ayuda porfa es de actividades fisicas
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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
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.
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]
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(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
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]
what is force? and give its formula
Force is a push or pull that changes or attempts to change the shape or size of an object's position.
Formula Of ForceF = m × a
Here
F= Force
m= mass of object
a= Acceleration
Hope This Helps You ❤️?????? ?????help please
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.
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
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]
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
13. When you say that something is a factor, you mean that it is an) (factor)
a. problem.
b. obstacle.
c. ingredient.
d. fact
Answer:
C.
A factor of something is an ingredient, or it makes up that thing. For example, in math, we always try to find the factors of a number. We are basically trying to find the "ingredients" of a number.
a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after 6 seconds
a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )
2 = ( v(2) - 10 ) ÷ ( 6 - 0 )
2 × 6 = v(2) - 10
v(2) = 12 + 10
v(2) = 22 m/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.
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 ball of 10kg falls from rest from a height of 150m, Neglating air resistance, calculate its kinetic energy after falling a distance of 40m. Take g=10m/s
Answer: [tex]3920\ J[/tex]
Explanation:
Given
mass of ball m=10 kg
It is placed at a height of 150 m
It is dropped from the height and allowed to free fall for 40 m
Velocity acquired by the ball during this fall is given by [tex]v^2-u^2=2as[/tex]
Insert u=0, a=g
[tex]\Rightarrow v^2-0=2\times 9.8\times 40\\\Rightarrow v=\sqrt{784}\\\Rightarrow v=28\ m/s[/tex]
Kinetic energy at this instant
[tex]K.E.=\dfrac{1}{2}\times 10\times 28^2\\\\\Rightarrow K.E.=3920\ J[/tex]
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
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.)
Please Help with thia question
Answer:
the types of force shown in the picture is
balanced force.
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
As the wavelength of light decreases,
What happens
Answer:
Waves. Refraction is an effect that occurs when a light wave, incident at an angle away from the normal, passes a boundary from one medium into another in which there is a change in velocity of the light. ... The wavelength decreases as the light enters the medium and the light wave changes direction.
Explanation:
As a wavelength increases in size, its frequency and energy (E) decrease. From these equations you may realize that as the frequency increases, the wavelength gets shorter. ... Mechanical and electromagnetic waves with long wavelengths contain less energy than waves with short wavelengths.
A ball of 5.2 kg is swings around on a rope with radius 0.86
meters. If the ball is traveling at 1.8 m/s, what is its angular
momentum? (Round to the nearest hundredth).
Answer:
Explanation:
Angular momentum is L = mvr:
L = (5.2)(1.8)(.86) so
L = 8.05 kg*m/s
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
PLEASE HELP WILL MARK BRAINLIEST *
What are the four components to a workout program?
Answer:
Aerobic Fitness. Aerobic fitness improves overall health and well-being. ...
Muscular Fitness. Strength training improves your muscle and bone health and helps with weight loss. ...
Flexibility. Flexibility allows you to move your body freely. ...
Stability and Balance
Explanation:
UP
Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by dE dt = q2a2 6pP0c3 where c is the speed of light. (a) Verify that this equation is dimensionally correct. (b) If a proton with a kinetic energy of 6.0 MeV is traveling in a particle accelerator in a circular orbit of radius 0.750 m, what fraction of its energy does it radiate per second? (c) Consider an electron orbiting with the same speed and radius. What fraction of its energy does it radiate per second?
Answer:
a) [tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]
b) [tex]x=\frac{6.0}{4.3*10^{-32}}[/tex]
c) [tex]x'=\frac{6.0}{2.22*10^{-3}}[/tex]
Explanation:
From the question we are told that:
Kinetic energy of Proton[tex]K.E_p= 6.0 MeV[/tex]
Radius [tex]r=0.750[/tex]
Energy [tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]
a)
[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]
b)
Generally the equation for acceleration of proton is mathematically given by
[tex]a_p=\frac{v^2}{r}[/tex]
Where
Speed of Proton particle is
[tex]V_p=2.12 *10^3 m/s[/tex]
[tex]a_p=\frac{(2.12 *10^3)^2}{0.750}\\a_p=1.27*10^{10}m/s^2[/tex]
Therefore
[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]
[tex]\frac{dE}{dt}=\frac{9*10^9*(1.6*10^{-19}^2*(1.27*10^{10})^2)}{(3*10^8)^3}[/tex]
[tex]\frac{dE}{dt}=6.881810^{-51}J/s[/tex]
Energy radiated per sec
[tex]mev=\frac{6.881810^{-51}}{1.6*10^{-19}}\\mev=4.3*10^{-32}ev[/tex]
Therefore the Fraction of its energy it radiates per second is given as
[tex]x=\frac{ K.E_p}{mev}\\[/tex]
[tex]x=\frac{6.0}{4.3*10^{-32}}[/tex]
c)
Generally the equation for acceleration of proton is mathematically given by
[tex]a_p=\frac{v^2}{r}[/tex]
Where
Speed of Proton particle is
[tex]V_p=2.5 *10^7 m/s[/tex]
[tex]a_p=\frac{(2.5 *10^7)^2}{0.750}\\a_p=0.33*10^{14}m/s^2[/tex]
Therefore
[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]
[tex]\frac{dE}{dt}=\frac{9*10^9*(1.6*10^{-19}^2*(0.33*10^{14})^2)}{(3*10^8)^3}[/tex]
[tex]\frac{dE}{dt}=59.26^{-25}J/s[/tex]
Energy radiated per sec
[tex]mev=\frac{59.26^{-25}}{1.6*10^{-19}}\\mev=2.22*10^{-3}ev[/tex]
Therefore the Fraction of its energy it radiates per second is given as
[tex]x'=\frac{ K.E_p}{mev}\\[/tex]
[tex]x'=\frac{6.0}{2.22*10^{-3}}[/tex]
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
convert 11 milliseconds into seconds
Answer:
0.011
Explanation:
Damian is texting while driving and ends up slamming into the back end of the car in front of him, which then strikes the car in front of it. By texting while driving, Damian has created a
The answer for this question is negative externality