Answer:
Explanation:
Find the diagram of the scenario in the attached file.
From the diagram, the weight = mass * acceleration due to gravity
mass of sled = 30.0kg
acc. due to gravity = 9.81m/s
weight of the sled = 30.0*9.81
weight of the sled = 294.3N
The normal reaction force R in the diagram will be gotten by resolving the 12N force along the vertical
Ry = 12sin45°
R = 12 * 1/√2
R = 12√2/√2
R = 6√2 N
R = 8.49N ≈ 9.0N
The normal force exerted on the sled is approximately is 9.0N
Get the acceleration
Using the formula [tex]\sum Fx = ma_x[/tex] where;
m is the mass of the sled = 30.0kg
[tex]a_x[/tex] is the acceleration of the sled
[tex]\sum Fx = 8 + 12cos45^0\\\sum Fx = 8 + 12(0.7071)\\\sum Fx = 8 + 8.49\\\sum Fx = 16.49N[/tex]
Substitute into the formula;
[tex]a_x = \frac{\sum Fx }{m} \\ a_x = \frac{16.49}{30}\\ a_x = 0.55 m/s^2[/tex]
Hence the acceleration of the sled rounded to the nearest hundredth is 0.55m/s²
Please help my test is going on. Please answer both (i) and (ii).
Answer:
i. 7.5 m
ii. 15000 N
Explanation:
Area under vt graph shows the displacement
so area of triangle ABE= 1/2×15×2
=7.5 m
ii) F=ma
here, m=1000kg
and a=v-u/t
=15-0/1
a=15
F=1000×15=15000N
Which expression can be used to calculate centripetal acceleration?
Answer:
B, 4π^2r/T^2
Explanation:
Im on the same quiz, and I'm pretty sure this is the right answer cause I found it on 4 other websites. Good luck
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Answer:
ok
Explanation:
find weight of 45 kg at height 800 km from earth's surface?
Answer:
The weight is 348.4 N
Explanation:
Effect of the Altitude on the Acceleration of Gravity
The effect of gravity decreases with altitude because greater altitudes mean greater distances from the Earth's center.
The practical formula to calculate the gravity as a function of the altitude h is:
[tex]{\displaystyle g_{h}=g_{0}\left({\frac {R_{\mathrm {e} }}{R_{\mathrm {e} }+h}}\right)^{2}}[/tex]
Where go is the conventional gravitational acceleration = [tex]9.80665\ m/s^2[/tex]
R is the Earth's mean radius at the equator = 6,378 Km
h is the altitude.
At a height of h=800 Km, the acceleration of gravity is:
[tex]{\displaystyle g_{h}=9.80665 m/s^2\left(\frac {6,378 Km }{6,378 Km+800Km}}\right)^{2}}[/tex]
Calculating:
[tex]g_h=7.74\ m/s^2[/tex]
The weight is the product of the acceleration of gravity by the mass of the object, thus:
[tex]W=45\ Kg\cdot 7.74\ m/s^2=348.4\ N[/tex]
The weight is 348.4 N
calculate your power in watts if you do 500 N*m of work picking up a dumbbell,and takes 5 s to perform that task?
Answer:
100watts
Explanation:
Given parameters:
Workdone = 500Nm
Time taken = 5s
Unknown:
Power in watts = ?
Solution:
Power is the rate at which work is done;
Power =[tex]\frac{workdone }{time}[/tex]
Input the parameters and solve;
Power = [tex]\frac{500}{5}[/tex] = 100watts
A 10 kg box is resting on the table. What is the normal force on the box?
Answer:
normal force and gravitational force are equal. change kg to newtons
98N
Explanation:
The normal force on the box resting on the table is 98 N.
The given parameters;
mass of the box, m = 10 kgThe normal force on the box resting on the table is the force acting perpendicular to the box and it is calculated as follows;
Fₙ = mg
where;
m is the mass of the boxg is the acceleration the to gravityFₙ = 10 x 9.8
Fₙ = 98 N
Thus, the normal force on the box resting on the table is 98 N.
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3. What are the two parts of a shadow?
O opaque area and dark area
the umbra and penumbra
O the umbra and
antunghaya
O the illuminated opaque object and the umbra
Answer:
the umbra and penumbra
Explanation:
The two parts of a shadow are the umbra and penumbra.
A shadow is an expression formed when light from a source is cut blocked by an opaque body.
Most shadows are made up of two parts, umbra and penumbra.
The umbra is the darkest, and the innermost part of a shadow. In the penumbra, only a portion of the light source is blocked. It is more of like a partial shadow.The umbra is directly formed by light which impinges on an opaque body and it completely cut off.
Select the correct answer.
Type O negative blood is considered the universal red blood cell donor because it
OA.
Has all 3 types of antigens.
OB.
Has all 3 types of antibodies.
Lacks all 3 types of antigens.
OD
Lacks all 3 types of antibodies.
Answer:
The answer is C.) Lacks all 3 types of antigens.
A sprinter can run 100.0 meters in 10.4 seconds. Find the velocity of this runner. (Please put answer into ladder cancellation format!!)
A person of 45 kg can accelerate at 1.5 m/s on a straight road. What is the force created?
Answer:
The answer is 67.5 NExplanation:
The force acting on an object can be found by using the formula
Force = mass × accelerationFrom the question
mass = 45 kg
acceleration = 1.5 m/s²
We have
Force = 45 × 1.5
We have the final answer as
67.5 NHope this helps you
An elevator ascends at a constant speed of 4 m/s. How much time is required for the elevator in order to travel 200m upwards?
Answer:
Explanation:
Hope this helps, cheers
Alguien ayudeme con estos problemas de fisica porfa 1. Una persona recorre 4200 m en 25 minutos, luego 4.54 Km en un hora y finalmente 9000 m en 55 minutos. Halle, el tiempo total empleado en minutos, su velocidad y rapidez. 2. Una partícula en una circunferencia de 40 cm de diámetro da 30 vueltas completas en dos minutos. Halle la velocidad y rapidez de la partícula. 3. Una persona da 500 pasos de 40 cm al oeste, luego 400 pasos de 35 cm al norte y finalmente 300 pasos de 30 cm al este utilizando en total 2 min y 120 segundos. Halle la distancia recorrida, desplazamiento, velocidad y rapidez. 4. Una persona recorre una pista circular de 250 metros de radio, ¿Cuál es su distancia, desplazamiento, velocidad y rapidez en dar media vuelta a la pista si el tiempo empleado es de 90 segundos
Answer:
1) t_total = 140 min , v = 126.7 m / min , 2) v = 0.628 m / s ,
3) v = 0.592 m / s , θ = -39.3 º
Explanation:
Here we have some short kinematics problems
1) d₁ = 4200 m at T1 = 25 min
d₂ = 4.54 km at t2 = 1 h
d₃ = 9000 m at t3 = 55 min
of the statement the direction of each route is the same
Let's reduce the distances to the SI system
d₂ = 4.54 km (1000 m / 1 km) = 4540 m
t₂ = 1 h (60 min / 1 h) = 60 min
Total time is
t_total = t1 + t2 + t3
t_total = 25 + 60 +55
t_total = 140 min
the distance is
d_total = 4200 + 4540 +9000
d_total = 17740 m
with movement it is in one dimension and all displacement is in the same direction, distance equals displacement.
Speed is
v = 17740/140
v = 126.7 m / min
in the direction of travel
Speed is the modulus of speed
| v | = 126.7 m / min
2) the angular velocity is
w = θ / t
In this exercise
t = 2 min (60 s / 1min) = 120 s
θ = 30 rev (2π rad / 1 rev) = 60π rad
w = 60π / 120
w = 1.57 rad / s
linear velocity is
v = w r
v = 1.57 0.40
v = 0.628 m / s
velocity is the speed tangential to the trajectory
3) the data for this exercise are
distance = (number of steps) (distance of one step)
d₁ = 500 0.40 m = 200 m West
d₂ = 400 0.35 m = 140 m North
d₃ = 300 0.30 m = 90 m East
in a total time of t = 2 min and 120 s
t = 2 60 + 120 = 240 s
the distance traveled is
x axis we assume the direction towards + x (East) positive
x_total = d₃ -d₁
x_total = 90-200 = -110 m
y axis
y_total = d₂
y_total = 140 m
we use the Pythagorean theorem to find the distance0
D = √ (x_total² + y_total²)
D = √ (110² + 90²)
D = 142.13 m
The speed is
v = D / t
v = 142.13 / 240
v = 0.592 m / s
speed is
v = 0592 m / s
tan θ = Y / X
θ = tan -1 y / x
θ = tan -1 (90 / (- 110))
θ = -39.3 º
4) This angle is measured on the positive side of the x axis in a clockwise direction
θ = 1.5 rev (2π rad / 1 rev) = 3π rad
t = 90 s
the angular velocity is
w = θ / t
w = 3πi / 90
w = 0.1047 rad / s
the speed is
v = w R
v = 0.14047 250
v = 26.17 m / s
velocity is tangent to path
many atoms of each element are in one molecule of 2-heptanone?
A chemical structure of a molecule includes the arrangement of atoms and the chemical bonds that hold the atoms together. The 2-HEPTANONE molecule contains a total of 21 bond(s) There are 7 non-H bond(s), 1 multiple bond(s), 4 rotatable bond(s), 1 double bond(s) and 1 ketone(s) (aliphatic).
:/ hope this helps
A truck is traveling at a constant speed of 20 m/s through a school zone. At time t = 0 seconds, he passes a hidden police car that is at rest. Five seconds after the truck passes, the police car begins accelerating at a constant rate of 2 m/s in order to catch the truck.
Determine the time at which the speed of the truck is equal to the speed of the police car. Mark this time as t↓1. How will the positions of the police car and the truck compare when they have the same speed and why?
Answer:
Let's define:
The position zero will be in the first point where the truck and police car meet. (So the initial position of both vehicles is zero)
Then, the position equation for the truck is:
T(t) = 20m/s*t
Where t is our variable, time in seconds.
Now, at t = 5s, the car starts accelerating.
Ac(t) = 2m/s^2 for( t ≥ 5s)
For the velocity of the car, we must integrate that.
V(t) = (2m/s^2)*(t - 5) for (t ≥ 5s)
Where i introduced a little change in the variable because the velocity of the car starts to increase for t larger than 5 seconds.
For the position of the car we integrate again.
C(t) = (1m/s)*(t - 5)^2 for (t ≥ 5s)
Now, let's answer the questions:
Determine the time at which the speed of the truck is equal to the speed of the police car.
Then we must have:
V(t) = (2m/s^2)*(t - 5) = 20m/s. (Remember that we only can use times requal or larger than 5 seconds).
2m/s^2*t - 10m/s = 20m/s
2m/s^2*t = 30m/s
t = (30/2) s = 15s
The velocities of both vehicles will be the same after 15 seconds.
t1 = 15s
How the positions will compare at this time?
The easier thing will be to evaluate the position equation of each vehicle in this time:
T(15s) = 20m/s*15s = 300m
C(15s) = (1m/s^2)*(15s - 5s)^2 = (1m/s^2)*(10s)^2 = 100m.
Then we can see that the truck is 200m ahead of the car.
But remember that the police car is accelerating, so the velocity will keep increasing meaning that eventually, the car will catch the truck.
You and your friends are going on a picnic in a very small car. On your way, the car breaks down and you try to push it to get it to move but in vain. You ask your friends to help you and the car starts moving and revs up. How are you and your friends applying Newton's second law of motion here? What if the car you were traveling in was a large SUV?
Answer:
The pushing of the car by you and your friends is the applied force and when the car moves, it means that the velocity has changed thus causing the movement or acceleration.
Explanation:
Newton's work on forces regarding motion can never be neglected by scientists. Sir Isaac Newton when he was alive, among several of his works he proposed the three laws guiding the forces of motion. In this question we are only going to be treating only one out of the three Newton's Law of motion and that is the second Law Of Newton's laws of motion.
The second Law Of Newton's laws of motion states that the acceleration of an object is directly proportional to the applied force and inversely proportional to the object's mass.
(1). Now, to the question: " How are you and your friends applying Newton's second law of motion here? "
The pushing of the car by you and your friends is the applied force and when the car moves, it means that the velocity has changed thus causing the movement or acceleration.
According to the law, the more the Force, the more the acceleration.
(2). For the second part of the question, " What if the car you were traveling in was a large SUV?"
From the law stated above we see that the acceleration is inversely proportional to the mass, thus if the car is a large SUV, It means that more force is needed to change the car's velocity.
Answer:
When a friend joins you in pushing the small car, the net force on it increases. Therefore, it starts to move. The mass of a SUV is much greater than that of a small car. You would thus need the help of more friends to help push it.
Explanation:
Plato/Edmentum
how big can a whale get
Answer:
Whale is the largest mammal in the world
Explanation:
you you know that the DNA of parent depend upon her babies.
so due to the DNA characteristics will get big one to another
Answer:
For example, in the North Atlantic and North Pacific, blue whales can grow up to about 90 feet, but in the Antarctic, they can reach up to about 110 feet and weigh more than 330,000 pounds. Like other baleen whales, female blue whales are generally larger than male Blue whales are the largest animals ever known to have lived on Earth. These magnificent marine mammals rule the oceans at up to 100 feet long and upwards of 200 tons.
Explanation:
6. The race car in the previous problem slows from
36 m/s to 15 m/s over 3.0 s. What is its average
acceleration?
Answer:
a_AV - 15 m/s – 36 m/s =-70 m/s2. At. 3.0 s.
Explanation:
The average acceleration of the car as it slows down is -5 m/s².
Given is a race car that slows down from 36 m/s to 15 m/s in 3 seconds.
initial velocity [u] = 36 m/s
final velocity [v] = 15 m/s
time taken [Δt] = 3 s
We can calculate the acceleration as follows -
a = Δv/Δt
a = (v - u) / Δt
a = (15 - 30)/3
a = -15/3
a = -5 m/s²
Therefore, the average acceleration of the car as it slows down is
-5 m/s².
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what is a force that always resists the relative motion of objects or surfaces?
Friction is the resistance to motion of one object moving relative to another. It is not a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.
6. Potential energy that depends on height is called *
O kinetic energy
O gravitational potential energy.
elastic potential energy.
O mechanical energy
The potential energy that depends on height is called gravitational potential energy. (Option B)
What is gravitational potential energy?Gravitational potential energy is a form of potential energy that an object possesses due to its position relative to a gravitational field.
When an object is raised to a certain height above the ground, it gains gravitational potential energy. The higher the object is raised, the greater its potential energy.
This potential energy is a measure of the work that would be required to move the object against gravity and bring it back down to its original position.
The formula for gravitational potential energy is;
Potential Energy (PE) = mg h
where;
PE is the gravitational potential energym is the mass of the objectg is the acceleration due to gravityh is the height or elevation of the objectLearn more about gravitational potential energy here: https://brainly.com/question/15896499
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lists the 4 states of matter from most amount of thermal energy to least amount of thermal energy?
Answer:
Explanation:
plasma, gas, liquid, solid
Consider a 400g baseball thrown from a height of 1.8m at 35m/s:
a. What is the gravitational potential energy of the baseball?
b. What is the elastic potential energy of the baseball?
c. What is the kinetic energy of the baseball?
d. What is the total mechanical energy of the baseball?
Answer:hhahe
Explanation:
If the x-component of velocity is 27m/s and the y-component of velocity is -23 m/s, what is the resultant vector?
Answer:
35.47 m/s
Explanation:
[tex]r = \sqrt{( - 23)^{2} + (27) ^{2} }[/tex]
=35.57
How are Covalent bonds formed?
I believe it is introduced when the variation between the electronegativities of only two atoms is way too small for any transfer to take place to produce new ions.
Considering the definition of covalent bond, Covalent bonds are formed when valence electrons of non-metallic elements are shared and hold this elements together, forming a molecule.
Covalent bondThe covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule.
Covalent bonds are established between non-metallic elements. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.
SummaryCovalent bonds are formed when electrons in the outermost level of non-metallic elements are shared and hold this elements together, forming a molecule.
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What type of System interact with its environment
Answer:
System management
Explanation:
If a train goes at constant speed of 40 miles per hour for one hour, then at constant
speed of 55 miles per hour for two hours, what is its average speed at the train?
Answer: because the train goes 72 mph for 2 hours
Explanation:
does the train go 72mph for 2 hours as well? if so the answer would be 324 because you have to add 90 twice scene it goes at that speed for 2 hours thats 180. assuming the train goes 72 mph for 2 hours it would be 144. then add 180 and 144 and you get 324 merry christmas and a happy newyear to you and your family
4. Calculate the momentum for each scenario:
a. .02 kg mass moving at 300 m/s
b. 2 kg mass moving at 40 m/s
c. 200 kg mass moving at 4 m/s
a) Answer:
Momentum = 6 kgm/s
Explanation:
Momentum = mass × velocity
Given that,
mass = 0.02 kg velocity = 300 m/s
Therefore, momentum = (0.02 × 300)kgm/s
Momentum = 6 kgm/s
b) Answer:
Momentum =80 kgm/s
Explanation:
Momentum = mass × velocity
Given that,
mass = 2 kg velocity = 40 m/s
Therefore, momentum = (40 × 2)kgm/s
Momentum = 80 kgm/s
c) Answer:
Momentum = 800 kgm/s
Explanation:
Momentum = mass × velocity
Given that,
mass = 200 kg velocity = 4 m/s
Therefore, momentum = (200 × 4)kgm/s
Momentum =800 kgm/s
Classify each of the following chemical reactions. Upper S plus upper O Subscript 2 right arrow upper S upper O subscript 2. Upper C a upper C l subscript 2 plus 2 upper A g upper N upper O subscript 3 right arrow upper C a (upper n upper O subscript 3) subscript 2 plus 2 upper A g upper C l. Upper Z n plus upper C u upper S upper O subscript 4 right arrow upper C u plus upper Z n upper S upper O subscript 4. 2 upper N a subscript 2 upper O right arrow 4 upper N a plus upper O subscript 2.
Answer:
synthesis and combustion
double replacement
single replacement
decomposition only
in that exact order edge 2020
The presented reactions are:
S + O₂ ⇒ SO₂ Synthesis or combination
CaCl₂ + AgNO₃ = Ca(NO₃)₂ + 2 AgCl Double displacement
Zn + CuSO₄ = Cu + ZnSO₄ Single displacement
2 Na₂O = 4 Na + O₂ Decomposition
Let's consider and classify the following reactions.
S + O₂ ⇒ SO₂
This is a synthesis or combination reaction. 2 compounds combine to form another.
CaCl₂ + AgNO₃ = Ca(NO₃)₂ + 2 AgCl
This is a double displacement reaction. Both compounds exchange their cations and anions.
Zn + CuSO₄ = Cu + ZnSO₄
This is a single displacement reaction. A metal displaces other metal from its salt.
2 Na₂O = 4 Na + O₂
This is a decomposition reaction. A compound decomposes into 2 other compounds.
The presented reactions are:
S + O₂ ⇒ SO₂ Synthesis or combination
CaCl₂ + AgNO₃ = Ca(NO₃)₂ + 2 AgCl Double displacement
Zn + CuSO₄ = Cu + ZnSO₄ Single displacement
2 Na₂O = 4 Na + O₂ Decomposition
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why can't you run from momentum ?
Which characteristic is related to kinetic energy but not potential energy?
A. An object's shape
B. An object's mass
C. An object's speed
D. An object's height
Answer:
c
Explanation:
An object's speed is the characteristic which is related to kinetic energy but not the potential energy. Thus, the correct option is C.
What is Kinetic energy?The kinetic energy of an object is the energy which it possesses due to the motion of the object. Kinetic energy is defined as the work which is needed to accelerate an object of a given mass from rest position to its stated velocity. Having gained this amount of energy during the acceleration of the object, the object maintains this kinetic energy unless its speed changes or it experiences acceleration.
Kinetic energy is directly proportional to the mass of the object and to the square of the velocity of the object, the expression is:
K.E. = 1/2 m × v²
where, KE = Kinetic energy,
m = mass of the object,
v = velocity of the object.
Therefore, the correct option is C.
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A driver of a car traveling at -15m/s applies the brakes, causing a uniform acceleration of +2.0m/s2. If the brakes are applied for 2.5s, what is the velocity of the car at the end of the braking period? How far has the car moved during the braking period?
Given :
Initial velocity, u = -15 m/s.
Acceleration , a = 2 m/s².
Time taken to applied brake, t = 2.5 s.
To Find :
The velocity of the car at the end of the braking period.
How far has the car moved during the braking period.
Solution :
By equation :
[tex]v = u+at\\\\v=-15 + 2\times 2.5\\\\v=-10 \ m/s[/tex]
Now, distance covered by car is :
[tex]s=ut+\dfrac{at^2}{2}\\\\s=(-15)(2.5)+\dfrac{2(2.5)^2}{2}\\\\s=-31.25\ m[/tex]
Hence, this is the required solution.