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3 years ago

What is Circular Motion?

2 answers:

5 0

Circular motion is what an object has if it is moving around and around and around and around and around and around and around and around and around in a path that is a circle.

andre [41]3 years ago

3 0

It is a movement of an object along the circumference of a circle

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A homing pigeon starts from rest and accelerates uniformly at +4.00 m/s squared for 10.0 seconds. What is its velocity after the

zavuch27 [327]

' +4 m/s² ' means that the pigeon's speed is 4 m/s greater every second.

Starting from zero speed, after 10 seconds, its speed is

(10 x 4m/s) = 40 m/s.

We can't say anything about its velocity, because we have
no information regarding the direction of its flight.

5 0

4 years ago

What is the mass of an object that experiences a gravitational force of 510 N near Earth's surface?

sineoko [7]

Answer:The mass of an object is 52 kg.

Explanation:

Gravitational force on the object ,F=510 N

Acceleration due to gravity = g = $9.8 m/s^2$

Mass of the object = m

Force = mass × acceleration

$510 N=m\times 9.8 m/s^2$

$m=52.04 kg\approx 52 kg$

The mass of an object is 52 kg.

8 0

3 years ago

Read 2 more answers

When the voltage and current have _____ polarities in a pure capacitive circuit, the capacitor is discharging and the energy is

tangare [24]

Answer:opposite

Explanation:for a capacitor to discharge (after charging) the polarities of the current and voltage have to be reversed

6 0

3 years ago

Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot

Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² = 2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ = m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁ = 4.427 - 2v₂ ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂ = 8.854

v₂ = 8.854 / 3

v₂ = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁ = - 1.477 m/s

v₁ = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

$mgh_{max} = \frac{1}{2}mv_{max}^2$

(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max} = 0.11 \ m$

(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max} = 0.44 \ m$

6 0

3 years ago

There are infinite black and white dots on a plane. Prove that the distance between one black dot and one white dot is one unit.

OlgaM077 [116]

Answer:this is confusing and what subject is this

Explanation:

6 0

3 years ago

Read 2 more answers