Answer:
X axis F=0
Y axis Fg = - 2.94 j ^
Explanation:
The motion of a ball in air where the air residence is indicated to be negligible can be analyzed using Newton's second law.
We set a reference system, where the x-axis is horizontal and the y-axis vertical.
X axis
There are no forces on this axis, therefore the ball goes at constant speed.
Force is zero
Y axis
In this axis it is subjected to the acceleration of gravity that creates a force equal to the weight of the body, in a vertical direction.
Fg = m g
Fg = 0.3 9.8
Fg = 2.94 N
Fg = - 2.94 j ^
the boold are vectors; negative sign indicates that the force eta directed vertically downward
Sarah took 204 seconds to bicycle to their grandmother's house, a total of
430 meters. What was their velocity in m/s?
Answer:
2.1m/s towards your grandmother's house
Explanation:
Given parameters:
Time taken = 204s
Distance = 430m
Unknown:
Velocity = ?
Solution:
The velocity is determined by:
Velocity = [tex]\frac{displacement}{time}[/tex]
Velocity = [tex]\frac{430}{204}[/tex] = 2.1m/s towards your grandmother's house
Pluto has a radius of 1.15 x 10^6 m, and its acceleration due to gravity is 0.61 | m/s^2. What is Pluto's mass?
so we will use Newton's gravitational law :
gravitational acceleration = G*m/r^2
G is the gravitational constant = G = 6.673×10^-11 N m^2 kg^-2
after substitution :
6.673×10^-11 * m / (1.15 x 10^6)^2 = 0.61
= 5.04575*10^-23 * m = 0.61
dividing over 5.04575*10^-23 :
m = 1.20894*10^22 kg
pls give me brainliest
noooooooooooooooooooooooo
Answer:
yes
Explanation:
Answer:yeeeeeeeeeeeeeeeeeeeeeees
Explanation:
Rank the four fundamental forces from strongest to weakest. Use 1 to indicate the strongest force and 4 to indicate the weakest force. The gravitational force: The electromagnetic force: The strong nuclear force: The weak nuclear force:
Answer:
4
2
1
3
Explanation:
Be safe, lovelies <3
plsss answer this plsss answer this plsss answer this plsss answer this
Answer:
I dont see file
dndndndbnfbfbfbfbfbf
Answer:
ye ek rod h or electric ⚡ field h P point
Can you explain how newtons second law of force can be used in football and explain if possible
Answer:
Newtons 2nd Law of motion affects football as well. This law states the greater the mass of an object, the less the acceleration. This means that the more the objects weighs, the more difficult it is to move the object.
Explanation:
this is my best answer i can give you and mark me brainliest plz
A hiker walks 7.7 miles to the east in 3.6 hours, then turns around and walks 2.1 miles to the west in 2.4 hours. What was the magnitude of her average velocity during the trip?
Answer:
2.18 miles per hour
Explanation:
Given:
A hiker walks 7.7 miles to the east in 3.6 hours.
A hiker walks 2.1 miles to the west in 2.4 hours
To find: magnitude of average velocity during the trip
Solution:
Total distance = 7.7 + 2.1 = 9.8 miles
Total Time = 2.1 + 2.4 = 4.5 hours
Average velocity = Total distance ÷ Total Time =[tex]\frac{9.8}{4.5}=2.18[/tex] miles per hour
An 80-kg clown sits on a 20-kg bike on a tightrope attached between two trees. The center of mass of the clown is 1.6 m above the rope, and the center of mass of the bike is 0.70 m above the rope.
A) A load of what mass should be fixed onto the bike and hang 2.10 m below the rope so that the center of mass of the clown-bike-load system is 0.50 m below the rope?
B) What is the force that the rope exerts on each tree if the angle between the rope and the horizontal is 15 (degrees)?
Note: I have tried the center of mass formula Xcm= [(m1*x1) + (m2*x2) + (m3*x3)]/(m1+m2+m3) with no luck. Any help is much appreciated.
Answer:
Explanation:
the center of mass formula
Ycm= [(m₁y₁) + (m₂y₂) + (m₃y₃)] / (m₁+m₂+m₃)
Rope forms the x axis and position of centre of different massses are above or below it so they represent their location on y - axis.
y₁ = 1.6 , y₂ = .7 and y₃ = - 2.1
Ycm ( given ) = - .5
Putting the values of masses and positions
- .5 = 80 x 1.6 + 20 x .7 + m₃ x - 2.1 / ( 80 + 20 + m₃ )
- .5 = 128 + 14 + m₃ x - 2.1 / ( 100+ m₃ )
- 50 - .5 m₃ = 142 - 2.1 m₃
1.6 m₃ = 192
m₃ = 120 kg .
B )
Total downward force is weight of total mass = 80 + 20 + 120
= 220 kg
weight = 220 x 9.8 = 2156 N .
component of weight perpendicular to rope
= 2156 cos 15 = 2082.53 N
This force will be equally distributed over each tree , so force on each tree = 2082.53 / 2 = 1041.26 N .
A fan is set on a desk next to a stack of paper. The fan is turned on and then turned
on HIGH SPEED. Which of the following would best apply Newton's First Law to this
example.
The papers accelerated due to the force of the fan.
The papers are acted upon by an unbalanced force from the fan.
O The papers exerted an equal force on the air blown by the fan.
O The papers that were the heaviest were blown the closest.
Answer:
The second option - the papers are acted upon by an unbalanced force from the fan.
Explanation:
PLEASE HELP GIVING BRAINLIEST ANSWER
Explain why your PE and KE are usually not both high at the same time (If PE is high then usually KE is low)
50 points Two waves combine with constructive interference. What must be true of the
combined wave that forms?
A. It has a lower frequency than that of the original waves.
B. It has a higher amplitude than that of the original waves.
C. It has a higher frequency than that of the original waves.
D. It has a lower amplitude than that of the original waves.
Answer:
it has a higher amplitude than that of the original waves
Explanation:
trust me its right
The two waves combined with constructive interference have a higher amplitude than the original wave.
Which wave has the highest amplitude?High energy waves are characterized by high amplitude. Low energy waves are characterized by their small amplitude. As explained in Lesson 2, wave amplitude is the maximum amount of particle movement on a medium from a resting position.
What are superposition and interference?Superposition is a combination of two waves in the same place. Constructive interference occurs when two identical waves interfere in phase. Destructive interference occurs when two identical waves are exactly out of phase and overlap.
Learn more about higher amplitude waves at
https://brainly.com/question/19036728
#SPJ2
A car starts from rest and accelerates uniformly at a rate of 2.0 meter per second squared for 4.0 seconds. During this time interval, the car traveled ________ meters.
Answer:
16
Explanation:
[tex] \frac{1}{2} \times 2 \times {4}^{2} = 16[/tex]
A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is falling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target?
Answer:
Answer:
15.67 seconds
Explanation:
Using first equation of Motion
Final Velocity= Initial Velocity + (Acceleration * Time)
v= u + at
v=3
u=50
a= - 4 (negative acceleration or deceleration)
3= 50 +( -4 * t)
-47/-4 = t
Time = 15.67 seconds
We have that the speed must be at the speed below if they want to just hit their target
From the Question we are told that
Distance [tex]d=50m[/tex]
Height [tex]h=70m[/tex]
Constant Velocity [tex]v= 7.0 m/s[/tex]
Generally the equation for the time is mathematically given as
[tex]T=\frac{h}{v}\\\\T=\frac{70}{7}\\\\T=10s[/tex]
Therefore
The velocity required to make horizontal movement is
[tex]V=\frac{d}{T}\\\\V=\frac{50}{10}\\\\V=5m/s[/tex]
Given that
Velocity on the Vertical axis is
[tex]v_y=7m/s[/tex]
Velocity on the horizontal axis is
[tex]v_x=5m/s[/tex]
Therefore resultant speed
[tex]v_r=\sqrt{v_x^2+V_y^2}\\\\v_r=\sqrt{(5)^2+(7)^2}[/tex]
[tex]v_r=8.6023m/s[/tex]
In conclusion
[tex]v_r=8.6023m/s[/tex] must be their total speed if they want to just hit their target
For more information on this visit
https://brainly.com/question/17127206
Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. What will its angular velocity be after 3t?
Answer:
θ = 225 rad
Explanation:
given data
angle = 25 rad
to find out
angular velocity after 3t?
solution
let angular acceleration α in t
θ = ω × t + 0.5 × α × t² ........................1
here ω = 0 (initial velocity )
so put this value here
25 = 0 + 0.5 × α × t² ..........................2
α = 25 ÷ (0.5 t²)
α = 50 ÷ t² .........................3
now here we take in 3t
θ = ω × 3t + 0.5 × α × (3t)²
for ω = 0
θ = 0 + 0.5 × α × 9t²
now put value in eq 2
so
θ = (0.5) × (50 ÷ t²) × (3t)²
θ = 25 × 9
θ = 225 rad
A pendulum has a period of 5.14s and a length of 0.25m. What is the acceleration
due to gravity? *
Answer:
Acceleration due to gravity, g = 2.68m/s²
Explanation:
Given the following data;
Period = 5.14s
Length = 0.25m
To find acceleration due to gravity, g;
[tex] Period, T = 2 \pi \sqrt {lg} [/tex]
Substituting into the equation, we have;
[tex] 5.14 = 2*3.142 \sqrt {0.25g} [/tex]
[tex] 5.14 = 6.284 \sqrt {0.25g} [/tex]
[tex] \frac {5.14}{6.284} = \sqrt {0.25g} [/tex]
[tex] 0.8180 = \sqrt {0.25g} [/tex]
Taking the square of both sides
[tex] 0.8180^{2} = 0.25g [/tex]
[tex] 0.6691 = 0.25*g[/tex]
[tex] g = \frac {0.6691}{0.25} [/tex]
Acceleration due to gravity, g = 2.68m/s²
A toy car accelerates uniformly from rest at a constant rate. The car travels 1.0 meters in 1.0 seconds. The acceleration of the car is ________ meters per second squared.
Answer:
a=1m/s^2
Explanation:
1÷1÷1=1m/s^2
5. An astronaut has a mass of 65kg where the gravitational field strength is 10N/kg
a. Calculate the weight of the astronaut on earth
[3]
Answer: a) weight on Earth = mass of the object and gravity n the Earth. = 65*10 = 650 kg.
Explanation:
An astronaut has a mass of 65 kg on Earth where the gravitational field strength is 10 N kg A calculate the astronaut's weight on Earth
hope this helps :)
Answer:
650
Explanation:
use the equation
weight = gm
I love buying physics toys. I recently broke out my new electromagnetic field meter and started playing with it. After turning it on, I noticed the device kept showing an electric field value of 200 N/C towards the ground, without being near anything obvious (e.g., an electronic device) that would be producing the electric field. I even took a long walk to check if the reading was somehow localized to my house, but I got the same result. How might you explain the reading (assuming the device is working properly)
Answer:
ionized particles from the sun.
* interactions in radiation belts.
* the friction of the planet in the solar wind
q = +9 10⁵ C
Explanation:
Due to being made up of matter, the planet Earth has a series of positive and negative charges, in general these charges should be balanced and the net charge of the planet should be zero, but there are several phenomena that introduce unbalanced charges, for example:
* ionized particles from the sun.
* interactions in radiation belts.
* the friction of the planet in the solar wind
This creates that the planet has a net electrical load
We can roughly calculate the charge of the planet
E = k q / r²
q = E r² / k
let's calculate
q = 200 (6.37 10⁶)²/9 10⁹
q = +9 10⁵ C
What school did Ronald McNair go to and what kind of science did he work in
Answer:
McNair graduated as valedictorian of Carver High School in 1967. In 1971, he received a Bachelor of Science degree in engineering physics, magna cu.m laude, from the North Carolina Agricultural and Technical State University in Greensboro, North Carolina.
17. During a game of tug of war, two teams of students pull on opposite sides of a rope. During the
game, the rope begins to accelerate towards the left. What must be true about the forces acting on the
rope at the time of the acceleration towards the left?
A. The team pulling towards the right is pulling with a force greater than the team pulling towards the left.
B. The team pulling towards the right is pulling with a force equal to the team pulling towards the left.
C. The team pulling towards the right is pulling with a force less than the team pulling towards the left.
D. The team pulling towards the right stopped pulling the rope while the team pulling towards the left
continued
Answer:
c
Explanation:
Pls help!!
1 example of a conductor and 1 example of a insulator in your EVERYDAY world.
Answer:
here
Explanation:
Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.
Examples of conductors include metals, aqueous solutions of salts
Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Falling off of a skateboard after you run into a curb A ball hits the ground and the ground pushes up on it with the same force
Answer:
A ball hits the ground and the ground pushes up on it
Explanation:
Newton's third law basically states that for every action, there's a reaction.
a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.
Hope this Helps!!! :)
A π meson of rest energy 139.6 MeV moving at a speed of 0.921c collides with and sticks to a proton of rest energy 938.3 MeV that is at rest. (a) Find the total relativistic energy of the resulting composite particle. (b) Find the total linear momentum of the composite particle. (c) Using the results of (a) and (b), find the rest energy of the composite particle.
Answer:
A) 1268 MeV
B) 299MeV/c
C) 1268 MeV
Explanation:
Given :
π meson rest energy = 139.6 MeV
Speed = 0.921c
proton at rest energy = 938.3 MeV
a) Find the total relativistic energy of resulting composite particle
E = E(meson) + E(proton)
= [tex]\frac{(mc^2)_{meson} }{\sqrt{1-\frac{v^2}{c^2} } } + (mc^2)_{proton}[/tex]
= [tex]\frac{139.6MeV}{\sqrt{1-\frac{(0.906c)^2}{c^2} } } + 938.3[/tex]
E = 1268 MeV
B) determine the total linear momentum of the composite particle
= 299MeV/c
attached below is the detailed solution
C) Determine the rest energy of the composite particle
E = 1268 MeV
Match the measurements with the proper SI unit.
Acceleration:
A. Meters
B. Meters per second
C. Meters per second squared
Velocity:
A. Meters
B. Meters per second
C. Meters per second squared
Distance:
A. Meters
B. Meters per second
C. Meters per second squared
Explanation:
C. meter per second squared
B. meter per second
A. meter
Answer:
b. meters per second
c.meters per second squared
c.meters
Explanation:
ginawa ko na rin KC toh
To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the conductor. A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
Answer:
Explanation:
The electric field inside of a conductor is 0 because the conduction electrons are pushed to the outer edges of the conductor. The surface of the conductor still has charge.
In traveling a distance of 2.3 km between points A and D, a car is driven at 83 km/h from A to B for t seconds and 41 km/h from C to D also for t seconds. If the brakes are applied for 4.4 seconds between B and C to give the car a uniform deceleration, calculate t and the distance s between A and B.
Answer:
- time t taken for car to travel is 64.57 s
- distance travelled between A and B is 1.4887 km
Explanation:
Given the data in the question;
[tex]U_{BC}[/tex] = 83 km/h = ( 83×1000 / 60×60) = 23.0555 m/s
[tex]U_{CD}[/tex] = 41 km/h = ( 41×1000 / 60×60) = 11.3888 m/s
now, we calculate the acceleration;
a = ( [tex]U_{BC}[/tex] - [tex]U_{CD}[/tex] ) / t
we substitute
a = ( 23.0555 - 11.3888 ) / 4.4
a = 11.6667 / 4.4
a = 2.6515 m/s²
Now equation for displacement from BC
[tex]S_{BC}[/tex] = [tex]U_{BC}[/tex]t + 1/2.at²
we substitute
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×a×(4.4)²
we substitute -2.6515m/s² for a
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×(-2.6515)×(4.4)²
= 101.4442 - 25.6665
[tex]S_{BC}[/tex] = 75.7792 m
Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m
so
[tex]S_{AB}[/tex] + [tex]S_{BC}[/tex] + [tex]S_{CD}[/tex] = 2300 m
we substitute substitute
[tex]S_{AB}[/tex] + 75.7792 m + [tex]S_{CD}[/tex] = 2300 m
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2300 - 75.7792
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2224.2208 m
so we substitute 23.0555t for [tex]S_{AB}[/tex] and 11.3888t for [tex]S_{CD}[/tex]
23.0555t + 11.3888t = 2224.2208
34.4443t = 2224.2208
t = 2224.2208 / 34.4443
t = 64.57 s
Therefore, time t taken for car to travel is 64.57 s
Distance Between A to B
[tex]S_{AB}[/tex] = t × [tex]U_{AB}[/tex]
we substitute
[tex]S_{AB}[/tex] = 64.57 s × 23.0555
[tex]S_{AB}[/tex] = 1488.69 m
[tex]S_{AB}[/tex] = 1.4887 km
Therefore, distance travelled between A and B is 1.4887 km
The deepest part of the ocean is the Challenger Deep, at 10,900 m. The depth was first measured in 1875 by the HMS Challenger by depth sounding (which does not involve sound waves). If you were to measure the depth by echo sounding (which does involve sound), what would you expect the time for a sound pulse at the surface to return in s, naively assuming a constant sound velocity throughout the ocean
Answer:
t = 14.53 s
Explanation:
The speed of a wave is constant and is given by
v = [tex]\sqrt{ \frac{B}{ \rho} }[/tex]
in this exercise they indicate that we assume the constant velocity, therefore we can use the uniform motion relations
v = x / t
t = x / v
in this case the sound pulse leaves the ship and must return so the distance is
x = 2d
where d is the ocean depth d = 10900m and the speed of sound in seawater is v = 1500 m / s
let's calculate
t = 2 10900/1500
t = 14.53 s
If a solid metal sphere and a hollow metal sphere of equal diameters are each given the same charge, the electric field (E) midway between the center and the surface is...A. greater for the solid sphere than for the hollow sphere.B. greater for the hollow sphere than for the solid sphere.C. zero for bothD. equal in magnitude for both, but one is opposite in direction from the other.
Answer:
C. zero for both
Explanation:
In case of solid metal sphere , when it is given any charge , all the charges are transferred on the surface and within surface no charge exists . In case of hollow metal sphere , all charges reside on surface . In this way , in both solid and hollow sphere , all charge reside on the surface and no charge resides inside it . Hence due to absence of any charge inside , there is no electric field inside the sphere in both the cases .
Hence in both the case electric field is zero .
option C is correct .
If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the initial displacement is 2.25 m, and the change in momentum is 0.58 kg*m/s, how far will it coast back up the ramp before changing directions
Answer:
l = 0.548 m
Explanation:
For this exercise we compensate by finding the speed of the car
p = m v
v = p / m
v = 0.58 / 0.2
v = 2.9 m / s
this is how fast you get to the ramp, let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ m v²
final point. Point where it stops on the ramp
[tex]Em_{f}[/tex] = U = m g h
mechanical energy is conserved
Em₀ = Em_{f}
½ m v² = m g h
h = [tex]\frac{m v^2}{2 g}[/tex]
let's calculate
h = [tex]\frac{0.2 \ 2.9^2}{2 \ 9.8}[/tex]
h = 0.0858 m
to find the distance that e travels on the ramp let's use trigonometry, we look for the angle
tan θ = y / x
tan θ = 12/75 = 0.16
θ = tan⁻¹ 0.16
θ = 9º
therefore
sin 9 = h / l
l = h / sin 9
l = 0.0858 / sin 9
l = 0.548 m
. Why is it harder to stop an elephant accelerating at 1m/s2 than a rabbit accelerating at 1m/s2
(10 Points)
the elephant has more mass
the rabbit is faster
the rabbit has more mass
the elephant is faster
Answer:
this is about momentum p=mv
A, the elephant has more mass