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2 years ago

Visible light energy.

Physics

2 answers:

Ipatiy [6.2K]2 years ago

6 0

What about it??

Please explain and I will help.

I don’t know what you’re asking.

Leto [7]2 years ago

6 0

Answer:

No se pregunta a alguien mas carnalito.

Explanation:

Lo siento:(

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How many structures do prokaryotic cells have in common with the eukaryotic cells?

Feliz [49]

Two. The both have a plasma membrane and cytoplasm.

6 0

2 years ago

Read 2 more answers

Problem 6: A bullet in a gun is accelerated from rest from the firing chamber to the end of the barrel at an average rate of 6.3

solong [7]

Answer:

v = 5.166 10² m / s

Explanation:

We can solve this exercise using the kinematics equations

v = v₀ + at

as the bullet starts from rest its initial velocity is zero

v = a t

let's calculate

v = 6.3 10⁵ 8.2 10⁻⁴

v = 5.166 10² m / s

3 0

2 years ago

Work-Energy Theorem: A 4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects en

allsm [11]

The same braking force does work on these objects to slow them down. The work done is equal to their change in kinetic energy:

FΔx = 0.5mv²

F = force, Δx = distance traveled, m = mass, v = speed

Isolate Δx:

Δx = 0.5mv²/F

Calculate Δx for each object.

Object 1: m = 4.0kg, v = 2.0m/s

Δx = 0.5(4.0)(2.0)²/F = 8/F

Object 2: m = 1.0kg, v = 4.0m/s

Δx = 0.5(1.0)(4.0)²/F = 8/F

The two objects travel the same distance before stopping.

4 0

2 years ago

What is the ultimate energy for all life on earth

professor190 [17]

The sun. Is the answer

7 0

2 years ago

Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet and legs.

Step2247 [10]

Answer:

Part a)

$F = 15600 N$

Part b)

force is 21.2 times more than the weight of the person

Explanation:

Part a)

As it is given that the distance after which he stopped is given as

d = 1.50 cm

here finally it stops so final speed is given as

$v_f = 0$

initial speed is given as

$v_i = 6 m/s$

now by the equation of kinematics we know that

$v_f^2 - v_i^2 = 2a d$

$0^2 - 6^2 = 2(a)(0.015)$

$a = -1200 m/s^2$

Now the force on the leg is given as

$F = ma$

m = mass of leg = 13 kg

$F = (13 kg)(-1200 m/s^2)$

$F = 15600 N$

Part b)

Force due to weight of the object

$F_g = Mg$

here we know

M = 75.0 kg

$F_g = 75(9.81)$

$F_g = 735.75 N$

Now we know that the ratio of the weight with the force on leg is given as

$\frac{F}{F_g} = \frac{15600}{735.75} = 21.2 N$

so force is 21.2 times more than the weight of the person

5 0

3 years ago