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vagabundo [1.1K]
3 years ago
14

The amount of work done against friction to slide a box in a straight line across a uniform, horizontal floor depends most on th

e
Physics
2 answers:
Ivenika [448]3 years ago
7 0
Amount of force that is applied to the box
ozzi3 years ago
3 0

Answer: Distance the box is moved.

Explanation:

Work is defined as force per distance moved by the force.

Work is done when there is relative change in position of an object exerted by the force.

Mathematically W = force × distance.

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a student is pushing a 50 kilogram cart with a force of 500 newtons another students measures the speed of the cart and finds th
Irina-Kira [14]

The force of friction is 300 N

Explanation:

We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. So we can write

\sum F = ma

where

\sum F is the net force acting on the object

m is its mass

a is its acceleration

For the cart in this problem, we have two forces acting on it:

- The force of push, F = 500 N, forward

- The force of friction, F_f, backward

So Newton's second law can be rewritten as

F-F_f = ma

where

m = 50 kg

a=4 m/s^2 is the acceleration of the cart

And solving for F_f, we find the force of friction:

F_f = F-ma=500-(50)(4)=300 N

Learn more about force of friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

4 0
3 years ago
The element carbon has 3 naturally occurring isotopes. About 99% of carbon isotopes are C-12, about 1% are C-13, and a tiny amou
mezya [45]
Average atomic mass of carbon is 12.01 amu
3 0
3 years ago
An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 ​7 ​​ m/s over
stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :  

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron  

• a is the acceleration of the electron  

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

          a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

  = 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .  

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

  t=√2(xf-xi)/a

 t=3.765×10^-10 s

now we can use the power P Eq.  

 P=W/Δt => ΔK/Δt  

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0
3 years ago
What is the centripetal force
Lelechka [254]

Answer:90N

Explanation:

Mass=30kg

Centripetal acceleration=3m/s^2

centripetal force=mass x centripetal acceleration

Centripetal force=30 x 3

Centripetal force =90

Centripetal force =90N

4 0
3 years ago
The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th
Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

E_i = E_f

0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

PART B) Replacing the values given as,

h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}

v_f = 1.8486m/s

Therefore the speed of the masses would be 1.8486m/s

6 0
3 years ago
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