Answer:
work done = mgh
350×10×2
7000J
The work done by the forklift truck as it lifts the box to the given height is 6860J.
WorkWork is simply referred to as the displacement of an object when a push or pull force is applied to the object. It is the energy transferred from or to an object when force is applied to it along a displacement.
It is expressed as;
W = F × s
Where F is force and s is displacement
Given the data in the question;
Mass of box m = 350kgDisplacement s = 2mWork done W = ?We substitute our given values into the expression above.
W = F × s
But F = Weight = mass × acceleration due to gravity = mg
acceleration due to gravity ( g = 9.8m/s²)
Hence,
W = mg × s
W = 350kg × 9.8m/s² × 2m
W = 6860kgm²/s²
W = 6860J
Therefore, the work done by the forklift truck as it lifts the box to the given height is 6860J.
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4. When setting goals, you should do everything EXCEPT which of the following?
Take into account your current level of activity.
Ask all of your friends what they think your goals should be..
Set a time limit.
Be realistic.
An oscillator completes 240 cycles in 5.2 minutes.
Calculate its period (in seconds) and frequency (in Hz).
Answer:
I. Period = 1.3 seconds
II. Frequency = 0.769 Hertz
Explanation:
Given the following data;
Number of oscillation = 240 cycles
Time = 5.2 minutes.
Conversion:
1 minute = 60 seconds
5.2 minutes = X seconds
X = 60 * 5.2
X = 312 seconds
To find the following;
I. Period
Mathematically, the number of oscillation of a pendulum is given by the formula;
[tex] Number \; of \; oscillation = \frac {Time}{Period} [/tex]
Making period the subject of formula, we have;
[tex] Period = \frac {Time}{Number \; of \; oscillation} [/tex]
Substituting into the formula, we have;
[tex] Period = \frac {312}{240} [/tex]
Period = 1.3 seconds
II. Frequency
[tex] Frequency = \frac {1}{Period} [/tex]
Substituting the values into the formula, we have;
[tex] Frequency = \frac {1}{1.3} [/tex]
Frequency = 0.769 Hertz
Một vật không mang điện sẽ bị nhiễm điện dương khí
Answer:
không có điện
Explanation:
Question in the picture please help me...
Show all steps please....
Answer: i)A to B : (ice) freezing
ii) B to C (water) boiling
C to D (steam) evaporating
explanation: 0° is the freezing point of water when temperature increases from 0° the water starts melting. As 100° is the boiling point of water so at 100° the water completely melts and it starts boiling during boiling water changes into steam(water vapour) and it evaporates
Which element would have the lowest electronegativity? (1 point)
an element with a small number of valence electrons and a large atomic radius
an element with a small number of valence electrons and a small atomic radius
an element with a large number of valence electrons and a small atomic radius
an element with a large number of valence electrons and a large atomic radius
Answer:
an element with a small number of valence electrons and a large atomic radius
Numerical Problems:
a. A man runs 1200 m on a straight line in 4 minutes. Find his velocity.
can anyone tell me with process
[tex]\boxed{\sf Velocity=\dfrac{Displacement}{Time}}[/tex]
[tex]\\ \sf\longmapsto Velocity=\dfrac{1200}{240}[/tex]
[tex]\\ \sf\longmapsto Velocity=5m/s[/tex]
what does a resistor in an electrical dp
Answer:
A resistor is a passive two-terminal electrical component that implements electrical resistance as a circuit element. In electronic circuits, resistors are used to reduce current flow, adjust signal levels, to divide voltages, bias active elements, and terminate transmission lines, among other uses.
Explanation:
Resistors are common elements of electrical networks and electronic circuits and are ubiquitous in electronic equipment. Practical resistors as discrete components can be composed of various compounds and forms. Resistors are also implemented within integrated circuits.
The electrical function of a resistor is specified by its resistance: common commercial resistors are manufactured over a range of more than nine orders of magnitude. The nominal value of the resistance falls within the manufacturing tolerance, indicated on the component.
The velocity ratio of a pulley system is 4. What does it means
Answer:
Velocity ratio = distance travelled by effort / distance travelled by load. Which is equal to effort arm / load arm. When velocity ratio (VR) is 4, the effort arm is greater than load arm.
Can someone pls help, thank you in advance!
What is an example of a force applied at an angle to displacement
Answer:
an object sliding down hill
Explanation:
On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.
A object of mass 200kg is pushed from rest by a force of 500N along a horizontal plane for 5.0 seconds. Calculate the acceleration of the object
Answer:
force=mass×acceleration
hence
acceleration is given by force÷mass
(500÷200)*5=12.5
Which of the following phenomena can be used to support the particle theory of light? Select all that apply.
The Photoelectric Effect
Polarization
Interference Patterns
Refraction
Reflection
một oto đang chạy với vận tốc là 10m/s thì tăng tốc và chuyển động nhanh dần đều sau 20s thì đạt vận tốc 14m/s
a) tính gia tốc của xe
b) tính vận tốc xe sau 40s kể từ khi tăng tốc và quãng đường xe đi được trong thời gian đó
a) gia tốc = vf-vi / t
a = 14-10 / 20
a = 0,2ms⁻²
b) dưới dạng a = Δv / t
v = lúc
v = 0,2 × 40
v = 8ms⁻¹
như v = d / t
do đó d = vt
d = 8 × 40
d = 320m
hãy đánh dấu là trí óc nhất
The Weight of an object is more at pole and less at equator of the earth. Give reason
The Weight of an object is more at pole and less at equator of the earth because the distance of the pole is less than the distance of the equator from the centre of the earth.
1) The position of an object to the north of a flagpole is given by x(t) = bt2 – c , where b and c are constants.
a) What is v(t), the velocity of the object as a function of time?
b) What is a(t), the acceleration of the object as a function of time?
c) At some time t the object is located at the flagpole. What is the velocity of the
object at that instant?
Answer:
a) The velocity of the object as a function of time, v(t) is 2·b·t
b) The acceleration of the function of time, a(t) is 2·b
c) The time at which the object is at the flagpole is t = √(c/b)
Explanation:
The function that gives the position of the object north of the flagpole, x(t) is presented as follows;
x(t) = b·t² - c (b and c are constants)
a) The velocity of the object as a function of time, v(t), is derived as follows
v(t) = x'(t) = d(b·t² - c)/dt = 2·b·t
The velocity of the object as a function of time, v(t) = 2·b·t
b) The acceleration of the function of time, a(t) = v'(t) = d(2·b·t)/dt = 2·b
c) The time at which the object is at the flagpole is given by the x-intercept of the function, where x(t) = 0, as follows;
At the x-intercept, we have, x(t) = 0 and x(t) = b·t² - c
∴ 0 = b·t² - c, which gives
b·t² = c
t² = c/b
t = ±√(c/b), we reject the negative value to get;
The time at which the object is at the flagpole, t = √(c/b).
An element is highly conductive, highly reactive, soft, and lustrous. The element most likely belongs to which group?(1 point)
transition metals
noble gases
metalloids
alkali metals
Answer:
Alkali metals
Explanation:
Elements in this group are highly reactive, soft, lustrous and highly conductive.
An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.
Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.
This causes them to be soft and highly reactive because:
The single valance electron leads to weak bonds amongst the element's atoms which makes them softThe elements want to lose the single valance electron so as to become stable so they will react with other elements to give away the electron.Examples of alkali electrons include:
Lithium Sodium Potassium etcIn conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.
Find out more at https://brainly.com/question/18722874.
What is your wheel and axle
Explanation:
The wheel and axle is a type of simple machine used to make tasks easier in terms of manipulating force by applying the concept of mechanical advantage.
an electron is moving at 3.6 x 10^3 m/s. a photon with what wavelength would have the same momentum????
Answer:
The wavelength of the photon that would have the same momentum as the electron is 202.2180996 nm
Explanation:
The velocity of the electron, v = 3.6 × 10³ m/s
The momentum of an electron, [tex]p_e[/tex] = m × v
Where;
v = The mass of the electron = 9.109 × 10⁻³¹ kg
∴ [tex]p_e[/tex] = 9.109 × 10⁻³¹ kg × 3.6 × 10³ m/s = 3.27924 × 10⁻²⁷ kg·m/s
According to the de Broglie equation, the momentum of a photon, p, is given as follows;
p = h/λ
Where;
h = 6.63 × 10⁻³⁴ J·s
λ = The wavelength of the photon
∴ λ = h/p
According to the question, we have;
p = [tex]p_e[/tex] = 3.27924 × 10⁻²⁷ kg·m/s
∴ λ = 6.63 × 10⁻³⁴ J·s/(3.27924 × 10⁻²⁷ kg·m/s) = 2.02180993 × 10⁻⁷ m
The wavelength of the photon, λ = 2.02180993 × 10⁻⁷ m = 202.2180993 × 10⁻⁹ m = 202.2180993 nm.
If a force of 1250 N acts on an area of 25 metres squared, what will be the pressure acting on the surface? Show your working as well.
Answer: 50 Pa
Explanation:
Pressure = Force/Area
Pressure = N/m^2
Pressure = 1250/25
Pressure = 50
Pressure = 50 Pa
please click thanks and mark brainliest if you like :)
Answer:
50 pascal
Explanation:
Force=1250N
Area=25m^2 [ pressure=force/area]
P=F/A
1250/25=50
can someone pls help me calculate this?
Answer:
10N is the answerrrerreer
Answer:
f= 100N
Explanation: F=m×(v₀-vf/t)
=0.05ₓ(200-0/0.1)
=0.05ₓ2000
=100N
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a box being pushed with a force of 85N right slides along the floor with a constant speed of 0.5m/s. What is the force of sliding friction on the box?
Answer:
[tex]85\; \rm N[/tex].
Explanation:
The box is sliding with a constant speed in a fixed direction (to the right.) In other words the velocity of this box is constant. Hence, this box would be in a translational equilibrium. The acceleration of this box would be zero.
By Newton's Second Law of motion, the net force on this box would be [tex]0[/tex]. In other words, forces on this box are balanced.
The question is asking for the size of the friction on the box. Assuming that the floor is horizontal. The friction on this box would also be horizontal,
The only other force that could balance that friction would be the [tex]85\; \rm N[/tex] push to the right. The direction of this push is horizontal (to the right.) Hence, the entirety of that [tex]85\; \rm N\![/tex] would be in the horizontal direction.
Thus, forces on this box in the horizontal direction would be:
The [tex]85\; \rm N[/tex] push to the right.Friction that opposes the rightward motion of the box (that is, to the left.)Since these two forces must balance each other, the size of the friction would also be [tex]85\; \rm N[/tex].
Two football players run towards each other along a straight path in Penrith Park in the clash between the Melbourne storms and the Penrith Panthers a month ago. Melbourne's Justin Olam who is about 95kg and ran towards Viliame Kikau at 3.75m/s. Viliame Kikau is 111kg and moves towards Justin Olam at 4.10m/s. They end up in a head-on collision and are stuck together.
A) What is their velocity immediately after the collision?
B) What are the initial and final kinetic energies of the system?
Answer:
a) v = 0.4799 m / s, b) K₀ = 1600.92 J, K_f = 5.46 J
Explanation:
a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.
Initial instant. Before the crash
p₀ = m v₁ + M v₂
where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive
Final moment. After the crash
p_f = (m + M) v
as the system is isolated, the moment is preserved
p₀ = p_f
m v₁ + M v₂ = (m + M) v
v =[tex]\frac{m v_1 + M v_2}{m+M}[/tex]
let's calculate
v = [tex]\frac{ -95 \ 3.75 \ + 111 \ 4.10}{95+111}[/tex]
v = 0.4799 m / s
b) let's find the initial kinetic energy of the system
K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2
K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2
K₀ = 1600.92 J
the final kinetic energy
K_f = ½ (m + M) v ^ 2
k_f = ½ (95 + 111) 0.4799 ^ 2
K_f = 5.46 J
An object is thrown from the ground with an initial velocity of 30 m/s. What is the velocity at the point 25 m above the ground?
Answer:
It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.
V^2=U^2+2*a*x
V=0 (as it is at max height)
U=30ms^-1 (initial speed)
a=-g /-9.8ms^-2 (as it is moving against gravity)
x is the variable you want to calculate (height)
0=30^2+2*(-9.8)*x
x=-30^2/2*-9.8
x=45.92m
Answer:
35
Explanation:
dfddffffffffffffdddfr
State examples of a transverse wave.
ripples on the surface of water.
vibrations in a guitar string.
a Mexican wave in a sports stadium.
electromagnetic waves – eg light waves, microwaves, radio waves.
seismic S-waves.
Answer
Light and other types of electromagnetic radiation are examples of transverse waves. Some other examples of transverse waves include a ripple on a pond and a wave in a string.
what do you mean by scattering of light
Answer:
Scattering of light is the phenomenon in which light rays get deviated from its straight path on striking an obstacle like dust or gas molecules, water vapours etc. Scattering of light gives rise to many spectacular phenomena such as Tyndall effect and the “red hues of sunrise and sunset”.
If 20N force produces an acceleration of 5ms^-2 In a body then the mass of the body will be:
A.4kg
B.5kg
C.1/4kg
D.1/5kg
7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ball h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.
Answer:
8392
Explanation:
d=s/t
Động vật nào sau đây máu đi nuôi cơ thể không pha trộn giữa máu giàu O2 và máu giàu CO2?
A.
Bò sát, chim, thú
B.
Cá, bò sát, chim
C.
Cá, lưỡng cư, bò sát
D.
Cá, chim, thú
Answer:
D
Explanation:
Cá xương, chim, thú, cá sấu không có sự pha trộn máu giàu O2 và máu giàu CO2 ở tim vì tim cá có 2 ngăn, tim các loài chim, thú, cá sấu có 4 ngăn
what is Newton's first law of motion?
EXPLAIN WITH SOME EXAMPLES
Answer:
The tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. This is why, the first law of motion is also known as the law of inertia.
Example.: A ball at rest on the ground continues to be at rest unless someone kicks it or any external force acts on it.
Explanation:
I hope this will help you buddy
A cyclist goes round a circular path of circumference 343 m in s. The angle made by him, with the vertical is
WITH STEPS PLZ
you need the number of seconds to calculate the distance. I'm assuming that the seconds were written in the question but you forgot to write the here, you can calculate the distance then and find the exact point where the cyclist stopped since you have the circumference. then you can find the angle (after finding the distance he cycled) by subtracting it from 360 since the circular path will have an angle of 360 degrees.
Although your question lacks some data A general answer is provided :
The angle made vertically = 360° - x°
where x = angle at which the the cyclist stopped
and circumference of the circular path = Total Distance travelled by the cyclist
First step : determine the distance travelled by the cyclist vertically
circumference = 2*π*r
343 = 2*π * r
∴ r = 343 / ( 2π ) = 54.59
therefore distance travelled by the cyclist vertically = 54.59 * 2 = 109.18 m
Given that the time travelled is missing
assuming the angle to the distance travelled by the cyclist before it stopped vertically = 360° - x°
This is because the Total angle of a circular path = 360°
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write down any 5 example of conservation of momentum?
Answer:
1) Motion of air mass moving from equator northward (closer to earth axis)
2) Motion of object in orbit
3) Collision of 2 objects
4) Skater changing rotation by extension of arms
5) Motion of rocket due to velocity of expelled gas