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3 years ago

Which type of disease cannot be spread from one person to another

Physics

1 answer:

aniked [119]3 years ago

7 0

I believe that would be D.

Cardiovascular disease generally refers to conditions that involve narrowed or blocked blood vessels that can lead to a heart attack, chest pain, or stroke. Which it has nothing to do with being infectious.

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A 3.0 kg pendulum swings from point A of height ya = 0.04 m to point B of height yb = 0.12 m, as seen in the diagram below.

lakkis [162]

i dont know soory pls give brainlyist

5 0

3 years ago

Read 2 more answers

A basketball is tossed upwards with a speed of 5.0 m/s. We can ignore air resistance.

FinnZ [79.3K]

We have that the maximum height reached by the basketball from its release point is

$S=1.3m$

From the question we are told

A basketball is tossed upwards with a speed of 5.0 m/s. We can ignore air resistance.
What is the maximum height reached by the basketball from its release point?

Generally the Newtons equation for Motion is mathematically given as

$V^2=u^2+2as\\\\Therefore\\\\0=25-19.6*S\\\\S=\frac{25}{19.6}\\\\S=1.276$

$S=1.3m$

Therefore

The maximum height reached by the basketball from its release point is

$S=1.3m$

For more information on this visit

brainly.com/question/23366835

4 0

3 years ago

A rope pulls a 20 kg box up. If the tension in the rope is 275 N, what is the acceleration of the box?

Nana76 [90]

Answer:

<h3>The answer is 13.75 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{275}{20} = \frac{55}{4} \\$

We have the final answer as

Hope this helps you

4 0

4 years ago

A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the ce

Dmitry [639]

Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad

6 0

3 years ago

A ball on a string makes 25.0 revolutions in 9.37 s, in a circle radius 0.450 m. What is it’s velocity?

Sauron [17]

The velocity of the ball is 12.5 m/s