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Anton [14]
2 years ago
9

Hmmm... Who can answer this question?How did life begin?Just a practice

Physics
1 answer:
Schach [20]2 years ago
8 0

Answer:

The earliest known life-forms are putative fossilized microorganisms, found in hydrothermal vent precipitates, that may have lived as early as 4.28 Gya (billion years ago), relatively soon after the oceans formed 4.41 Gya, and not long after the formation of the Earth 4.54 Gya.

Explanation:

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You might be interested in
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. She finds the pendulum m
forsale [732]

Answer:

Acceleration due to gravity will be g=5.718m/sec^2

Explanation:

We have given length of pendulum l = 55 cm = 0.55 m

It is given that pendulum completed 100 swings in 145 sec

So time taken by pendulum for 1 swing =\frac{145}{100}=1.45sec

We have to find the acceleration due to gravity at that point

We know that time period of pendulum;um is given by

T=2\pi \sqrt{\frac{l}{g}}

So 1.45=2\times 3.14\times \sqrt{\frac{0.55}{g}}

\sqrt{\frac{0.55}{g}}=0.230

Squaring both side

{\frac{0.3025}{g}}=0.0529

g=5.718m/sec^2

So acceleration due to gravity will be g=5.718m/sec^2

3 0
3 years ago
Consider the two-body situation at the right. A 300kg crate rests on an inclined plane and is connected by a cable to a 100 kg m
trasher [3.6K]

Answer:

a= 0.578 m/s

T = 1037.8 N

Explanation:

Data

m₁= 300 kg

m₂= 100 kg

inclined plane, θ =  30°

μk = 0.120

Newton's second law to m₁:

We define the x-axis in the direction parallel to the movement of the 300kg (m₁) crate on the ramp and the y-axis in the direction perpendicular to it.

∑F = m₁*a Formula (1)

Forces acting on m₁

W₁: m₁ weight : In vertical direction

N : Normal force : perpendicular to the inclined plane

f : Friction force: parallel to the inclined plane

T:  cable tension : parallel to inclined plane

Calculated of the W₁

W₁=m₁*g

W₁= 300kg* 9.8 m/s² = 2940 N

x-y weight components

W₁x= W₁sin θ =2940 N*sin(30)° =1470 N

W₁y= W₁cos θ =2940 N *cos(30)° =2156.4 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - W₁y = 0

N = W₁y

N = 2156.4 N

Calculated of the f

f = μk* N= (0.120)*(2156.4 N)

f = 258.77 N

Newton's second law to m₁ in direction  x-axis :

∑Fx = m₁*ax   ,ax  =a

We assume that m₁ descends on the inclined plane and we positively take the direction of movement:

wx-f-T = m*a

wx - f - m*a =T

1470  -258.77 -300*a =T

T= 1211.23-300*a   Equation (1)

Newton's second law to m₂

∑Fy = m₂*ay   ,ay  =a

Forces acting on m₂

W₂: m₂ weight : In vertical direction

T:  cable tension:In vertical direction

Calculated of the W₂

W₂=m₂*g

W₂= 100kg* 9.8 m/s² = 980 N

∑Fy = m₂*a

Because we assume that m₁ descends on the inclined plane, then, m₂ ascends  vertically, we take positive the direction of movement:

T-W₂ = m₂*a

T-980 = 100*a

T = 980 + 100*a Equation (2)

Problem development

Equation (1) =  Equation (2) = T

1211.23-300*a= 980  + 100*a

1211.23- 980 = 100*a + 300*a

231.23 = 400*a

a= 231.23 / 400

a= 0.578 m/s

Because the acceleration tested positive then effectively m₁ descends on the inclined plane and m₂ ascends  vertically.

We replace a= 0.578 m/s in the equatión (2)

T = 980 + 100* (0.578 )

T = 1037.8 N

5 0
3 years ago
Can someone pleaseeee answer this !!!!!!
LenaWriter [7]

Answer:

The person with locked legs will experience greater impact force.

Explanation:

Let the two persons be of nearly equal mass (say m)

The final velocity of an object (person) dropped from a height H (here 2 meters) is given by,

v=\sqrt{2gH}

(g = acceleration due to gravity)

which can be derived from Newton's equation of motion,

v^2=u^2+2aS

Now, the time taken (say t ) for the momentum ( mv ) to change to zero will be more in the case of the person who bends his legs on impact than who keeps his legs locked.

We know that,

Force=\frac{\Delta(mv)}{t}

Naturally, the person who bends his legs will experience lesser force since t is larger.

3 0
3 years ago
Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a
Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and  t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

      Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

      Vf = √( 2 g X)

      Vf = √(2 9.8  1)

      Vf = 4.43 m/s

This is the speed with which it reaches the ground

 

Having the final speed we can find the time

      Vf = Vo + g t

       t = Vf / g

       t = 4.43 / 9.8

       t = 0.45 s

This is the time of fall of the body to touch the ground

3 0
3 years ago
If a group of workers can apply a force of 1,275 newtons to move a crate 26 meters, what amount of work will they have accomplis
JulijaS [17]
Work=applied Force x distance

= 1275 x 26
=33150 Joules
8 0
3 years ago
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