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alex41 [277]
3 years ago
10

What does balanced and unbalanced forces have to do with newtons laws of motion

Physics
1 answer:
Ivahew [28]3 years ago
5 0
Newton's second law of motion involves the effect of force acting on a mass.
It has to be understood that this actually means the combination of any and
all individual forces acting on it, with their individual strengths and directions
all added together.

When all the individual forces acting on an object add up to zero, then the
whole group of forces is said to be 'balanced'.

When all the individual forces acting on an object don't add up to zero, then
the whole group of forces is said to be 'unbalanced'.

Notice that there's no such thing as 'a balanced force' or 'an unbalanced force'.
It's a <u>group</u> of two or more forces that's balanced or unbalanced.



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Answer the question with step.​
Maru [420]

Answer:

f1/f2 =W1/W2 = 1/3

.0 f2 = 3f1

As ,

1/F= 1/f1 +1/f2

...1/40 = 1/f1 - 1/3f1

f1=> 80/3 cm

... f2 = 2f1 = 3 x 80/3 = 80 cm

7 0
3 years ago
From the top of a cliff, a person tosses a pebble straight downward with an initial velocity of -9.0 meters/second. After 0.50 s
Irina-Kira [14]
Y - yo = Vo*t - g * (t^2) / 2

Vo = - 9.0 m/s
t = 0.50 s

=> y - yo = -9.0 m/s * 0.5 s - 9.8 m/s^2 * (0.5s)^2 / 2 = - 4.5m - 1.225m = - 5.725 m.

Answer: option c) - 5.7
8 0
3 years ago
Number of waves that pass a given point in one second
Studentka2010 [4]
<em>number of waves that pass a given point in one second is called <u>frequency..</u></em>
5 0
3 years ago
When light of wavelength 240 nm falls on a potassium surface, electrons having a maximum kinetic energy of 2.93 eV are emitted.
olchik [2.2K]

A) We want to find the work function of the potassium. Apply this equation:

E = 1243/λ - Φ

E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function

Given values:

E = 2.93eV, λ = 240nm

Plug in and solve for Φ:

2.93 = 1243/240 - Φ

Φ = 2.25eV

B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:

E = 1243/λ - Φ

0 = 1243/λ - Φ

0 = 1243/λ - 2.25

λ = 552nm

C) We want to find the frequency associated with the threshold wavelength. Apply this equation:

c = fλ

c = speed of light in a vacuum, f = frequency, λ = wavelength

Given values:

c = 3×10⁸m/s, λ = 5.52×10⁻⁷m

Plug in and solve for f:

3×10⁸ = f(5.52×10⁻⁷)

f = 5.43×10¹⁴Hz

7 0
3 years ago
The chart show the masses and velocities of two colliding objects that stick together after a collision.
finlep [7]

Answer:

<u><em></em></u>

  • <u><em>1,500 kg.m/s</em></u>

Explanation:

First, arrange the information in a table:

Object        Mass (kg)           Velocity (m/s)

  A                    200                      15

  B                     150                    - 10

After the collision, the two objects are stick together, thus you talk aobut one object and one momentum.

According to the law of convervation of momentum, the momentum after the collision is equal to the momentum before the collision.

<u>Momentum before the collision, P₁</u>:

          P_1=m_{A,1}\times v_{A,1}+m_{B,1}\times v_{B,1}

         P_1=200kg\times 15m/s+150kg\times (-10m/s)\\\\  P_1=3000kg.m/s-1500kg.m/s=1500kg.m/s

<u>Momentum after the collision</u>:

  • As stated, it es equal to the momentum before the collision: 1,500 kg . m/s
6 0
3 years ago
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