Acar is traveling in a race. The car went from the initial velocity of 35 m/s to the final velocity of 65 m/s in 5 seconds.

Mathematics

1 answer:

Ray Of Light [21]3 years ago

4 0

Answer:

6 m/s^2

hope this helped!<3

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A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .

7 0

3 years ago

Please help with geometry

Margarita [4]

Step-by-step explanation:

a) gradient= y2-y1/x2-x1

gradient of AB:

gradient=-4-1/2-3

=-5/-1

gradient of BC:

=-5+4/7-2

=-1/5

b) as the priduct of both gradient is -1 ,both lines are perpendicular.

c) as we have seen lines are perpendicular so it is a right angled triangle.

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3 years ago

HELP MUST BE DONE BY 4:00

Leya [2.2K]

Answer:

the antelope is 5280 feet the cougar is 4224 feet the hare is 4136 feet the kangaroo is 3520 feet and the coyote is 3773 feet the ostrich is 3773 feet hope this helps the<em> answer is the </em><em>antelope</em>