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3 years ago

An object moves in uniform motion at 25 m/s and takes 1.0 seconds to go a quatar circle. What is the radius of the circle

1 answer:

3 0

If the it took the object travelling at 25m/s 1 second to travel a quarter of the circle, it means a quarter of the circle is 25m (it travels 25m in one second).

If the quarter of the circle is 25m than the circumference of the circle is 100m.

We want to calculate the radius of the circle.

Remember that the equation for finding the circumference contains the radius. We can use that equation to solve for the radius.

Circumference = 2πr

π = pi or 3.1415...... ( a number )

r = radius

Since we know the circumference, plug it in.

100m = 2πr

solve for the radius by isolating r.

100m/2π = r

r = radius = 100m/2π = 50m/π = 50/π meters

<h2>Answer: Radius = 50/π meters</h2>

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3 years ago

Read 2 more answers

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli

Pie

Answer:

a) v = 3,843 m / s, b) 46.7º North- East

Explanation:

Moment is a vector quantity, so one of the best ways to solve this problem is to solve each component separately.

The system is formed by the two vehicles so that the moment is preserved during the crash

Direction to the East

initial instant. Before the crash

p₀ = mₐ vₐ₀

final insttne. After the crash

p_f = (mₐ + m_b) vₓ

p₀ = p_f

mₐ vₐ₀ = (mₐ + m_b) vₓ

vₓ = $\frac{m_a}{m_a + m_b} \ v_{ao}$

let's calculate

vₓ = $\frac{16.7}{16.7 + 29.3} \ 7.26$

vₓ = 2,636 m / s

direction north

initial p₀ = m_b v_{bo}

final p_f = (mₐ + m_b) v_y

p₀ = p_f

m_b v_{bo} = (mₐ + m_b) v_y

v_y = $\frac{m_b}{m_a+m_b} \ v_{bo}$

let's calculate

v_y = $\frac{29.3}{16.7 + 29.3} \ 4.39$

v_y = 2.796 m / s

the final speed of the two two vehicles is

v = (2,636 i ^ + 2,796 j ^) m / s

a) the magnitude of the velocity

let's use the Pythagorean theorem

v = $\sqrt{v_x^2 + v_y^2}$

v = $\sqrt{2.636^2 + 2.796^2}$

v = 3,843 m / s

b) let's use trigonometry to find the direction

tan θ = v_y / vₓ

θ = tan⁻¹ v_y / vₓ

θ = tan⁻¹ (2,796 / 2,636)

θ = 46.7º

This direction is 46.7º North East

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3 years ago