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Arlecino [84]
2 years ago
10

. Calculate the efficiency of a bicycle if the input work to turn the pedal is 45J and the output work is 20J. * 1 point 2.25 2.

25% 44 44.4%
Physics
1 answer:
cestrela7 [59]2 years ago
8 0

20/45=0.4*100= 44.4 so the answer is..................................................

Answer: 44.4%

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A point charge q1=−4.00nc is at the point x=0.600 meters, y=0.800 meters, and a second point charge q2=+6.00nc is at the point x
Katen [24]

electric field between mid point of two charges is given by

E = E_1 + E_2

here we have

E_1 = \frac{kq_1}{r^2}

E_1 = \frac{9*10^9 * 4 * 10^{-9}}{0.4^2}

E_1 = 225 N/c

now similarly for other charge

E_2 = \frac{kq_2}{r^2}

E_2 = \frac{9*10^9 * 6 * 10^{-9}}{0.4^2}

E_2 = 337.5 N/C

so the net electric field will be given as

E = 225 + 337.5

E = 565.5 N/C

8 0
2 years ago
Read 2 more answers
Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. otherwise, we keep
kvv77 [185]

This pair of pucks will rotate at the same rate.

Further Explanation:

Hover Puck:  

The Hover Puck coasts effectively on a self-produced pad of air until followed up on by an outside power.  

Employments of hover puck:  

The Hover Puck coasts effectively on a self-produced pad of air until followed up on by an outside power.  

• Use one Hover Puck to show themes from Newton's First Law to impacts to reflection.  

• Use at least two pucks to examine the protection laws in two measurements.  

• Set up a bowling alley utilizing plastic beverage bottles.  

Innovation of hover puck:  

The First Rubber Hockey Pucks Were Made From Sliced-Up Lacrosse Balls. At the point when the game moved inside, entire balls were initially utilized, yet arena proprietors before long thought that it was desirable over cut them into thirds and keep the center segment. This fundamental plan was the standard by 1885.  

hockey puck was developed:  

The hockey puck appeared in 1875. It's hazy who really imagined it. Specialists accept the main hockey puck was likely only an elastic ball cut down the middle. This gave players an article with a level side that would slide over the ice.

Subject: physics

Level: High School

Keywords: Hover Puck, Employments of hover puck, Innovation of hover puck, hockey puck was developed.  

Related links:  

Learn more about evolution on

brainly.com/question/727976

brainly.com/question/6953278

5 0
3 years ago
Read 2 more answers
A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
Tasya [4]
<h3><u>Answer;</u></h3>

<em>Work = 125 joules </em>

<h3><u>Explanation and solution</u>;</h3>
  • Work is the product of force and the distance covered. Therefore, Work = force × distance.
  • Work is measured in joules.
  • Work is also a change in energy, such that work is done when energy changes, so when kinetic energy, or potential energy changes the there is work being done.

Thus; kinetic energy = work done

Kinetic energy = 1/2mv²

                       = 1/2 × 10× 5²

                       = 5 × 25

                       = 125 joules

Hence, work done is 125 joules.

3 0
2 years ago
Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2
Alona [7]

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

<em>Given: W = 12 J</em>

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

8 0
2 years ago
A 1.0 kg object moving at 4.5 m/s has a wavelength of:
Lisa [10]
By wave particle  duality.

Wavelength , λ = h / mv

where h = Planck's constant = 6.63 * 10⁻³⁴ Js,  m = mass in kg,  v = velocity in m/s.
m = 1kg,  v = 4.5 m/s

λ = h / mv

λ = (6.63 * 10⁻³⁴) /(1*4.5)

λ ≈  1.473 * 10⁻³⁴  m

Option D.
7 0
2 years ago
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