All categories

3 years ago

Describe your acceleration as you skate down your neighborhood sidewalk.

Physics

1 answer:

Paraphin [41]3 years ago

6 0

Answer: when you increase or decrease your speed.

Explanation:

Moving a skate at rest, you need to apply force in order to cause acceleration.

F = ma Where

F = force applied

m = mass of the skate

a = acceleration

The initial velocity u will be equal to zero and the skate will acceleration to a certain velocity.

as you skate down your neighborhood sidewalk, you will accelerate when you increase your speed. Because

Acceleration is the rate of change of velocity. That is,

Acceleration = change in velocity/ time.

And also, you will decelerate when you reduce the speed or velocity down your neighborhood sidewalk.

You might be interested in

A ball thrown in the air will never go as far as physics ideally would predict because?

katrin2010 [14]

Air resistance fefrgtdnjrcbrdg

6 0

3 years ago

A 2100 kg car starts from rest and accelerates at a rate of 2.6 m/s2 for 4.0 s. Assume that the force acting to accelerate the c

Reika [66]

When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.

a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s

Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ

8 0

3 years ago

Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 197

Leno4ka [110]

(a) 3.58 km 45° south of east

The total displacement is given by:

$d=vt$

where

v is the average velocity

t is the time

The average velocity is:

v = 3.53 m/s

While we need to convert the time from minutes to seconds:

$t=169 min \cdot 60 s/min = 10140 s$

Therefore, the magnitude of the displacement is

$d=(3.53)(10140)=35794 m = 3.58 km$

And the direction is the same as the velocity, therefore 45° south of east.

(b) 5.53 m/s 90° south of east

The velocity of the air relative to the ground is

$v_a = 2.00 m/s$

and the direction is exactly opposite to that of Allen, so it is 45° north of west. Allen's velocity relative to the ground is

v = 3.53 m/s

So this must be the resultant of Allen's velocity relative to the air (v') and the air's speed ( $v_a$ ). Since these two vectors are in opposite direction, we have

$v= v'-v_a$

Therefore we find v', Allen's velocity relative to the air:

$v'=v+v_a = 3.53 + 2.00 = 5.53 m/s$

The direction must be measured relative to the air's reference frame. In this reference frame, Allen is moving exactly backward, so his direction will be 90° south of east.

(c) 56.1 km at 90° south of east.

Since Allen's velocity relative to the air is

v' = 5.53 m/s

Then the displacement of Allen relative to the air will be given by

$d'=v't$