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Tanzania [10]
2 years ago
9

A sister spins her brother in a circle of radius R at angular speed wi. Then the sister decreases her angular speed

Physics
1 answer:
Igoryamba2 years ago
6 0

Answer:

The change in the centripetal acceleration of the brother,

                               Δa = V₂²/R - V₁²/R

Explanation:

Given data,

A sister spins her brother in a circle of radius, R

The angular velocity of the brother, ω₁ = V₁/R

The angular velocity of the brother, ω₂ = V₂/R

The centripetal acceleration is given by the relation

                                 a = V²/R

Therefore change in the centripetal acceleration of the brother,

                              Δa = V₂²/R - V₁²/R                                    

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Molodets [167]

b) between poles M1 and M2

Explanation:

From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.

The law of attraction between two magnetic poles states that:

<em>  the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s

 

    Mathematically:

            FM = K \frac{M1 M2}{r^{2} }

 here r is the distance between the poles

  FM is the magnetic force between the poles

   M1 is the strength of the first magnetic pole

   M2 is the strength of the second pole

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8 0
2 years ago
A child pushes her toy across a level floor at a steady velocity of 0.50 m/s using an applied force of 2.0 N. If the weight of t
choli [55]
Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.

Rate of speed that toy is moving is irelevant.

childs force is:
Fc = 2N
Fc = Ff  (Ff -friction force)

Ff = a*Q

where Q is weight of the toy and a is friction

if we express a we get
a = F/Q = 2/8 = 0.25
8 0
2 years ago
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Consider a 4-mg raindrop that falls from a cloud at a height of 2 km. When the raindrop reaches the ground, it won't kill you or
docker41 [41]

Answer:

The work done by the air resistance is -0.0782 J

Explanation:

Hi there!

The energy of the raindrop has to be conserved, according to the law of conservation of energy.

Initially, the raindrop has only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the raindrop.

g = acceleration due to gravity (9.8 m/s²)

h = height.

Let´s calculate the initial potential energy of the drop:

(convert 4 mg into kg: 4 mg · 1 kg / 1 × 10⁶ mg = 4 × 10⁻⁶ kg)

PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000  m

PE = 0.0784 J

When the drop starts falling, some of the potential energy is converted into kinetic energy and some energy is dissipated by the work done by the air resistance. On the ground all the initial potential energy has been either converted into kinetic energy or dissipated by the resistance of the air:

initial PE = final KE + W air

Where:

KE = kinetic energy.

W air = work done by the air resistance.

The kinetic energy when the raindrop reaches the ground is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass

v = velocity

Then:

KE = 1/2 ·  4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

Now, we can calculate the work done by the air resistance:

initial PE = final KE + W air

0.0784 J = 2 × 10⁻⁴ J + W air

W air = 0.0784 J - 2 × 10⁻⁴ J

W air = 0.0782 J

Since the work is done in the opposite direction to the displacement, the work is negative, then, the work done by the air resistance is -0.0782 J.

5 0
2 years ago
Sachi wants to throw a water balloon to knock over a target and win a prize. The target will only fall over if it is hit with a
posledela

Answer:

3.1 m/s²

Explanation:

Given:

Mass of the balloon (m) = 11.4 g = 0.0114 kg ( 1 kg = 1000 g)

Force acting on the balloon (F) = 0.035 N

Acceleration with which the balloon must be hit (a) = ?

Now, we know that, from Newton's second law, net force acting on an object is equal to the product of its mass and acceleration.

Therefore, framing in equation form, we have:

F=ma

Rewriting in terms of acceleration 'a', we get:

a=\frac{F}{m}

Now, substitute the given values and solve for 'a'. This gives,

a=\frac{0.035\ N}{0.0114\ kg}\\\\a=3.07\approx 3.1\ m/s^2(Nearest\ tenth)

Therefore, the acceleration of the water balloon to reach the target must be equal to 3.1 m/s².

7 0
2 years ago
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