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3 years ago

Fan object moves in uniform circular motion in a circle of radius R=200 meters, and the objectes 5.00 seconds to

Physics

1 answer:

Leviafan [203]3 years ago

6 0

Answer:

The centripetal acceleration of the object is $31550.72\ m/s^2$ .

Explanation:

We have,

Radius of a circular path is 200 m

It takes 5 seconds to complete 10 revolutions. The angular velocity of the object is given by the rate of change of angular displacement per unit time :

$\omega=\dfrac{\Delta \theta}{\Delta t}\\\\\omega=\dfrac{2\pi \times 10}{5}\\\\\omega=12.56\ rad/s$

The centripetal acceleration of the object is given by :

$a=\omega^2 r\\\\a=(12.56)^2 \times 200\\\\a=31550.72\ m/s^2$

So, the centripetal acceleration of the object is $31550.72\ m/s^2$ .

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Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s

enot [183]

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:

$p_i = p_f$

$m_1u_1 + m_2v_2 = (m_1 + m_2)v$

$v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}$

$v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s$

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}$

Here, initial velocity is the final velocity from the first stage. Therefore:

$v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s$

3. Use conservation of momentum to find the combined speed of Gayle and her brother.

Given:

Initial velocity of Gayle and sled is, $u_1(i)=10.5$ m/s

Initial velocity of her brother is, $u_2(i)=0$ m/s

Mass of Gayle and sled is, $m_1=55.0$ kg

Mass of her brother is, $m_2=30.0$ kg

Final combined velocity is given as:

$v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}$

$v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79$ m/s

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s$

6 0

3 years ago

A star rotates with a period of 37 days about an axis through its center. The period is the time interval required for a point o

vlabodo [156]

Answer:

T = 1.1285 10⁻² day

Explanation:

For this exercise the forces in the premiere are internal, so the angular momentum is conserved

L₀ = I₀ w₀

L = I w

L₀ = L

I₀ w₀ = I w

Angular velocity and period are related

w₀ = 2π / T₀

w = 2π / T

The moment of inertia of a sphere is

I₀ = 2/5 M R²

I = 2/5 m r²

If we assume that the mass of the star does not change in the transformation

We substitute

2/5 M R² 2π/T₀ = 2/5 M r² 2π/ T

R² /T₀ = r² / T

T = (r / R)² T₀

T = (6.1 / 2.0 104) 37

T = 1.1285 10⁻² day

5 0

3 years ago

point A is at the bottom of a rough plane which is inclined at angle tita to the horizontal . A body of mass M is projected from

ziro4ka [17]

<h2>The work done , when body moves along the plane </h2>

Explanation:

A body is projected from bottom of the inclined plane . When body is going up the plane .

The downward force = m g sinθ is developed due to its weight

As body is moving upwards , the force of friction will act downwards

The force of friction = μ R

here μ is the coefficient of friction ans R is the normal reaction

Thus force of friction f = μ mg cosθ

Let the acceleration upwards is a

The upward force required = m a

Thus m a = mg sinθ + μ mg cosθ

or acceleration a = g ( sinθ + μ cosθ )

The work done in moving upwards W = F S

Thus W = mg ( sinθ + μ cosθ ) S

here S is the displacement on the plane

When body moves down , the force of friction acts upwards

Thus m a = m g ( sinθ - μ cosθ )

The work done W = m g ( sinθ - μ cosθ ) S

As the body is projected with velocity u

which can be calculated by the relation v² - u² = - 2 a X

Here v = 0 at the highest point

Thus u = $\sqrt{2ax}$

here a = g ( sinθ + μ cosθ )

Similarly , when it moves down , the initial velocity u = 0

Thus v² - 0 = 2 a x

or v = $\sqrt{2ax}$

here a = g ( sinθ - μ cosθ )

3 0

3 years ago

Sudhir walk .40 km in a direction 60 degrees west of north then goes .50km due west. what is his displacement?

DENIUS [597]

consider east-west direction along x-axis and north-south direction along y-axis

from the diagram

A = Ax i + By j = - (0.40 Sin60) i + (0.40 Cos60) j = - 0.35 i + 0.20 j

B = Bx i + By j = - 0.50 i + 0 j

Net displacement is given as

D = A + B

D = (- 0.35 i + 0.20 j ) + (- 0.50 i + 0 j )

D = - 0.85 i + 0.2 j

magnitude of displacement is given as

|D| = sqrt((- 0.85)² + (0.2)²)

|D| = 0.87 km

direction of displacement is given as

θ = tan⁻¹(0.2/0.85)

θ = 13.5 deg north of west

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