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1 year ago

Find the work done on a 50 Kg student by the elevator in 2 seconds, if the elevator is :

1 answer:

5 0

The work done on a 50 Kg student by the elevator in 2 seconds if the elevator is accelerating upwards from rest at a rate of 2 ms⁻² would be

The total amount of energy transferred when a force is applied to move an object through some distance.

The work done is the multiplication of applied force with displacement.

Work Done = Force × Displacement

As given in the problem we have to find the work done on a 50 Kg student by the elevator in 2 seconds if the elevator is accelerating upwards from rest at a rate of 2 ms⁻²,

Lets us first calculate the displacement of the elevator in 2 seconds if it is accelerating upwards from rest at a rate of 2 ms⁻².

s = ut + 0.5at²

= 0 + 0.5×2×2²

= 4 meters

The work done on the elevator = mgh

=50×9.81×4

=1962 joules

Thus, the work done on the elevator would be 1962 joules.

To learn more about work done, refer to the link;

brainly.com/question/13662169

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A circular orbit would have an eccentricity of

vagabundo [1.1K]

<h2>Answer: zero (0)</h2>

Explanation:

The orbit of a body around another in space, is described by six orbital elements that determine its orientation, position, size and shape.

In the specific case of the shape of the orbit, this is determined by its <u>eccentricity</u>, which varies between 0 and 1 in the case of closed orbits (circle and ellipse). When the eccentricity is 0, the shape of the orbit is circular, when this value begins to vary until approaching 1 (without reaching 1), the shape of the orbit becomes more elliptical.

In this sense, a circular orbit will have an eccentriciy of zero.

6 0

3 years ago

Two fans are watching a baseball game from different positions. One fan is located directly behind home plate, 18.3 mfrom the ba

hjlf

Answer:

Δt = 0.315s

Explanation:

To calculate the time difference, in which both fans hear the batterstrike, you first calculate the time which takes the sound to travel the distances to both fans:

$t_1=\frac{d_1}{v_s}$

$t_2=\frac{d_2}{v_s}$

d1: distance to the first fan = 18.3 m

d2: distance to the second fan = 127 m

vs: speed of sound = 345 m/s

You replace the values of the parameters to calculate t1 and t2:

$t_1=\frac{18.3m}{345m/s}=0.053s\\\\t_2=\frac{127m}{345m/s}=0.368s$

The difference in time will be:

$\Delta t =t_2-t_2=0.368s-0.053s=0.315s$

Hence, the time difference between hearing the sound at the location s of both fans is 0.315s

4 0

3 years ago

Every surface has different ____________________ of friction.

dedylja [7]

Answer:

amounts

Explanation:

every surface has different amounts of friction

hope this helps :) plz brainliest?

3 0

3 years ago

5. You're examining some of the tiny printing on one of the newer twenty-dollar bills. A 1.5 mm tall letter appears 3 mm tall an

kow [346]

Answer:

80mm or 8cm

Explanation:

According to the lens formula,

1/f = 1/u+1/v

If the object distance u = 4cm = 40mm

Object height = 1.5mm

Image height = 3mm

First, we need to get the image distance (v) using the magnification formula Magnification = image distance/object distance = Image height/object height

v/40=3/1.5

1.5v = 120

v = 120/1.5

v = 80mm

The image distance is 80mm

To get the focal length, we will substitute the image distance and the object distance in the mirror formula to have;

1/f = 1/40+1/-80

Note that the image formed by the lens is an upright image (virtual), therefore the image distance will be negative.

Also the focal length of the converging lens is positive. Our formula will become;

1/f = 1/40-1/80

1/f = 2-1/80

1/f = 1/80

f = 80mm

The focal length of the lens 80mm or 8cm

5 0

4 years ago

Read 2 more answers

An 870 N firefighter, F_ff, stands on a ladder that is 8.00 m long and has a weight of 355 N, F_ladder. The weight of the ladder

castortr0y [4]

Answer:

Explanation:

Weight of the ladder is 355N

WL = 355N

The weight of the ladder acts at center of the ladder I.e at 4m from the bottom

Weight of firefighters is 870N

Wf = 870N

The fire fighter is at 6.3m from the bottom of the ladder.

N1 is the normal force exerted by the wall

N2 is the normal force exerted by the ground

Using Newton law

Check attachment

ΣFy = 0, since body is in equilibrium

N₂ - WL - Wf = 0

N₂ = WL + Wf

N₂ = 870 + 355

N₂ = 1225 N

This the normal force exerted by the ground on the wall.

Now to get N₁, let take moment about point A

So, before we take the moment we need to make sure that the forces are perpendicular to the plane(ladder), we need to resolve the weight of the ladder, firefighter and the normal of the wall to be perpendicular to the plane.

ΣMa = 0

Clockwise moment is equal to anti-clockwise moment

Moment Is the produce of force and perpendicular distance.

M = F×r

So,

WL•Cos50 × 4 + Wf•Cos50 × 6.3 —N₁•Sin50 × 8 = 0

355•Cos50 × 4 + 870•Cos50 × 6.3 =

N₁•Sin50 × 8

1825.52 + 3523.12 = 6.13N₁

6.13N₁ = 5348.64

N₁ = 5348.64/6.13

N₁ = 872.54 N.

The normal force exerted by the wall is 872.54N

5 0

3 years ago