Work is done by a force on an object. It can causes an object to move. It requires motion (no motion = no work). Object moving in the same direction as the force = work. Object moving in a different direction than the force= no work. Work is equal to the force exerted on an object times the distance it moves. The metric unit for work is the joule (J). 1 joule= 1 newton-meter. The amount of work done is not affected by the time it takes to do the work. Power is the rate which work is done. Power equals work divided by time. The metric unit for power is the watt (W). 1 watt= 1 joule per second.
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When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the period of oscillation, T , neglecting the mass of the spring itself.
The period of oscillation is 1.33 sec.
Mass = 275.0 g
Suppose value of spring constant is 6.2 N/m.
We need to calculate the angular frequency
Using formula of angular frequency
Where, m = mass
k = spring constant
Put the value into the formula
We need to calculate the period of oscillation,
Using formula of time period
Put the value into the formula
Hence, The period of oscillation is 1.33 sec.
You cause a particle to move from point A, where the electric potential is 11.3 V, to point B, where the electric potential is −25.9 V. Calculate the change that occurs in the particle's electrostatic potential energy, when the particle is an electron, a proton, a neutral hydrogen atom, and a singly ionized helium atom (i.e., lacking one electron from its neutral state).
The electric potential is the electric potential energy per unit of charge
Using this definition, we can calculate the electrostatic potential energy change between point A and B:
Neutral hydrogen atom:
Singly ionized helium atom:
3. Someone at the third-floor window (12.0 m above the ground) hurls a ball downward at an angle of 45 degrees with a speed of 25 m/s. How fast will the ball be traveling when it strikes the sidewalk
Now we find the vertical component of the initial velocity downwards:
Now the final vertical velocity of the ball when it hits the ground:
using the equation of motion,
Since the horizontal component of the motion is uniform since no force acts in the horizontal direction:
Now the resultant final velocity:
The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the center of the wire's cross section as J(r) = Br, where r is in meters, J is in amperes per square meter, and B = 2.35 ✕ 105 A/m3. This function applies out to the wire's radius of 2.00 mm. How much current is contained within the width of a thin ring concentric with the wire if the ring has a radial width of 11.5 μm and is at a radial distance of 1.20 mm?
The current contained within the width of a thin ring concentric is 18.1 x 10⁻⁶AWhat is Current?
This is defined as electric charges moving through an electric conductor or space.Parameters
Current density of J(r) = Br, where B = 2.35 x 10⁵ A/m³.
I = Jₓ A
where I is current, A is area and J is current density
where 2r = circumference, Δr = width,
Substitute the values into the equation.
I= 2(2.35 x 10⁵)(1.2 x 10⁻³)^2(11.5 x 10⁻⁶)
= 18.1 x 10^-6 A
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18.1 x 10^-6 A
A cylindrical wire carries a current density of J(r) = Br, where B = 2.35 x 10^5 A/m^3, to find the current within a certain area we multiply the current density with the are of this area:
I = J*A
for a ring with r distance from the center and width Δr, where Δr
D=v^0 t+1/2 at^2 the 0 is supposed to be really small right next to the v under it
3. Suppose that an airplane flying 60 m/s, at a height of 300m, dropped a sack of flour (pere the effect of air resistance). How far from the point of release would the sack have traveled when it stack the
469.24m. An airplane flying 60m/s at a height of 300m dropped a sack of flour that stack the ground 469.24m from the point of release.
This is a example of horizontal parabolic projectile motion,and we represents this motion in the coordinate axis, which means that the velocity has components in x axis and y axis.
The equation of components on the x axis.
, where x is the distance and Vox the initial velocity before the drop
The equation of components on the y axis.
, where y is the height, and the velocity in y component before the drop is 0, reducing the equation to
Clear t from both the equation of components on the x axis and the y axis:
Equating both equations and clearing the distance x:
Substituting the values:
What effect does increasing mass have on the amount of friction generated
A uniform diameter rod is weighted at end and is floating in a liquid. Determine if the liquid is (a) lighter than water, (b) water, or (c) heavier than water. Justify your answer.
(C) Heavier than water
Diameter of the rod is uniform
The road is floating in a liquid.
Let the area of cross-section of the rod = A
Let the weight density of water = γw
Let the weight density of liquid = γl
Let the density of water = ρw
g is the gravitational force
So, weight of rod = density of the rod X volume
Wrod = ρw X g X ( A X 2L) + 2 ρwg (A X L)
Wrod = 4ρwgAL
Wrod = 4γwAL
We can see that the weight of the rod is more than the density of water. So, in water, the rod will sink.
If the rod is floating on the liquid then the liquid is heavier than the water.
Therefore, The liquid is heavier than the water.
a 20 hp electric motor pumps water from a walk that is 10m deep into a water tank 3m above the ground level. how long does it take the pump to fill a 2000 L water tank.