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3 years ago

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___________ is a type of angiosperm characterized by having two cotyledons, branching leaf veins, and flower parts that are in f

ours and fives.

1 answer:

lara31 [8.8K]3 years ago

5 0

The answer to this question is dicot

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Linda has a cup of coffee sitting in the cup holder in her car. She has to suddenly slam on her brakes and stop. The cup remaine

dybincka [34]

This would be Newton's first law, which states that an object at rest will stay at rest and an object in motion will remain in motion unless acted on by an unbalanced force. Hope this helped.

7 0

4 years ago

Read 2 more answers

4. A 1,000-kilogram satellite completes a uniform circular orbit of radius 8.0 x 10 meters as

lesantik [10]

Answer:

Zero

Explanation:

The work done by a force on an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement of the object

In this situation, the force is the force of gravity acting on the satellite. This force always points towards the centre of the trajectory, so it is always perpendicular to the direction of motion of the satellite (since the orbit is circular), so $\theta=90^{\circ}$ and $cos \theta =0$ . Therefore, the work done by gravity is also zero.

5 0

3 years ago

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altidude

Serjik [45]

Answer:

a) 252 ft/s

b) 1076 ft

Explanation:

The equation for motion for uniform acceleration (we can use it because the rocket is affected only by gravity) is as follows:

Y(t) = Y0 + V0 * t + 1/2 * a * t^2

Where

Y(t): altitude at a given time

Y0: initial altitude

V0: initial speed

a: acceleration, in this case -32.2 ft/s^2 (negative because gravity points down)

We set a 1 dimensional coordinate system with Y pointing up and the origin of coordinates at ground level.

We consider t=0 as the moment where powered flight ended (motor ran oou of fuel), at this moment the altitude was

Y(0) = 89.6 ft

Therefore:

Y0 = 89.6 ft

We also know that the rocket fell to ground 16 seconds later, therefore

Y(16) = 0 ft

So we can write

Y(t=16) = Y0 + V0 * t + 1/2 * a * t^2

V0 * t = Y(t=16) - Y0 - 1/2 * a * t^2

V0 =( Y(t=16) - Y0 - 1/2 * a * t^2 )/t

V0 =( 0 - 89.6 - 1/2 * (-32.2) * 16^2 )/16 = 252 ft/s

In the highest point of flight the rocket will have a speed = 0

The first derivative of the equation of motion is the equation of speed:

V(t) = V0 + a * t

If we equate this to zero we eill find the time at which the rocket achieved it's highest altitude.

0 = V0 + a * t

a * t = 0 - V0

t = -V0/a

t = -252/(-32.2) = 7.83 s

Now, we can take this time value andd plug it back into the position equation

Y(7.83) = 89.6 + 252 * 7.83 + 1/2 * (-32.2) * 7.83^2 = 1076 ft

6 0

3 years ago

At the bottom of a frictionless ramp, a 2kg mass reaches a speed of 9 m/s, how high is the top of the ramp? *

noname [10]

V=√(2gs)
9=√(2×9.8×s)
s=4.1m

3 0

3 years ago

Suppose the earth suddenly came to halt and ceased revolving around the sun. The gravitational force would then pull it directly

AleksAgata [21]

Answer:

613373.65233 m/s

Explanation:

M = Mass of Sun = $1.989\times 10^{30}\ kg$

m = Mass of Earth

v = Velocity of Earth

r = Distance between Earth and Sun = $147.12\times 10^{9}\ m$

$r_e$ = Radius of Earth = $6.371\times 10^6\ m$

$r_s$ = Radius of Sun = $695.51\times 10^6\ m$

In this system it is assumed that the potential and kinetic energies are conserved

$\dfrac{1}{2}Mv_2-\dfrac{GMm}{r_e+r_s}=0-\dfrac{GMm}{r}\\\Rightarrow v=\sqrt{2GM(\dfrac{1}{r_e+r_s}-\dfrac{1}{r})}\\\Rightarrow v=\sqrt{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}(\dfrac{1}{6.371\times 10^6+695.51\times 10^6}-\dfrac{1}{147.12\times 10^{9}})}\\\Rightarrow v=613373.65233\ m/s$

The velocity of Earth would be 613373.65233 m/s

3 0

3 years ago