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OLga [1]
3 years ago
10

A 12.0 kg object is attached to a cord that is wrapped around a wheel of radius 10.0 cm. the acceleration of the object down the

frictionless incline is measured to be 2.00 m/s2. assuming the axis of the wheel to be frictionless, determine
a.the tension in the rope,
b.the moment of inertia of the wheel, and
c.the angular speed of the wheel 2.00 s after it begins rotating, starting from rest.
Physics
1 answer:
olganol [36]3 years ago
6 0

 a) We use the formula:

T = [g sinθ – a] * m

where g is gravity, θ is the angle, a is acceleration, m is mass

T = [9.81 sin37 - 2.00] * 12 = 46.85 N <span>

b) We use the formula for moment of interia:</span>

I = T * r² / a

I = 46.85 * 0.10² / 2.00 = .2343 kg∙m² <span>

c) The formula we can use here is:</span>

w = α * t = (a/r) * t

w = (2/.10) * 2

<span>w = 40 rad/sec</span>

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The particles ejected from the sun during a coronal mass ejection is called
ipn [44]

Answer:

Plasma

Explanation:

A Coronal Mass Ejection (CME) is an outburst of energy that occurs near the outer part of the sun's atmosphere which causes a production of plasma along with a magnetic field.

The outermost part of the sun's atmosphere is called the Solar Corona Although difficult to see, the corona can be seen during a total solar eclipse.

Plasma from CME are clouds of magnetized electrically charged particles which the solar wind causes to travel at a speed of 1.6 million km/hr.

7 0
3 years ago
How does force do positive work on an object?​
julia-pushkina [17]

Answer:

It can do work on an object by moving it or changing it.

Explanation:

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5 0
3 years ago
Read 2 more answers
Calculate the potential energy of a body of a mass 2kg held 4 meters above the ground if g=10m/s?​
Drupady [299]

Answer:

16000

Explanation:

Mass(m)=2Kg (1kg= 1ooo g then 2 kg=2000 g)

Velocity(v)= 4 meter

Acceleration due to gravity (g)=10m/s

We know that,

P.E= 1/2 mv^2

or, 1/2 × 2ooo × 4^2

or, 1/2×2000 ×16

or, 2000×8

Therefore= 16000

7 0
2 years ago
The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about
nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

6 0
3 years ago
A school bus has a mass of 18,200 kg. The bus moves at 13.5 m/s. How fast must a 0.142-kg baseball move in order to have the sam
nikitadnepr [17]

Answer:

bus momentum

p_bus= m_bus x v_bus

=18,200 x 16.5

basball momentum

pball=mball x vball

=0.142 x v

p_bus = pball

18200 x 16.5 = 0.142 x v

v=(18200 x 16.5)/0.142

v is the answer for baseball

Explanation:

⚠️not my answer tryna be honest here⚠️

3 0
3 years ago
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