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3 years ago

A ball thrown straight up takes 1.89s to reach a height of 41.6m.

Physics

1 answer:

yulyashka [42]3 years ago

8 0

Answer:

Explanation:

average velocity is 41.6/1.89 = 22.0 m/s

initial velocity u is

22.0 = (u + u - 9.81(1.89))/2

44.0 = 2u - 18.5

62.56 = 2u

u = 31.3 m/s

max height is

v² = u² + 2as

s = (v² - u²) / 2a

s = (0² - 31.3²) / (2(-9.81)) = 49.9 m

49.9 = 41.6 = 8.27 m higher

In process numbers have been rounded to 3 s.d. for reporting purposes.

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A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of

miskamm [114]

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= $\frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}$

= 1.72 x 10³ N.

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4 years ago

What types of sound waves diffract more than others

cricket20 [7]

Answer:

Infrasound

but the longer the wavelength of a soundwave, the more it diffracts

Explanation:

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3 years ago

A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass M which is

Charra [1.4K]

Answer:

M is equal to m

Explanation:

In case we say that the green block's mass m is less than red block's mass M, then the green block would have bounced and moved back to the left instead of coming to rest. The other case where if mass of green block's mass m would have been greater than the red block's mass M, the green block would have kept moving to the right instead of coming to rest. After collision, the red block moves to the right because of exchange of velocities. Therefore, m=M since m comes to rest and M moves to the right

In any collision, as it is asumed that no external forces can act during the collision, momentum must be conserved.

So, if we call p₁ to the momentum before collision, and p₂ to momentum after it, taking into account the information above, we can write the following:

p₁ = mv₁ + M.0 = p₂ = m.0 + Mv₂ ⇒ mv₁ = Mv₂

From the question, we also know that it was an elastic collision.

In elastic collision, added to the momentum conservation, it must be conserved the kinetic energy also.

So, if we call k₁ to the kinetic energy prior the collision, and k₂ to the one after it, we can write the following:

k₁ = 1/2 m(v₁)² + 1/2 M.0 = k₂ = 1/2m.0 + 1/2M(v₂)² ⇒ m(v₁)² = M(v₂)²

Mathematically, the only way in which both equations be true, should be with v₁ = v₂, which is only possible if m=M too.

In this type of collision, it is said that the energy transfers from one mass to the other.

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3 years ago

What is nuclear power

katrin2010 [14]

Answer:

<h2>D. All of the statements describe nuclear power</h2>

What is NUCLEAR POWER? 

➡️ is the use of nuclear reactions to produce electricity.

➡️ it can be obtained from nuclear fission, nuclear decay and nuclear fusion reactions. 

Hope it helps

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2 years ago

5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal whil

Lelu [443]

Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.

By Newton's second law, the net vertical force is

• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0

where a is the acceleration of the wagon.

Solve for n (the magnitude of the normal force) :

n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N

Then

f = 0.500 (58.0 N) = 29.0 N

Meanwhile, the horizontal component of the applied force has magnitude

(80.0 N) cos(30.0°) ≈ 69.3 N

Now calculate the work done by either force.

• friction: -(29.0 N) (10.0 m) = -290. J

• pull: (69.3 N) (10.0 m) = 693 J

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3 years ago