A(n) _____ is a gap in the geologic record where some rock layers have been lost because of erosion.

When you enter a toll road, your ticket is stamped 1:00 p.m. When you leave, after traveling 55 miles, your ticket is stamped 2:

Viefleur [7K]

27.5 because of you divide the 55miles with the time you get your velocity which is the speed.

4 0

3 years ago

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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

Clarissa thought that learning to play the banjo was a waste of time. She didn't bother to practice after her lessons

emmasim [6.3K]

Answer:

The answer ro this question is fear of failure

7 0

3 years ago

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A plane travels down a runway 2750 m before it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1

ss7ja [257]

Answer: 4.236km

Explanation:

Let's define the point (x, y) as:

x = horizontal distance moved.

y = vertical distance moved.

If the plane starts in the point (0, 0) then:

"A plane travels down a runway 2750 m before it lifts off..."

At this time, the position will be:

P = (0 + 2750m, 0) = (2750m, 0).

"it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1.8 km since its wheels left the ground."

In this case, as the angle is measured from the horizontal, the components will be:

x = 1.8km*cos(37°) = 1.438km

y = 1.8km*sin(37°) = 1.083 km

Then the new position is:

P = (2750m + 1.438 km, 0 + 1.083 km)

Let's write it using the same units for all the quantities:

we know that

1km = 1000m

Then:

2750m = (2750/1000) km = 2.750 km.

Then we can write the new position as:

P = (2.750 km + 1.438km, 1.083km) = (4.188km, 1.083km)

Now, we define the displacement as the distance between the final position and the initial position.

The distance between two points (a, b) and (c, d) is:

D = √( (a c)^2 + (b - d)^2)

In this case the points are:

(0, 0) for the initial position

(4.188km, 1.083km) for the final position.

And the displacement will be:

D = √( (4.188km - 0)^2 + (1.083 - 0)^2) = 4.236km

5 0

3 years ago

A 1200 kg aircraft going 30 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and

Leno4ka [110]

Answer:

83.055 m

Explanation:

According to the given scenario, the calculation of skid distance is shown below:-

$S = \frac{1}{2} \times (u + v) \times t$

Where

u = 11.3

v = 0

t = 14.7

Now placing these values to the above formula,

So,

$S = \frac{1}{2} \times (11.3 + 0) \times 14.7$

= 83.055 m

Therefore for computing the skid distance we simply applied the above formula i.e by considering the all items given in the question

4 0

3 years ago