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just olya [345]
3 years ago
8

What net is required to push a sofa with a mass of 59 kilograms so that it accelerates at 9.75 meters/second^2 (assume a flat,fr

ictionless surface.)

Physics
1 answer:
stich3 [128]3 years ago
8 0

Newton's 2nd Law of Motion:  Force = (mass) · (acceleration)

Do I need to go any further ?

I get  575.25 Newtons .

Oh !  Look at that !  Right there at the top, it says "Using Newton's Second Law".  This might have suggested to you that maybe you could solve the problem by using Newton's Second Law.

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Which of the following statements about digestion is/are TRUE? (Select all that apply.)
MArishka [77]
1st one and fourth one are true.
7 0
3 years ago
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Light in vacuum is incident on the surface of a slab of transparent material. In the vacuum the beam makes an angle of 39.9° wit
wariber [46]

Answer:

n=2.053

Explanation:

We will use Snell's Law defined as:

n_{1}*Sin\theta_{1}=n_{2}*Sin\theta_{2}

Where n values are indexes of refraction and \theta values are the angles in each medium. For vacuum, the index of refraction in n=1. With this we have enough information to state:

1*Sin(39.9)=n_{2}*Sin(18.2)

Solving for n_{2} yields:

n_{2}=\frac{Sin(39.9)}{Sin(18.2)}=2.053

Remember to use degrees for trigonometric functions instead of radians!

6 0
3 years ago
Inertia is the resistance to change in motion so inertia depends solely on what
Alex Ar [27]
Inertia depends on mass, the more mass the more inertia.
6 0
3 years ago
How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00
laiz [17]

Answer:

Work= -7.68×10⁻¹⁴J

Explanation:

Given data

q₁=q₂=1.6×10⁻¹⁹C

r₁=2.00×10⁻¹⁰m

r₂=3.00×10⁻¹⁵m

To find

Work

Solution

The work done on the charge is equal to difference in potential energy

W=ΔU

Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\  Work=-7.68*10^{-14}J

4 0
3 years ago
A flea jumps straight up to a maximum height of 0.400 m . what is its initial velocity v0 as it leaves the ground?
timama [110]

For an object`s motion, the Kinematic equation is,

v^2=v_{0}^2+2ah

Here, v is the final velocity and h is stands for the height of the object and a is the acceleration of the object.

As according to question,

v=0m/s,a=g-9.8 m/s^2 and h = 0.400 m

Thus, putting these values in above equation, we get

0= v_{0}^2 -2gh

or

v_{0} =\sqrt{2 \times 9.8 \times 0.400 }

v_{0} = 2.8 m/s

Therefore, initial velocity is 2.8 m/s



8 0
3 years ago
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