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vodka [1.7K]
3 years ago
10

A force of 20 N acts toward the east. Another force of 27 N acts at the same point toward the west. What is the magnitude and di

rection of a third force that produces a resultant of 10 N toward the east?
Physics
1 answer:
Jet001 [13]3 years ago
8 0
Since we are working in one dimension (left right or East West), we don't need to worry about angles! It's just simply a matter of adding things up!

First list out all the forces and add negative (-ive) signs to each of the 'west' forces like this.

20 East + (-27 West) + ? = 10 East

so it's easy to see that 20 + (-27) = -7
So to get to 10 from -7 just do the sum to get 17.
Since 17 is not negative it must be in the direction of East.
So the answer is:
Magnitude = 17 N
Direction = toward the East

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For a projectile launched horizontally, which of the following best describes the downward component of a projectile's velocity?
Angelina_Jolie [31]

C. The downward component of the projectile's velocity continually increases

Explanation:

The motion of a projectile consists of two independent motions:  

- A uniform motion (with constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (equal to the acceleration of gravity) in the downward direction  

Here we want to study the downward component of the projectile's velocity. Since the vertical motion is a uniformly accelerated motion, the vertical velocity is given by:

v=u+at

where

u = 0 is the initial vertical velocity (zero since the projectile is fired horizontally)

a=g=9.8 m/s^2 downward is the acceleration of gravity

t is the time

So the equation becomes

v=gt

This means that

C. The downward component of the projectile's velocity continually increases

Because every second, it increases by 9.8 m/s in the downward direction.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0
3 years ago
How is it technically correct to say that a car making a u-turn can have a constant speed but cannot have a constant velocity?
saw5 [17]

During the "U" part of the turn, the car would follow an approximately circular path, and if it's moving at a constant speed, it would have to accelerate toward the center of the circle in order to change its direction.

5 0
2 years ago
Which is the best example of Newton's First Law of Motion? A small, lightweight ball and a large, heavy ball are dropped off the
Marizza181 [45]
I believe the best example of Newton's First Law of motion would be the example or illustration with the basketball player. An object will move in a straight line or a given direction at a constant speed unless or until another force acts upon the object, causing a change in speed and or direction.
4 0
3 years ago
Read 2 more answers
For a standard production car, the highest road-tested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution
Ann [662]

Answer:

a = 8.06 m/s²

Explanation:

The acceleration of this car can be found using the first equation of motion:

v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}

where,

a = acceleration = ?

vf = final speed = 26.8 m/s

vi = initial speed = 0 m/s

t = time = 3.323 s

Therefore,

a = \frac{26.8\ m/s-0\ m/s}{3.323\ s}

<u>a = 8.06 m/s²</u>

3 0
2 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a
Vesna [10]

Answer:

Tangential speed=5.4 m/s

Radial acceleration=88.6m/s^2

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution=2\pi rad

2.59 rev=2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s

By using \pi=3.14

Angular velocity=\omega=16.27rad/s

Distance from axis=r=0.329 m

Tangential speed=r\omega=16.27\times 0.329=5.4m/s

Radial acceleration=\frac{v^2}{r}

Radial acceleration=\frac{(5.4)^2}{0.329}=88.6m/s^2

6 0
3 years ago
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