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sattari [20]
2 years ago
6

Write the first law of motion​

Physics
2 answers:
Dima020 [189]2 years ago
7 0

Answer:

A object in rest will remain in rest and a object in motion will move at a constant pace and in a straight line unless a unbalanced force acts upon it

Explanation:

give brainliest if you'd like

Vaselesa [24]2 years ago
6 0

Newton's first law: the law of inertia

Newton's first law states that

if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

hope you understand thank you.

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Match the speed to the section that describes.
inn [45]

Answer:

a=2 ok do it and ........

4 0
3 years ago
HELP ME PLSSSS!!!!!!!
DedPeter [7]

Answer:

A

Explanation:

A. The molecules that make up olive oil are longer than those that

make up alcohol, so they have more resistance to sliding past one

another.

Olive oils are unsaturated fats with many carbons whereas, alcohols are typically short with very few carbons.

4 0
3 years ago
Read 2 more answers
The top of the pool table is 0.810 m from the floor. the placement of the tape is such that 0 m is aligned with the edge of the
8090 [49]
Compute first for the vertical motion, the formula is:

y = gt²/2 

0.810 m = (9.81 m/s²)(t)²/2 

t = 0.4064 s 


whereas the horizontal motion is computed by: 

x = (vx)t 

4.65 m = (vx)(0.4064 s) 

4.65 m/ 0.4064s = (vx)

(vx) = 11.44 m / s
So look for the final vertical speed. 

(vy) = gt 

(vy) = (9.81 m/s²)(0.4064 s) 

(vy) = 3.99 m/s 


speed with which it hit the ground: 

v = sqrt[(vx)² + (vy)²] 

v = sqrt[(11.44 m/s)² + (3.99 m/s)²] 

v = 12.12 m / s
6 0
3 years ago
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

3 0
3 years ago
Question 7 of 10
nevsk [136]

Answer:

a

Explanation:

you take 32,000kg ÷2.0m

4 0
3 years ago
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