A cheetah accelerates from rest to 30m/s in 3 seconds. Calculate the acceleration of the cheetah.

An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. An unbalanced force is an unopposed force that causes a change in motion.

7 0

3 years ago

You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh

lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

$F=\frac{kQq}{r^2}$

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

$\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}$

And the force on the charge due to the charge on the north is,

$\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}$

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of $45^\circ$ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

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3 0

1 year ago

Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1

miv72 [106K]

Answer:

w = 4,786 rad / s , f = 0.76176 Hz

Explanation:

For this problem let's use the concept of angular momentum

L = I w

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

Initial Before sticking

L₀ = 0 + I₂ w₂

Final after coupling

$L_{f}$ = (I₁ + I₂) w

The moments of inertia of a disk with an axis of rotation in its center are

I = ½ M R²

How the moment is preserved

L₀ = $L_{f}$

I₂ w₂ = (I₁ + I₂) w

w = w₂ I₂ / (I₁ + I₂)

Let's reduce the units to the SI System

d₁ = 60 cm = 0.60 m

d₂ = 40 cm = 0.40 m

f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz

Angular velocity and frequency are related.

w₂ = 2 π f₂

w₂ = 2π 3.33

w₂ = 20.94 rad / s

Let's replace

w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

w = 20.94 1.28 / 5.6

w = 4,786 rad / s

Angular velocity and frequency are related.

w = 2π f

f = w / 2π

f = 4.786 / 2π

f = 0.76176 Hz

7 0

3 years ago