constant applied force F across a horizontal frictionless table with constant acceleration such that the blocks remain in contact with each other, as shown above. The 1.0 kg block pushes the 2.0 kg block with a force of 2.0 N. The acceleration of the two blocks is
0
1.0 m/s2
1.5 m/s2
2.0 m/s2
3.0 m/s2
Answer:
1.0 m/s^2
Explanation: happy to help :)
Answer: [tex]1\ m/s^2[/tex]
Explanation:
Given
Masses of the block are [tex]m_1=1\ kg[/tex] and
[tex]m_2=2\ kg[/tex]
Force applied by [tex]1\ kg[/tex] block on [tex]2\ kg[/tex] block is [tex]2\ N[/tex]
From the free body diagram of [tex]2\ kg[/tex] block, the net force on
[tex]\therefore m_2a=2\\\\\Rightarrow 2\times a=2\\\\\Rightarrow a=\dfrac{2}{2}\\\\\Rightarrow a=1\ m/s^2[/tex]
Thus, the acceleration of two blocks is [tex]1\ m/s^2[/tex]
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Collision Lab
This activity will help you meet these educational goals:
You will explain or predict phenomena by exploring qualitative relationships between variables.
You will use positive and negative numbers to represent quantities in real-world contexts.
Directions
Read the instructions for this self-checked activity. Type in your response to each question, and check your answers. At the end of the activity, write a brief evaluation of your work.
Activity
Open this collision simulator and click Introduction. You’ll use the simulator to explore and compare elastic collisions and inelastic collisions. The mass and starting velocity of the colliding objects are kept constant. Follow the instructions in each part, and then answer the questions that follow. Use the math review if you need help with adding and subtracting negative numbers.
Question 1: Elastic Collisions
In this question, you will investigate elastic (bouncy) collisions. Be sure that the slider is to the extreme right (elasticity 100%).
Part A
Click Show Values in the upper-right corner. Study the boxes on the screen. What are the mass and initial velocity of ball 1 and ball 2?
I NEED HELP!
Part B
Part B
Click Play, and watch the balls collide. Then click Pause. What are the final velocities of ball 1 and ball 2?
The number line shows the starting and ending velocities for ball 1. What’s the change in velocity of ball 1? Calculate the value mathematically, and check it using the number line.
a number line showing an ending velocity of -0.50 meter/second and a starting velocity of 1.00 meter/second
Answer:
Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.
Explanation:
Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.
What is Collision?
A collision is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.
The following are a few instances of physical encounters that scientists might classify as collisions. Legs of an insect are said to collide with a leaf when it falls on one.
Every contact of a cat's paws with the ground while it strides across a lawn is seen as a collision, as is every brush of its fur with a blade of grass.
Therefore, Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.
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