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Soloha48 [4]
2 years ago
10

Looking for some help with this :)

Physics
1 answer:
garri49 [273]2 years ago
5 0
So the ting goes sssskkkkkkrrrrrrrraaaaaaa papakakaka skidiki pop pop and a Pom Pom pppddddrrrr pooom big Shaq mans not hot
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Find the resultant force of the following forces :
bezimeni [28]

The resultant of the given forces is; 6√2 N

<h3>How to find the resultant of forces</h3>

We are given the forces as;

10 N along the x-axis which is +10 N in the x-direction

6 N along the y-axis which is +6N in the y-direction

4 N along the negative x-axis which is -4N

Thus;

Resultant force in the x-direction is; 10 - 4 = 6N

Resultant force in the y-direction is; 6N

Thus;

Total resultant force = √(6² + 6²)

Total resultant force = 6√2 N

Read more about finding resultant of a force at; brainly.com/question/14626208

4 0
2 years ago
Read 2 more answers
What must the driver do when approaching an intersection and seeing the traffic light turn from green to yellow?
NISA [10]
The person should start to slow down but if close enough or in the intersection go threw. Otherwise come to a complete stop until the light turns green again
4 0
3 years ago
.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c
ohaa [14]

Answer:

 a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

          I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

        F t = 2 (m v)

therefore the average force is

         F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

        Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

        Em_f = K = ½ m v²

energy is conserved in this system

        Em₀ = Em_f

        m g h = ½ m v²

        v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

        F = 2m √(2gh)   /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

          F = m a

          a = F / m

 

          a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

         a = 2 √ (2 9.8 1.5) / 10⁻³

         a = 1.1 10⁵ m / s²

6 0
2 years ago
Hanna tosses a ball straight up with enough speed to remain in the air for several seconds?
olga_2 [115]
A) the velocity is 0 m/s
3 0
3 years ago
Read 2 more answers
A rock of mass m is thrown straight up into the air with initial speed |v0 | and initial position y = 0 and it rises up to a max
REY [17]

Answer:

Explanation:

Case 1:

mass = m

initial velocity = vo

final velocity = 0

height = y

Use third equation of motion

v² = u² - 2as

0 = vo² - 2 g y

y = vo² / 2g       ... (1)

Case 2:

mass = 2m

initial velocity = 2vo

final velocity = 0

height = y '

Use third equation of motion

v² = u² - 2as

0 = 4vo² - 2 g y'

y ' = 4vo² / 2g

y' = 4 y

Thus, the second rock reaches the 4 times the distance traveled by the first rock.

7 0
2 years ago
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