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irina [24]
3 years ago
13

What did Thomas Edison create that made him most famous today? I know it is something about light, but I do not believe he was f

irst. May someone tell me his most famous accomplishment in easy words to understand? Thanks
Physics
2 answers:
sp2606 [1]3 years ago
6 0

Answer:

One of the most famous and prolific inventors of all time, Thomas Alva Edison exerted a tremendous influence on modern life, contributing inventions such as the incandescent light bulb, the phonograph, and the motion picture camera, as well as improving the telegraph and telephone.

Explanation:

<em>Your </em><em>well-wisher</em>

Advocard [28]3 years ago
4 0

Explanation:

Thomas Edison invented the incandescent light bulb which is kind of like the light bulbs we use today. but they have been improved throughout the years.

hope this is simple and understandable

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The cylindrical head bolts on a car are to be tightened with a torque of 63.3 N · m. If a mechanic uses a wrench of length 18.2
Nataly_w [17]

Answer: The perpendicular force to be used to tighten the bolt is

F=347.8N

Explanation:

Mathematically the torque is expressed as

T=F*L

Where T= torque in Nm

F= force in N

And L = perpendicular distance

Now given

T=63.3Nm

L=18.2cm to meter 18.2/100

0.182m

F=?

Making F the subject of the formula in our equation

F=T/L

F=63.3/0.182

F=347.8N

What is torque?

Torque is the product of force and perpendicular distance of the line of action.

6 0
4 years ago
A pulley of radius 8.0 cm is connected to a motor that rotates at a rate 7000 rad s-1 and then decelerate uniformly at a rate of
zlopas [31]

Answer:

(a) α = - 1000 rad/s²

Negative sign represents deceleration.

(b) θ = 3581 rotations

(c) L = 1800 m

(d) a = - 80 m/s²  

Explanation:

(a)

using First equation of motion for angular motion:

ωf = ωi + αt

where,

ωf = Final Angular Speed = 2000 rad/s

ωi = Initial Angular Speed = 7000 rad/s

α = Angular Acceleration = ?

t = time = 5 s

Therefore,

2000 rad/s = 7000 rad/s + α(5s)

α = (2000 rad/s - 7000 rad/s)/5 s

<u>α = - 1000 rad/s²</u>

<u>Negative sign represents deceleration.</u>

(b)

Using second equation of motion:

θ = ωi t + (1/2)αt²

where,

θ = No. of Rotations = ?

Therefore,

θ = (7000 rad/s)(5 s) + (1/2)(- 1000 rad/s²)(5 s)²

θ = 35000 rad - 12500 rad

θ = (22500 rad)(1 rotation/2π rad)

<u>θ = 3581 rotations</u>

(c)

Length of String = L = (Circumference of Pulley)(θ)

L = [2π(0.08 m)][3581 rotations]

<u>L = 1800 m</u>

<u></u>

(d)

Tangential Acceleration = a = rα

a = (0.08 m)(-1000 rad/s²)

<u>a = - 80 m/s²</u>

4 0
3 years ago
A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler
Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

W=K_f - K_i

Since the cart was initially at rest, K_i = 0, so

W=K_f = \frac{1}{2}mv^2 (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, F_g = 99.5 N:

m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg

So solving eq.(1) for v, we find the final speed of the cart:

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s

2) 2.51\cdot 10^7 J

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

F=4.28 \cdot 10^5 N

d=586 m

So the work done is

W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J

3)  2.51\cdot 10^7 J

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

W=\Delta K = K_f - K_i

where

W is the work done

\Delta K is the change in kinetic energy

Therefore, the change in kinetic energy is

\Delta K = W = 2.51\cdot 10^7 J

4) 37.2 m/s

According to the work-energy theorem,

W=\Delta K = K_f - K_i

where

K_f is the final kinetic energy of the train

K_i = 0 is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

W=K_f = \frac{1}{2}mv^2

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s

7 0
4 years ago
What is the potential energy of a 20-kg safe sitting on a shelf 0.5 meters
ki77a [65]

Explanation:

P.E=MGH

Where m is mass

Where G is Acceleration Due to Gravity

Where h is Height

So the parameters are M = 20kg

G = 9.8m/s

H = 0.5meters

P.E= 20x9.8x0.5

=98J.

So Ans. is A= 98J

8 0
3 years ago
The orbital period of a satellite is 2 x 106 s and its total radius is 2.5 x 1012 m.
MrMuchimi

Explanation:

Orbital period = 2 x 10⁶s

Total radius = 2.5 x 10¹²m

Unknown:

Tangential speed of the satellite = ?

Solution:

The tangential speed simply expresses the rate at which the distance moves around its orbit with time.

   Since the orbit is taken to be circular;

Tangential speed = \frac{2 x pi x r}{t}

  r is the radius

  t is the time or orbital radius

  Tangential speed = \frac{2 x 3.14 x 2.5 x 10¹²}{2 x10[tex]^{6}}[/tex]

Tangential speed = 7.85 x 10⁶m/s

Learn more:

Circular motion brainly.com/question/2562955

#learnwithBrainly

 

7 0
3 years ago
Read 2 more answers
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