What kind of medium does ocean waves have

Can somebody help please ! a. -8.3 m/s b.-4.2 m/s c.-0.12 m/s d. 0 m/s

Gnom [1K]

The answer is a
$the \: answer \: is \: a.$

4 0

4 years ago

In tae-kwon-do, a hand is slammed down onto a target at a speed of 14 m/s and comes to a stop during the 3.0 ms collision. Assum

GrogVix [38]

Answer:

A) The impulse is 11,7 kg.m/s . B) The average force on the hand is 3900N

Explanation:

Givens:

Hand's mass (m)= 0,90kg

Hand's initial velocity (vi)= 14m/s

Time of collision (t)= 3ms

A) Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time. We use the formula:

/F/ = m. /Δv/

Where:

F= Force of Impulse

m= Mass

Δv= change in speed ( vf-vi)

Using that formula we get:

/F/= 0,90kg . / (0m/s-14m/s) /

/F/=0,90kg . 14m/s

/F/ = 11,7 kg. m/s

*note that (vf) is 0 because the hand stops in the action, so it's final

velocity = 0

B)The average force is equal to the change in the momentum over the change in time. We use the formula:

/F/= m. (Δv) : Δt

Where:

F= the average force from the target

m= the mass of the hand

Δv= hange in speed (vf/vi)

Δt= change in time

*if we look closely, we find the we alredy have half of the equation because we have alredy calculated m.(Δv) in point A.

We boil the equation down to:

/F/= Impulse : Δt

/F/= 11,7kg.m/s : 0,003s

/F/= 3900N

*we use 0,003s as our time because the given time was 3ms.

*the final result is expressed in Newtons because our final result ends up beeing kg.m/s² = N

8 0

4 years ago

A grocery cart with a mass of 15 kg is pushed at constant speed along an aisle by a force fp = 12 n which acts at an angle of 17

Korolek [52]

Given:

Mass of cart: 15kg

Aisle length = 14m

Angle = 17° below the horizontal

Force fp = 12 N

So, the solution would be like this for this specific problem:

1) W(by applied force) = F(applied) x s x cosθ 
=>W(a) = 12 x 14 x cos17* = 160.66 J 

2) By F(net) = F(applied) - F(friction) 
=>As v = constant => a = 0 => F(net) = 0
=>F(applied) = F(friction)
=>W(friction) = -F(friction) x s
=>W(friction) = -F(applied) x s
=>W(f) = -12 x 14 x cos17* = -160.56 J 

3) 0, as the displacement is perpendicular to Force 

4) 0, as the displacement is perpendicular to Force

To add, the force that is applied to an object by a person or another object is called the applied force.

8 0

3 years ago

Problem: The circular blad on a radial arm saw is turning at262

kobusy [5.1K]

Answer:

Net torque, $\tau=-0.033\ N-m$

Explanation:

It is given that,

Initial angular speed of the blade, $\omega_i=262\ rad/s$

Final angular speed of the blade, $\omega_f=85\ rad/s$

Time, t = 18 s

Radius of the disk, r = 0.13 m

Mass of the disk, m = 0.4 kg

We need to find the net torque applied to the blade. We know that in rotational mechanics the net torque acting on an object is equal to the product of moment of inertia and the angular acceleration such that,

$\tau=I\times \alpha$