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DIA [1.3K]
3 years ago
8

A force moves an object along an x axis. how is the work done by the force related to the force?

Physics
2 answers:
Effectus [21]3 years ago
6 0
When an object moves along the x axis through a particular distance (d) due to a force (F) applied on it, then the work done (W) and the applied force (F) are related by the formula below

W = Fd


Mnenie [13.5K]3 years ago
3 0
The work done is equal to the product of the force applied and the displacement produced in the direction of the force.
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A piece of transparent material that is used to focus light and form a image
Soloha48 [4]

Convex Lenses are used to focus light and form a image. They are transparent.

They are one type of lenses. Different type of lenses are used to focus light differently. The basic lenses are concave and convex. Convex lenses converge light falling on it whereas concave lenses diverge light falling on it.

4 0
3 years ago
Read 2 more answers
An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1
Svetlanka [38]

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

4 0
3 years ago
A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s.
wlad13 [49]

Answer:

12m/s

Explanation:

v_f=v_o+at

Let's call the velocity that the car maintains for 10 seconds v_f_1, and the final velocity v_f_2.

v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s

Hope this helps!

5 0
3 years ago
What is the speed of an object after falling (from rest) a distance of 9.80 meters near the surface of the Earth? Assume air
madam [21]

Hi there!

We can use the kinematic equation:

v_f^2 = v_i^2 + 2ad

vf = Final velocity (? m/s)

vi = initial velocity (0 m/s, dropped from rest)

a = acceleration (due to gravity, 9.8 m/s²)

d = distance (9.8 m)

Simplify the equation to solve for vf:

v_f^2 = 0 + 2ad\\\\v_f = \sqrt{2ad}

Substitute in the given values:

v_f = \sqrt{2(9.8)(9.8)} = \boxed{13.86 m/s}

8 0
3 years ago
A 2.61 g lead weight, initially at 11.1 ∘C, is submerged in 7.67 g of water at 52.6 ∘C in an insulated container. What is the fi
sleet_krkn [62]

Answer:

Equilibrium temperature will be T=52.2684^{\circ}C

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So 2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)

0.334T-3.67=1688-32.031T

T=52.2684^{\circ}C

5 0
3 years ago
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