If person A weights less then Person B, and they both push away from each other with 10N of force *
Answer:
The system will tend to person A
Explanation:
A force is defined as either a pull or a push, in this scenario person A weighs lass that person B, so the resultant effect of the 10N interactive force will tend towards person A.
This is solely because person A is less than person B in weigh
A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from the least amount of motion energy to the most.
Answer:
Train Bike Truck
Explanation:
If an electron (with a charge of 1.6 x10−19c) Experiences a force of 500 N at a certain point in an electric field, then find the strength of the electric field in that location
Answer:
3.125×10²¹ N/C
Explanation:
Electric Field: This can be defined as the force experienced per unit charge. The S.I unit of electric Field is N/C
Applying,
E = F/q.................. Equation 1
Where E = Electric Field, F = Force experienced, q = Charge of an electron.
From the question,
Given: F = 500 N, q = 1.6×10⁻¹⁹ C
Substitute these values into equation 1
E = 500/(1.6×10⁻¹⁹)
E = 312.5×10¹⁹
E = 3.125×10²¹ N/C
plzzzz urgent
solve this
Answer:
[tex] \large{ \tt{☄ \: EXPLANATION}} : [/tex]
Before solving , You'll have to know - When an object starts from the state of rest , in this case , initial velocity ( u ) = 0Notice that we're provided the time ( t ) in minutes. So , first thing we have to do is convert the minutes into seconds. It would be - Time ( t ) = 5 minutes = 5 × 60 sec = 300 sec [ 1 min = 60 sec ]Here , We're provided - Initial velocity ( u ) = 0 , Final velocity ( v ) = 60 m / s , Time taken ( t ) = 300 seconds & We're asked to find out the acceleration ( a ) & distance covered by the jeep ( s ) .[tex] \large{ \tt{♨ \:LET'S \: START}} : [/tex]
Acceleration is defined as the rate of change of velocity. We know :[tex] \large{ \boxed{ \tt{❁ \: ACCELERATION \: (a) = \frac{FINAL \: VELOCITY(v) - INITIAL \: VELOCITY(u)}{TIME \: TAKEN \: ( \: t \: )}}}} [/tex]
- Plug the values & then simplify !
[tex] \large{ \bf{↬a = \frac{60 - 0}{300} = \frac{60}{300} = \boxed{ \bold{ \bf{0.2 \: m {s}^{ - 2} }}} }}[/tex]
The acceleration of the jeep is 0.2 m/s²[tex] \large{ \tt{۵ \: AGAIN, \: USING\: SECOND \: EQUATION \: OF \: MOTION}} : [/tex]
[tex] \boxed{ \large{ \bf{✾ \: s = \frac{u + v}{2} \times t}}}[/tex]
- Plug the values & then simplify !
[tex] \large{ \bf{↦s = \frac{0 + 60}{2} \times 300 = \boxed{ \bold{ \bf{9000 \: m}}}}}[/tex]
The distance covered by the jeep is 9000 m .❃ The days that break you are the days that make you ! ♪
♡ Hope I helped! ツ
☃ Have a wonderful day / evening ! ☼
# StayInAndExplore ! ☂
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Answer:
What she said is right I know it is
If 150 Joules of work is needed to move a box 1000 cm, calculate the force that was used?
Answer:
[tex] \large{ \tt{☄ EXPLANATION}} : [/tex]
We're provided : Work ( W ) = 150 J & Displacement ( D ) = 1000 cm & We're asked to find out the force that was used. Firstly, Notice that we're provided the unit of Displacement as centimetre so we have to convert 1000 cm into m. Displacement ( D ) = [tex] \frac{1000}{ 100} = 10[/tex] m[tex] \large{ \tt{♡ \: LET'S \: GET \: STARTED}} : [/tex]
Work is defined as the product of force and Displacement. By definition ,[tex] \large{ \boxed{ \tt{❃ \: WORK ( \: W \: ) = FORCE\: ( \: F) \times DISPLACEMENT \: ( \: D \: )}}}[/tex]
- Plug the values and then simplify !
[tex] \large{ \bf{↦ \: 150 = F \times 10}}[/tex]
[tex] \large{ \bf{↦F\times 10 = 150}}[/tex]
[tex] \large{ \bf{↦ \: F = \frac{150}{10}}} [/tex]
[tex] \large {\bf{↦F = \boxed{ \bf{15} \: N}}}[/tex]
[tex] \boxed{ \boxed{ \large{ \tt{✤ \: OUR \: FINAL \: ANSWER : \boxed{ \bf{15 \: n}}}}}}[/tex]
KEEP READING , KEEP STUDYING , KEEP WORKING , KEEP PUSHING , KEEP TAKING CARE OF YOURSELF. YOUR HARD WORK WILL PAY OFF ! ♪ツ Hope I helped ! ۵
☪ Have a wonderful day / evening ! ☼
# StayInAndExplore ! ☂
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Two light waves of equal wavelength, lambda, are emitted in phase from separate sources and propagate to a common point P. Light wave 1 must travel a longer distance (d1) than light wave 2 (d2) to reach point P, where d1 – d2 is equal to the path difference between the two light waves. If the two waves interfere constructively at point P, what must be true about the path difference between the two light waves?
Answer:
The path difference must be equal to an integral (1 * lambda, 2 * lambda, -------n * lambda) number of wavelengths for constructive interference to occur.
A train accelerates from 30 km/h to 45 km/h in 15.0 second. Find its acceleration and the distance it travels during this time
Answer:
a. Acceleration, a = 0.28 m/s²
b. Distance, S = 156 meters
Explanation:
Given the following data;
Initial velocity = 30 km/h
Final velocity = 45 km/h
Time = 15 seconds
a. To find the acceleration;
Conversion:
30 km/h to m/s = 30*1000/3600 = 8.33 m/s
45 km/h to m/s = 45*1000/3600 = 12.5 m/s
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
Substituting into the equation;
[tex]a = \frac{12.5 - 8.3}{15}[/tex]
[tex]a = \frac{4.2}{15}[/tex]
Acceleration, a = 0.28 m/s²
b. To find the distance travelled, we would use the second equation of motion given by the formula;
[tex] S = ut + \frac {1}{2}at^{2}[/tex]
Where;
S represents the displacement or height measured in meters.
u represents the initial velocity measured in meters per seconds.
t represents the time measured in seconds.
a represents acceleration measured in meters per seconds square.
Substituting into the equation, we have;
[tex] S = 8.3*15 + \frac {1}{2}*(0.28)*15^{2}[/tex]
[tex] S = 124.5 + 0.14*225[/tex]
[tex] S = 124.5 + 31.5 [/tex]
S = 156 meters
convert 11 milliseconds into seconds
Answer:
0.011
Explanation:
1. If you use an applied force of 45N to slide a 12Kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?
Answer:
Coefficient of kinetic friction = 0.38 (Approx.)
Explanation:
Given:
Applied force = 45 N
Mass of wooden crate = 12 kg
Find:
Coefficient of kinetic friction
Computation:
Coefficient of kinetic friction = Applied force / (Mass)(Acceleration due to gravity)
Coefficient of kinetic friction = 45 / (12)(9.8)
Coefficient of kinetic friction = 45 / 117.6
Coefficient of kinetic friction = 0.3826
Coefficient of kinetic friction = 0.38 (Approx.)
what is the direction of acceleration due to gravity ?
The direction of acceleration due to gravity is always towards earth, going downwards.
mark me brainliesttt :))
The Big Bang Theory states that
A. the universe is continuously
expanding
B. all of the above
C. the universe was created by an
explosion
D. all matter and energy in the
universe was created
Answer:
All of above
Explanation:
A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time did the skateboarder spend in the air and how far will be his final position from the bridge?
Answer:
t = 0.714 s and x = 5.0 m
Explanation:
This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed
vₓ = 7.0 m / s
Let's find the time it takes to get to the river
y = y₀ + v_{oy} t - ½ g t²
the initial vertical speed is zero and when it reaches the river its height is zero
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{\frac{2y_o}{g} }[/tex]
t = ra 2 2.5 / 9.8
t = 0.714 s
the distance traveled is
x = vₓ t
x = 7.0 0.714
x = 5.0 m
the ability to speak does not say you intelligent essay
Answer:
Change is for better and is necessary too. Everyone knows it but only the ones who apply it are the ones whom we call as intelligent. When a person changes in an order to make his life better, he chooses the right thing because staying put at a place in life doesn't take anyone anywhere.
Taking the example of a few infamous personalities, whom we consider as intelligent, we see a common thing that each one of them had a certain vision in life and in order to achieve it, they were ready to accept any kind of change and still are. So, this makes it clear that change is what leads us to our goal and in order to achieve it, we need to change with time.
30. A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pushing the box with a horizontal force of 35N. What is the magnitude of the force of static friction between the box and floor
Answer:
84.05
Explanation:
F=mg×0.25F=20x9.81×0.25f=49.05NF=35NF=f+F
F=49.05+35
=84.05
An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object then slides at a constant velocity for 6.0 s until it reaches a rough section that causes the object to stop in 2.5 s.
Answer:
[tex]D_T=18.567m[/tex]
Explanation:
From the question we are told that:
Acceleration [tex]a=8.0 m/s^2[/tex]
Displacement [tex]d=1.05 m[/tex]
Initial time [tex]t_1=6.0s[/tex]
Final Time [tex]t_2=2.5s[/tex]
Generally the equation for Velocity of 1.05 travel is mathematically given by
Using Newton's Law of Motion
[tex]V^2=2as[/tex]
[tex]V=\sqrt{2*6*1.05}[/tex]
[tex]V=4.1m/s[/tex]
Generally the equation for Distance traveled before stop is mathematically given by
[tex]d_2=v*t_1[/tex]
[tex]d_2=3.098*4[/tex]
[tex]d_2=12.392[/tex]
Generally the equation for Distance to stop is mathematically given by
Since For this Final section
Final velocity [tex]v_3=0 m/s[/tex]
Initial velocity [tex]u_3=4.1 m/s[/tex]
Therefore
Using Newton's Law of Motion
[tex]-a_3=(4.1)/(2.5)[/tex]
[tex]-a_3=1.64m/s^2[/tex]
Giving
[tex]v_3^2=u^2-2ad_3[/tex]
Therefore
[tex]d_3=\frac{u_3^2}{2ad_3}[/tex]
[tex]d_3=\frac{4.1^2}{2*1.64}[/tex]
[tex]d_3=5.125m[/tex]
Generally the Total Distance Traveled is mathematically given by
[tex]D_T=d_1+d_2+d_3[/tex]
[tex]D_T=5.125m+12.392+1.05 m[/tex]
[tex]D_T=18.567m[/tex]
(only answer if you know the answer or I'll report) Solve it w the steps tysm
Answer:
v_f = 20 m / s
Explanation:
For this exercise let's use the relationship between momentum and moment
I = Δp
F t = m v_f - m v₀
as the body starts from rest v₀ = 0
F t = m v_f
v_f = [tex]\frac{F t}{m}[/tex]
let's calculate
v_f = 4 2 / 0.4
v_f = 20 m / s
Please Help with thia question
Answer:
the types of force shown in the picture is
balanced force.
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, causing it to reverse its direction and giving it a velocity of +25 m/s the impulse the stick applies to the park is most nearly
Answer:
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Explanation:
The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:
[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)
Where:
[tex]I[/tex] - Impulse, in kilogram-meters per second.
[tex]m[/tex] - Mass, in kilograms.
[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.
[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.
If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:
[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]
[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Which of these describes a real image?
An iron nail having threads along its cylindrical surface is
Answer:
A screw is a mechanism that converts rotational motion to linear motion, and a torque (rotational force) to a linear force.[1] It is one of the six classical simple machines. The most common form consists of a cylindrical shaft with helical grooves or ridges called threads around the outside.[2][3] The screw passes through a hole in another object or medium, with threads on the inside of the hole that mesh with the screw's threads. When the shaft of the screw is rotated relative to the stationary threads, the screw moves along its axis relative to the medium surrounding it; for example rotating a wood screw forces it into wood. In screw mechanisms, either the screw shaft can rotate through a threaded hole in a stationary object, or a threaded collar such as a nut can rotate around a stationary screw shaft.[4][5] Geometrically, a screw can be viewed as a narrow inclined plane wrapped around a cylinder.
Explanation:
A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same material and length stretch if its cross-sectional area is 8.00 mm2 and the same force is used to stretch it
Answer: [tex]0.05\ mm[/tex]
Explanation:
Given
Cross-sectional area of wire [tex]A_1=4\ mm^2[/tex]
Extension of wire [tex]\delta l=0.1\ mm[/tex]
Extension in a wire is given by
[tex]\Rightarrow \delta l=\dfrac{FL}{AE}[/tex]
where, [tex]E=\text{Youngs modulus}[/tex]
[tex]\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)[/tex]
for same force, length and material
[tex]\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)[/tex]
Divide (i) and (ii)
[tex]\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm[/tex]
what is the dimension formula of power and pressure
Answer:
Power = ML²T⁻³
Pressure = ML⁻¹T⁻²
Explanation:
Applying,
Power(P) = Work(W)/Time(t)
P = W/t..................... Equation 1
But
W = Fd............. Equation 2
Where F = force, d = distance
Also,
F = ma.............. Equation 3
Where m = mass, a = acceleration.
Also,
a = v/t................ Equation 4
Where v = velocity
Also,
v = d/t............... Equation 5
Where d = distance
Substitute equation 5 into equation 4
a = d/t²................. Equation 6
Substitute equation 6 into equation 3
F = m(d/t²)........... Equation 7
Susbtitute equation 7 into equation 2
W = m(d/t²)×d
W = md²/t²........... Equation 8
Substitute equation 8 into equation 1
P = (md²/t²)/t
P = md²/t³............ Equation 9
In dimension,
mass(m) = M, distance(d) = L, time(t) = T
Substitute into equation 9
P = ML²/T³
P = ML²T⁻³
And
Pressure(R) = Force(F)/Area(A)
R = F/A................ Equation 10
F = md/t²,
A = d²
Susbtitute into equation 10
R = (md/t²)/d²
R = m/t²d
Therefore,
R = ML⁻¹T⁻²
A stunt performer falls off a wall that is 1.6 m high and then lands on a mat.
What is his impact velocity?
A. 5.6 m/s
B. 1.1 m/s
C. 4.7 m/s
o
D. 2.9 m/s
According to law of conservation of energy,
= ½mv² = mgh
= mv² = 2mgh
= v² = 2mgh/m
= v = √2gh
So, now just put the values of g & h, abd you are done;
= v = √2×9.8×1.6
= v = √31.36
= v = 5.6 m/s
A forensics investigator discharged an assault rifle-replica such that the bullet fired at an angle of 30 (degrees) off the horizontal with an initial velocity
of 28
m/s northwest. What is the maximum height the bullet will reach?
O 14 m/s
10 m
O 30 km
O 0.4351 seconds
Answer:
Initial y-component of speed
Vy = 28 * sin 30 = 14 m/sec vertically
1/2 m Vy^2 = 2 m g h conservation of energy of y-component
h = Vy^2 / (2 * g) = 14^2 / (2 * 9.8) = 10 m
car moves a distance of 420 m. Each tire on the car has a diameter of 42 cm. Which shows how many revolutions each
tire makes as they move that distance?
Plzzzz help asap
Answer:
10 is the correct answer
Answer:
Total Distance: 420 meters
Diameter: 42 cm
Notice the units meters vs cm
420÷ 42 = 10 total revolutions
Give your answer to 2 dp
When taking off a plane accelerates at 2.7m/s2 down the runway. It accelerates from a stationary position for 25 seconds before leaving the ground. What
is the planes speed when it leaves the ground?
Answer:
67.5
Explanation:
The plane accelerates at 2.7m/s,^2
Time is 25 seconds
The velocity can be calculated as follows
= 25×2.7
= 67.5
Hence the speed f the plane is 67.5
58
74
3 points
A hummingbird beats its wings up and down with a frequency of 100 Hz. What is the period of the hummingbirds flaps? (D
YOUR ANSWER)
59
Answer:
T = 0.01 s
Explanation:
Given that,
The frequency of the beats of a hummingbird, f = 100 Hz
We need to find the period of the hummingbirds flaps. Let the time is t. We know that the relation between frequency and time period is given by :
T = 1/f
Put all the values,
T = 1/100 = 0.01 s
So, the time period of the humming bird is 0.01 s.
A skipper on a boat notices wave crests passing the anchor chain every 6.0 seconds. The skipper estimates the distance between crests at 30.0
m. What is the speed of the water waves?
Explanation:
given Time Period = 5 Sec
Amplitude = 1m, Wave length = 15 m
using time period = 1/ frequency , Frequency = 1/5 cycles per sec = 0.2 sec = 12 per min
wave speed= Frequency * Wave length
speed = 0.2*15 = 3m/s
When electrons flow through wires from a terminal to a terminal a/an _____
created
Answer:
When electrons flow through wires from a terminal to a terminal circuit is created.
What is the net force acting on the airplane?
740 N right -->
700 N right -->
100 N left <--
760 N right -->
[tex]\huge{ \mathrm{ \underline{ Answer }\: \: ✓ }}[/tex]
Total force acting on right side = 800 N
Total force acting on left side :
60 N + 40 N100 NNow, equivalent force acting on the plane is :
greater force - minor force 800 N - 100 N 700 NewtonsAnd the direction of equivalent force will be the direction of greater force, that is right direction.
Hence, Correct option is :
700 N right -->_____________________________
[tex]\mathrm{ \:TeeNForeveR\:}[/tex]