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ella [17]
3 years ago
5

Determine weather or not the equation below is balanced.

Physics
1 answer:
julia-pushkina [17]3 years ago
8 0

Answer:

I think the above equation is not balanced.

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At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the ear
Mandarinka [93]
We apply the gravity calculation expressed in the formula: g=GM/r2 
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1)       Radius = √6.674e-11*5.972e24/8             = 7058 kms    Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2     Radius=√6.674e-11*5.972e24/4.9                 = 9019 kms        Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.
4 0
3 years ago
A car has a mass of 1200 kg. A man can comfortably push down with a force of 400 N. If the small piston in a hydraulic jack has
kotegsom [21]
A) F1/A1=F2/A2
F2 = 12000 Newtons, while F1 = 400 Newtons and area A1= 6 cm²
400/6 = 12000/A2
A2 = (6 × 12000)/400
     =  180 cm²
 
b) How long does this lift the car
 d2 =F1/F2×d1
      = (400/12000)× 20
       = (1/30)×20
       = 20/30
       = 0.67 cm
This lifts the car a distance of 0.67 cm
        

4 0
3 years ago
Which of the following is an example or an orginasm mantaining homeostasis
Kryger [21]

a cat grooming itself

8 0
3 years ago
A white pool ball of mass 1.0 kg moving at 10 m/s collides with
solniwko [45]

The velocity of the red ball after the collision is 5.8 m/s

Explanation:

In absence of external forces on the system, we can apply the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision, so we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 1.0 kg is the mass of the pool ball

u_1 = 10 m/s is the initial velocity of the pool ball

v_1 = 3.0 m/s is the final velocity of the pool ball

m_2 = 1.2 kg is the mass of the red ball

u_2 = 0 is the initial velocity of the red ball

v_2 is the final velocity of the red ball

Solving the equation for v2, we find the final velocity of the red ball after the collision:

v_2 = \frac{m_1 u_1-m_1v_1}{m_2}=\frac{(1.0)(10)-(1.0)(3.0)}{1.2}=5.8 m/s

Learn more about collisions:

brainly.com/question/13966693#

brainly.com/question/6439920

#LearnwithBrainly

7 0
3 years ago
In tae-kwon-do, a hand is slammed down onto a target at a speed of 14 m/s and comes to a stop during the 3.0 ms collision. Assum
GrogVix [38]

Answer:

A) The impulse is 11,7 kg.m/s .  B) The average force on the hand is 3900N

Explanation:

Givens:

Hand's mass (m)= 0,90kg

Hand's initial velocity (vi)= 14m/s

Time of collision (t)= 3ms

<u>A) Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time. We use the formula:</u>

<em />

<em>/F/ = m. /Δv/</em>

<u>Where:</u>

F= Force of Impulse

m= Mass

Δv= change in speed ( vf-vi)

Using that formula we get:

/F/= 0,90kg . / (0m/s-14m/s) /

/F/=0,90kg . 14m/s

<em>/F/ = 11,7 kg. m/s</em>

*note that (vf) is 0 because the hand stops in the action, so it's final

velocity = 0

<u>B)The average force is equal to the change in the momentum over the change in time. We use the formula:</u>

<em>/F/= m. (Δv) : Δt</em>

<u>Where:</u>

F= the average force from the target

m= the mass of the hand

Δv= hange in speed (vf/vi)

Δt= change in time

*if we look closely, we find the we alredy have half of the equation because we have alredy calculated m.(Δv) in point A.

We boil the equation down to:

/F/= Impulse : Δt

/F/= 11,7kg.m/s : 0,003s

<em>/F/= 3900N</em>

*we use 0,003s as our time because the given time was 3ms.

*the final result is expressed in Newtons because our final result ends up beeing <em>kg.m/s²</em> = N

<u />

8 0
4 years ago
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