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Nookie1986 [14]
2 years ago
5

A 3.7-kg object is acted on by two forces. One of the forces is 11 N acting toward the

Physics
1 answer:
vitfil [10]2 years ago
7 0

Answer:

F₂ = -7.3 N

Explanation:

Given that,

The mass of an object, m₁ = 3.7 kg

First force, F₁ = 11 N

The net acceleration of the object is 1 m/s².

We know that,

F₁+F₂ = ma

11+F₂ = (3.7)(1)

F₂ = 3.7-11

F₂ = -7.3 N

so, the other force is 7.3 N and it is acting in west direction.

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3 years ago
The zero tolerance law applies to drivers _________.
s2008m [1.1K]

The zero tolerance law applies to drivers <u>under the age of 21.</u>

<h3>Zero Tolerance Law:</h3>

Drivers under the age of 21 who operate a motor vehicle with a BAC of between 0.02% and 0.07% are subject to what is known as the "Zero Tolerance Law." The Zero Tolerance Law aims to prevent underage drivers from driving after consuming alcohol.

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brainly.com/question/17257160

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4 0
2 years ago
In 3 meters a person running 0.5 m/s accelerates 1.2 m/s 2. How fast were they going afterward? Choose the right equation for th
Feliz [49]

Answer:

2.72 m/s

Explanation:

In 3 meters a person running 0.5 m/s accelerates 1.2 m/s².

It means,

Distance, s = 3 m

Initial velocity, u = 0.5 m/s

Acceleration, a = 1.2 m/s²

We need to find the final velocity of the person. Using equation of motion to find it as follows :

v^2-u^2=2as\\\\v^2=2as+u^2\\\\v^2=2\times 1.2\times 3+(0.5)^2\\\\v=2.72\ m/s

So, the final velocity of the person is 2.72 m/s.

7 0
2 years ago
14. a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s. a. how much later does the ball
Burka [1]

The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s

<h3>Motion Under Gravity</h3>

The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.

Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.

a. how much later does the ball hit the ground?

The time can be calculated by considering the vertical component of the motion with the use of formula below.

h = ut + 1/2gt²

Where

  • Height h = 75 m
  • Initial velocity u = 0 ( vertical velocity )
  • Acceleration due to gravity g = 9.8 m/s²
  • Time t = ?

Substitute all the parameters into the formula

75 = 0 + 1/2 × 9.8 × t²

75 = 4.9t²

t² = 75/4.9

t² = 15.30

t = √15.3

t = 3.9 s

b. how far from the building will it land?

The range can be found by using the formula

R = ut

Where u = 4.6 m/s ( horizontal velocity )

R = 4.6 × 3.9

R = 18 m

c. what is the velocity of the ball just before it hits the ground?

The final velocity will be

v = u + gt

v = 4.6 + 9.8 × 3.9

v = 4.6 + 38.22

v = 42.82 m/s

Therefore, the answers are 3.9 s, 18 m and 42.82 m/s

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7 0
1 year ago
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