What speed do all electromagnet waves travel

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

$d'=v't$

and substituting,

$d'=(5.53)(10140)=56074 m = 56.1 km$

And the direction is the same as that of the velocity, therefore will be 90° south of east.

3 0

3 years ago

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied

Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied on the book, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that the minimum force that must be applied on the book is

F=44N

From the question we are told

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied on the book, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the Force is mathematically given as

$F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\$

F=44N

Therefore

the minimum force that must be applied on the book is

F=44N

For more information on this visit

brainly.com/question/23379286

8 0

2 years ago

A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg

saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

$x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2$

$F =ma$

$V(t) = V_{o} +at$

$x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2$

(a) Calculating velocity right before going up the ramp.

Wooden block is going on a straightaway and has net for on it.

$F_{n} =F-F_{s} = F-uF_{n} = 100N-0.4*9.8m/s^2*5kg$ = $81N$

and this force produces acceleration of

$a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .$

With this acceleration, wooden block would reach at the foot of ramp in.

$x(t) = 12m = 16.2m/s^2*t^2$

$t = 1.217s$

and final velocity will be

$v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.$

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

$V= V_{h}cos(20) = 18.5288m/s$

and deceleration along the ramp is

$a = \frac{F_{s} }{m}$

Where $F_{s}$ force of friction along the inclined plane.

$F_s = uF_n = u*m*a$

$a = 9.8m/s^2*cos(20) = 9.2089m/s^2$

is a component of g along normal of the inclined plane.

$F_{s} = 0.25*5kg*9.2089m/s^2$

$= 11.5112N$

$a = \frac{11.5112N}{5kg} = 2.3022m/s^2$

And with this deceleration time needed to get wooded block to stop is.

$v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0$

$t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s$

and in that time wooden block would travel

$x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m$

This is how up wooden box will go before coming to stop.

3 0

3 years ago

We cannot consider something to be work unless

iren [92.7K]

force applied causes movement of an object in the same direction as the applied force.

5 0

3 years ago

Read 2 more answers

What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each

Bezzdna [24]

This question apparently comes after an EARLIER one,
where you were told either the voltage across the same
capacitor or the total charge stored in it. You can't answer
THIS one without that information.

3 0

3 years ago

What speed do all electromagnet waves travel​

What speed do all electromagnet waves travel