Answer:
[tex]W_F=127.64283 J[/tex]
Explanation:
Information Given:
[tex]m = 5kg[/tex] [tex]v=constant[/tex]
Key: μ = Kinetic Friction (Kf) θ = Theta α = 180° N = Normal Force
[tex]W_F=F_ydcos[/tex]θ
[tex]W_F=Fdsin[/tex]θ
[tex]_{net}F_y = sin[/tex]θ-μ[tex]N-mg=0[/tex]
[tex]_{net}F_x = 0[/tex]
[tex]N=Fcos[/tex]θ
[tex]Fsin[/tex]θ-μ[tex]N=mg[/tex]
[tex]Fsin[/tex]θ-μ[tex]Fcos[/tex]θ[tex]=mg[/tex]
[tex]F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}[/tex] →[tex]W_F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}[/tex]
[tex]W_F=\frac{(2)(9.81)(2)sin(30)}{sin(30)-(0.40)cos(30)}[/tex]
[tex]W_F=127.64283 J[/tex]
What causes friction between two solids?
Answer:
Friction is when 2 solids move against each other. The cause of friction is adhesion, and surface roughness. Surface roughness is when a surface is rough enough that is causes friction against another surface. Adhesion is when 2 surfaces collide because of thier molecular force.
3
4
Lucy runs 4 meters to the east, then 5 meters south. What is the magnitude of her displacement?
Show Your Work
Answer:
The displacement is 6.4m
Explanation:
Step one:
given
we are told that Lucy runs 4 meters to the east,
then 5 meters south.
let the distance east be the displacement in the x-direction, and south be the y-direction
Step two:
The resultant of the x and y displacement is the magnitude of the total displacement z
applying Pythagoras theorem we have
z=√x^2+y^2
z=√4^2+5^2
z=√16+25
z=√41
z=6.4m
a mass of 2.00 kg rest on a rough horizontal table. The coefficient of static friction between the block and the table is 0.60. The block is attached to a hanging mass by a string that goes over a smooth pulley,as shown in the diagram. Determine the largest mass that can hang in this way without forcing the block to slide.
Answer:
1.2 kg
__________________________________________________________
We are given:
Mass of the block = 2 kg
Coefficient of Static Friction = 0.6
__________________________________________________________
Friction Force on the Block:
Finding the Normal Force:
We know that the normal force will be equal and opposite to the weight of the 2 kg block
So, Normal Force = mg
replacing the variables with the given values
Normal Force = (2)(9.8) [Taking g = 9.8]
Normal Force = 19.6 N
Friction force on the Block:
We know that:
Coefficient of Static Friction = Static Friction Force/Normal Force
replacing the variables
0.6 = Static Friction force / 19.6
Static Friction force = 0.6*19.6 N [Multiplying both sides by 19.6]
Static Friction force = 11.76 N
__________________________________________________________
Largest Mass that can Hang:
We know that the Static Friction force is 11.76 N, this means that a force of 11.76 N will be applied to keep the object at rest
So, if the weight of the second block is less than the static friction force, it will hang
Weight of the second block ≤ 11.76
We know that weight = mg
mg ≤ 11.76
m(9.8) ≤ 11.76 [since g = 9.8]
m ≤ 1.2 kg [dividing both sides by 9.8]
From this, we can say that the maximum mass of the second block is 1.2 Kg
When weather predictions are incorrect what is the most likely cause
A: measurements of the initial conditions may have been very in accurate
B: small differences in models can lead to large differences in complex systems
C: The person predicting the weather may have had a bias
D: The elevation of different landforms I have been significantly in accurate
Answer:small differences in models can lead to large differences in complex systems
Explanation: this is the most accurate phrase
turns into _____ energy (when plants convert into food)
Answer:
its called photosynthesis
Explanation:
when plants turn food into energy. Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy and stored in the form of starch which can be used later.
Answer:
The answer would be photosynthesis
Explanation:
In this case plants convert light energy (1) into chemical energy, (in molecular bonds), through a process known as photosynthesis. Most of this energy is stored in compounds called carbohydrates. The plants convert a tiny amount of the light they receive into food energy. :)
Find the binding energy per nucleon for the plutonium isotope 239Pu. The mass of the neutral atom is 239.05216 u.
Answer:
The answer is "[tex]\bold{7.56 \ Me\ V}[/tex]".
Explanation:
calculating the binding energy on per nucleon:
calculating number of proton and neutrons:
proton [tex]P_u=94[/tex]
neutron[tex]= 239-94=145[/tex]
calculating mass:
proton mass [tex]\ m_P=1.007825 \ amu\\\\[/tex]
neutron mass [tex]\ m_n=1.008665 \ amu\\\\[/tex]
neutral atom mass [tex]m = 239.05216 \ amu\\\\[/tex]
mass of prtons[tex]= 94 \times 1.007825 = 94.73555 \ amu\\\\[/tex]
mass of neutrons[tex]= 145 \times 1.008665= 146.256425 \ amu\\\\[/tex]
Total nucleons mass formula:
[tex]\to m_n = (P+n)[/tex]
[tex]= 94.73555+ 146.256425\\\\= 240.991975 \ amu[/tex]
calculating the mass of defect:
[tex]\to \Delta m= m_n-m\\\\[/tex]
[tex]= 240.991975 - 239.05216\\\\= 1.939815 \ amu\\\\[/tex]
calculating the total of the binding energy:
[tex]\to BE=\Delta m\times 931.5 \ mev[/tex]
[tex]= 1.939815 \times 931.5\\\\=1806.938 \ Me \ V\\\\[/tex]
BE in per nucleon [tex]=\frac{BE}{239}= 7.56 \ Me\ V[/tex]
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
Bobbie is on a skateboard riding it down a hill. As he approaches the hill his velocity is 26 m/s. While he travels down the hill he realizes his acceleration is 6 m/s2. At the bottom of the hill he finds out that his velocity is 102 m/s. What time will be required to create this change in velocity?
Answer:
Bobbie will need 12.7 seconds.
Explanation:
Constant Acceleration Motion
It's a type of motion in which the velocity of an object changes uniformly in time. If the speed changes from vo to vf in a time t, then the acceleration is:
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
The question states Bobbie is riding on a skateboard down a hill, starting from vo=26 m/s with an acceleration of a=6\ m/s^2 until his speed is 102 m/s. We are required to find the time needed to create that change of velocity.
Solving for t:
[tex]\displaystyle t=\frac{v_f-v_o}{a}[/tex]
[tex]\displaystyle t=\frac{102-26}{6}=\frac{76}{6}[/tex]
t = 12.7 s
Bobbie will need 12.7 seconds.
one newton equals 0.225
in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s
Answer:25 N
Explanation:
in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s is 25N
what is speed ?Speed is the ratio of distance with respect to the time in which the distance was covered. Speed is a scalar quantity as it does not have magnitude only have direction
The formula of speed can be represented as s=d/t, Where, s is the speed in m.s-1, d is the distance traveled in m, t is the time taken in s
Uniform speed is defined when the object covers equal distance at equal time intervals, variable speed is defined as when the object covers a different distance at equal intervals of times.
Average speed is defined as the total distance travelled by an object to the total time taken by the object.
Instantaneous speed is defined as when the object is move with variable speed, then the speed at any instant of time is known as instantaneous speed.
For more details regarding speed, visit
brainly.com/question/13263542
#SPJ5
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
A disk with a rotational inertia of 2.0 kg m2 and a radius of 1.6 m is free to rotate about a frictionless axis perpendicular to the disk's face and passing through its center. A force of 5.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk as a result of this applied force?
a) 4.0 rad/s2
b) 1.0 rad/s2
c) 2.0 rad/s2
d) 0.40 rad/s2
e) 0.80 rad/s2
Answer:
a) 4.0 rad/s2
Explanation:
For rigid bodies, Newton's 2nd law becomes :τ = I * α (1)
where τ is the net external torque applied, I is the rotational inertia
of the body with respect to the axis of rotation, and α is the angular
acceleration caused by the torque.
At the same time, we can apply the definition of torque to the left side of (1), as follows:[tex]\tau = F*r*sin \theta (2)[/tex]
where τ = external net torque applied by Fnet, r is the distance
between the axis of rotation and the line of Fnet, and θ is the
angle between both vectors.
In this particular case, as Fnet is applied tangentially to the disk, Fnet
and r are perpendicular each other.
Since left sides of (1) and (2) are equal each other, right sides are equal too, so we can solve for the angular acceleration as follows:[tex]\alpha = \frac{F*r}{I} = \frac{5.0N*1.6m}{2.0 kg*m2} = 4.0 rad/s2 (3)[/tex]
How to calculate net radiation
Answer:
(1) R n = ( 1 − α ) R si − L ↑ + L ↓ where Rn is the net radiation (W m−2), Rsi is the solar radiation (W m−2), α is the soil surface albedo (α = 0–1
Explanation:
How high does a rocket have to go above the earth's surface to be subject to a gravitational field from the earth that is 50.0 percent of its value at the earth's surface?
A) 2.650 km
B) 3,190 km
C) 9.020 km
D) 12.700 km
Answer:
A) 2.650 km
Explanation:
The relationship between acceleration of gravity and gravitational constant is:
[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)
Where
[tex]R = 6,400 km[/tex] -- Radius of the earth.
From the question, we understand that the gravitational field of the rocket is 50% of its original value.
This means that:
[tex]g_{rocket} = 50\% * g[/tex]
[tex]g_{rocket} = 0.50 * g[/tex]
[tex]g_{rocket} = 0.5g[/tex]
For the rocket, we have:
[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]
Where r represent the distance between the rocket and the center of the earth.
Substitute 0.5g for g rocket
[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)
Divide (1) by (2)
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]
[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]
[tex]2 = \frac{r^2}{R^2}[/tex]
Take square root of both sides
[tex]\sqrt 2 = \frac{r}{R}[/tex]
Make r the subject
[tex]r = R * \sqrt 2[/tex]
Substitute [tex]R = 6,400 km[/tex]
[tex]r = 6400km * \sqrt 2[/tex]
[tex]r = 6400km * 1.414[/tex]
[tex]r = 9 049.6\ km[/tex]
The distance (d) from the earth surface is calculated as thus;
[tex]d = r - R[/tex]
[tex]d = 9049.6\ km - 6400\ km[/tex]
[tex]d = 2649.6\ km[/tex]
[tex]d = 2650\ km[/tex] --- approximated
Determine the distance between a newly discovered planet and its single moon if the orbital period of the moon is 1.2 Earth days and the mass of the planet it orbits is 9.38E24 kg.You may assume the orbit to be circular.
Answer:
The distance is [tex]r = 55430496 \ m[/tex]
Explanation:
From the question we are told that
The period of the moon [tex]T = 1.2 days = 1.2 * 24 * 3600 = 103680 \ s[/tex]
The mass of the planet is [tex]m_p = 9.38*10^{24} kg[/tex]
Generally the period of the moon is mathematically represented as
[tex]T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }[/tex]
Here G is the gravitational constant with value
[tex]G = 6.67 *10^{-11} \ N \cdot m^2/kg^2[/tex]
=> [tex]T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }[/tex]
=> [tex]103680 = 2 * 3.142 * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }[/tex]
=> [tex]272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}[/tex]
=> [tex]r = \sqrt[3]{ 1.7031241*10^{23}}[/tex]j
=> [tex]r = 55430496 \ m[/tex]
find the vector parallel to the resultant of the vector A=i +4j-2k and B=3i-5j+k
Answer:
2008
Explanation:
2000+3+5======2008
Answer:
[tex]8\hat i-2\hat j-2\hat k[/tex]
Explanation:
Vectors in 3D
Given a vector
[tex]\vec P = P_x\hat i+P_y\hat j+P_z\hat k[/tex]
A vector [tex]\vec Q[/tex] parallel to [tex]\vec P[/tex] is:
[tex]\vec Q = k.\vec P[/tex]
Where k is any constant different from zero.
We are given the vectors:
[tex]\vec A = \hat i+4\hat j-2\hat k[/tex]
[tex]\vec B = 3\hat i-5\hat j+\hat k[/tex]
It's not specified what the 'resultant' is about, we'll assume it's the result of the sum of both vectors, thus:
[tex]\vec A +\vec B = \hat i+4\hat j-2\hat k + 3\hat i-5\hat j+\hat k[/tex]
Adding each component separately:
[tex]\vec A +\vec B = 4\hat i-\hat j-\hat k[/tex]
To find a vector parallel to the sum, we select k=2:
[tex]2(\vec A +\vec B )= 8\hat i-2\hat j-2\hat k[/tex]
Thus one vector parallel to the resultant of both vectors is:
[tex]\mathbf{8\hat i-2\hat j-2\hat k}[/tex]
7. DRAW A PICTURE TO SHOW WORK.
Brandon buys a new Seadoo. He goes 12
km north from the beach. He jumps
wakes for 6 km to the east. Then chases
a boat 12 km south. He then turns and
goes 3 km to the West. What distance
did he cover? What was his
displacement?
What is the initial vertical velocity of the ball?
A.
0 m/s
B.
9.81 m/s
C.
20.0 m/s
D.
60.0 m/s
It takes a truck 3.56 seconds to slow down from 112 km/h to 87.4 km/h. What is its average acceleration?
Answer:
1.92 m/s2
Explanation:
What do we call the material such as air that light travels through
Answer:
Transparent or Translucent
Explanation:
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:
Two ice-skaters are skating in circles on a frozen pond. Maria is making large circles with a radius of 12 m and skating 4.5 m/s. Her friend, Samantha, is making smaller circles with a radius of 6 m but is not skating as quickly, going only 3.8 m/s. ii. How could each skater increase her centripetal acceleration without changing the size of her path? Explain your reasoning.
Answer:
v_maria > 4.5 m/s
v_samantha > 3.8 m/s
Explanation:
Formula for centripetal acceleration is;
a_c = v²/r
Where;
v is speed
r is radius
For Maria;
r = 12 m
v = 4.5 m/s
Thus;
a_c = 4.5²/12
a_c = 1.6875 m/s²
For Samantha;
r = 6
v = 3.8 m/s
Thus;
a_c = 3.8²/6
a_c = 2.41 m/s²
We want to find how could each skater increase her centripetal acceleration without changing the size of her path.
From the centripetal acceleration formula, since the size of path can't be changed it means the radius can't be changed and so the only thing that can now increase the centripetal acceleration is when the speed increases.
Thus;
For Maria, she has to move with a faster speed. Thus: v_maria > 4.5 m/s
For Samantha, she also has to move with a faster speed. Thus; v_samantha > 3.8 m/s
A bullet with an initial kinetic energy of 400 J strikes a wooden block where a 8000 N resistive force stops the bullet. What is the distance the bullet travels into the block?
How do you answer this question?
Answer:
d = 0.05 [m] = 50 [mm]
Explanation:
We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.
[tex]E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm][/tex]
Hey guys this is Ap physics please help I need this to pass i will mark brainliest for a good attempt
Split up the forces into components acting parallel to and perpendicular to the slope. See the attached picture for the reference axes.
The box stays on the surface of the plane, so that the net force acting perpendicular to it is 0, and the only acceleration is applied in the parallel direction.
Let m be the mass of the box, θ the angle the plane makes with the ground, and a the acceleration of the box. By Newton's second law, we have
• net parallel force
∑ Force (//) = W (//) - F = m a
(that is, the net force in the parallel direction is the sum of the parallel component of the weight W and the friction F which acts in the negative direction)
• net perpendicular force
∑ Force (⟂) = W (⟂) + N = 0
Notice that
W (//) = W sin(θ) … … … which is positive since it points down the plane
W (⟂) = -W cos(θ) … … … which is negative since it points opposite the normal force N
So the equations become
W sin(θ) - F = m a
-W cos(θ) + N = 0
Solving for a gives
a = (W sin(θ) - F ) / m
which is good enough if you know the magnitude of the friction force.
If you don't, you can write F in terms of the coefficient of kinetic friction between the box and plane, µ, as
F = µ N
so that
a = (W sin(θ) - µ N ) / m
and the normal force itself has a magnitude of
N = W cos(θ)
so that
a = (W sin(θ) - µ W cos(θ) ) / m
The weight W has magnitude m g, where g is the magnitude of the acceleration due to gravity, so
a = (m g sin(θ) - µ m g cos(θ) ) / m
a = g (sin(θ) - µ cos(θ))
Light travels from a
laser across an 8 m
room where it reflects
off a mirror. Uniform or non uniform velocity
Answer:
Non-uniform velocity as the laser light beam has got reflected by the mirror
And as the light got reflected there is a change in velocity making it non-uniform velocity
Two objects of equal mass are a distance of 5.0 m apart and attract each other with a gravitational force of 3.0 x 10^-7 N find their mass.
A) 150 kg
B) 9.8 kg
C) 11.000 kg
D) 340 kg
Answer
I Think Its 150
How long would it take a train traveling at 18 m/s to travel 7,500 m? Answer in minutes.
Please I need help !!
Answer:
6.94446 min
Explanation:
t=s/v
In order to prevent injury in a car crash, it is recommended that you _______.
A) Increase the time of the collision.
B) Increase the change in momentum of the collision.
C) Increase the force in the collision.
D) Increase the initial velocity of the collision.
Increase of momentum of the collision will have the car in a unsettling position, creating an unsettling spot for it. Moreover increasing this would most likely take worse effect, it is better than increasing the time, which will only create the car faster. Let me show you what I mean...
A) Creates the car go faster, creating an even worse tragedy. B) save this for later...C) Increasing force will only result in worse damage.And finally, D) Increasing the velocity is basically increasing speed, once again, making things worse.So overall in this piece, the answer may very well end up being B. I sincerely hope this helped you by whatever means possible. It's logic that helps in real life situations, so take this as a little lesson- I guess :3E=6.63E-34J-s*4.6E14 hz
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: